output
stringlengths 52
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#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
#define INF INT_MAX
#define LINF 1LL<<60
struct SegTree {
int N;
ll init_v = INF;
vector<ll> node, lazy;
SegTree(int _N) {
N = 1;
while (N < _N) N *= 2;
node.resize(2 * N - 1, init_v);
lazy.resize(2 * N - 1, -1);
}
void lazy_evaluate(int k) {
if (lazy[k] == -1) return;
node[k] = lazy[k];
if (k < N - 1) {
lazy[2 * k + 1] = lazy[k];
lazy[2 * k + 2] = lazy[k];
}
lazy[k] = -1;
}
/* [a,b) */
void update(int a, int b, int x) { update(a, b, 0, 0, N, x); }
void update(int a, int b, int k, int l, int r, int x) {
if (r <= a || b <= l) return;
if (a <= l && r <= b) {
lazy[k] = x;
lazy_evaluate(k);
}
else {
lazy_evaluate(k);
update(a, b, 2 * k + 1, l, (l + r) / 2, x);
update(a, b, 2 * k + 2, (l + r) / 2, r, x);
}
}
/* [a,b) */
ll query(int a, int b) { return query(a, b, 0, 0, N); }
ll query(int a, int b, int k, int l, int r) {
if (r <= a || b <= l) return init_v;
if (a <= l && r <= b) {
lazy_evaluate(k);
return node[k];
}
else {
lazy_evaluate(k);
ll vl = query(a, b, 2 * k + 1, l, (l + r) / 2);
ll vr = query(a, b, 2 * k + 2, (l + r) / 2, r);
return min(vl, vr);
}
}
};
int main() {
cin.tie(0); ios::sync_with_stdio(false);
int n, q; cin >> n >> q;
SegTree ST(n);
for (int i = 0; i < q; i++) {
int com; cin >> com;
if (com == 0) {
int s, t, x; cin >> s >> t >> x;
ST.update(s, t + 1, x);
}
else {
int j; cin >> j;
cout << ST.query(j, j + 1) << endl;
}
}
} | ### Prompt
Please create a solution in cpp to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
#define INF INT_MAX
#define LINF 1LL<<60
struct SegTree {
int N;
ll init_v = INF;
vector<ll> node, lazy;
SegTree(int _N) {
N = 1;
while (N < _N) N *= 2;
node.resize(2 * N - 1, init_v);
lazy.resize(2 * N - 1, -1);
}
void lazy_evaluate(int k) {
if (lazy[k] == -1) return;
node[k] = lazy[k];
if (k < N - 1) {
lazy[2 * k + 1] = lazy[k];
lazy[2 * k + 2] = lazy[k];
}
lazy[k] = -1;
}
/* [a,b) */
void update(int a, int b, int x) { update(a, b, 0, 0, N, x); }
void update(int a, int b, int k, int l, int r, int x) {
if (r <= a || b <= l) return;
if (a <= l && r <= b) {
lazy[k] = x;
lazy_evaluate(k);
}
else {
lazy_evaluate(k);
update(a, b, 2 * k + 1, l, (l + r) / 2, x);
update(a, b, 2 * k + 2, (l + r) / 2, r, x);
}
}
/* [a,b) */
ll query(int a, int b) { return query(a, b, 0, 0, N); }
ll query(int a, int b, int k, int l, int r) {
if (r <= a || b <= l) return init_v;
if (a <= l && r <= b) {
lazy_evaluate(k);
return node[k];
}
else {
lazy_evaluate(k);
ll vl = query(a, b, 2 * k + 1, l, (l + r) / 2);
ll vr = query(a, b, 2 * k + 2, (l + r) / 2, r);
return min(vl, vr);
}
}
};
int main() {
cin.tie(0); ios::sync_with_stdio(false);
int n, q; cin >> n >> q;
SegTree ST(n);
for (int i = 0; i < q; i++) {
int com; cin >> com;
if (com == 0) {
int s, t, x; cin >> s >> t >> x;
ST.update(s, t + 1, x);
}
else {
int j; cin >> j;
cout << ST.query(j, j + 1) << endl;
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
#define rep(i, N) for(ll i=0; i<(N); ++i)
//sqrtN -> 最大値のみに対応できればいいので，それを使う
const int sqrtN = 512;
struct SqrtDecomposition {
int N, K;
vector<int> data;
vector<int> bucketData;
const int INIT_VALUE = 0;
SqrtDecomposition(int n) : N(n) {
//分割数
K = (N + sqrtN - 1) / sqrtN;
data.assign(K * sqrtN, 2147483647);
bucketData.assign(K, -1);
}
//ここまで共通
void update(int x, int y, int v) {
for(int k = 0; k < K; ++k) {
int l = k * sqrtN;
int r = (k + 1) * sqrtN;
if(y <= l || x >= r) continue; //完全に区間外
if(l >= x && y >= r) { //完全に区間内をUPDATE（遅延bucketのみ）
bucketData[k] = v;
} else {
//一部区間内
if(bucketData[k] != -1) {
//まず，bucketを伝搬
for(int i = l; i < r; ++i) {
data[i] = bucketData[k];
}
bucketData[k] = -1;
}
for(int i = max(x, l); i < min(y, r); ++i) {
data[i] = v;
}
}
}
}
// [x, y)
int find(int x) {
int k = x / sqrtN;
//伝搬
if(bucketData[k] != -1) {
for(int i = k * sqrtN; i < (k + 1) * sqrtN; ++i) {
data[i] = bucketData[k];
}
bucketData[k] = -1;
}
return data[x];
}
};
int main() {
//range sum query
int n, q;
cin >> n >> q;
SqrtDecomposition dec(n);
vector<string> res;
rep(_, q) {
int c;
cin >> c;
if(c == 0) {
int s, t, x;
cin >> s >> t >> x;
++t;
dec.update(s, t, x);
} else {
int i;
cin >> i;
cout << dec.find(i) << endl;
}
}
}
| ### Prompt
Please create a solution in cpp to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
#define rep(i, N) for(ll i=0; i<(N); ++i)
//sqrtN -> 最大値のみに対応できればいいので，それを使う
const int sqrtN = 512;
struct SqrtDecomposition {
int N, K;
vector<int> data;
vector<int> bucketData;
const int INIT_VALUE = 0;
SqrtDecomposition(int n) : N(n) {
//分割数
K = (N + sqrtN - 1) / sqrtN;
data.assign(K * sqrtN, 2147483647);
bucketData.assign(K, -1);
}
//ここまで共通
void update(int x, int y, int v) {
for(int k = 0; k < K; ++k) {
int l = k * sqrtN;
int r = (k + 1) * sqrtN;
if(y <= l || x >= r) continue; //完全に区間外
if(l >= x && y >= r) { //完全に区間内をUPDATE（遅延bucketのみ）
bucketData[k] = v;
} else {
//一部区間内
if(bucketData[k] != -1) {
//まず，bucketを伝搬
for(int i = l; i < r; ++i) {
data[i] = bucketData[k];
}
bucketData[k] = -1;
}
for(int i = max(x, l); i < min(y, r); ++i) {
data[i] = v;
}
}
}
}
// [x, y)
int find(int x) {
int k = x / sqrtN;
//伝搬
if(bucketData[k] != -1) {
for(int i = k * sqrtN; i < (k + 1) * sqrtN; ++i) {
data[i] = bucketData[k];
}
bucketData[k] = -1;
}
return data[x];
}
};
int main() {
//range sum query
int n, q;
cin >> n >> q;
SqrtDecomposition dec(n);
vector<string> res;
rep(_, q) {
int c;
cin >> c;
if(c == 0) {
int s, t, x;
cin >> s >> t >> x;
++t;
dec.update(s, t, x);
} else {
int i;
cin >> i;
cout << dec.find(i) << endl;
}
}
}
``` |
#include<bits/stdc++.h>
using namespace std;
#define inf INT_MAX
#define INF LLONG_MAX
#define ll long long
#define ull unsigned long long
#define M (int)(1e9+7)
#define P pair<int,int>
#define FOR(i,m,n) for(int i=(int)m;i<(int)n;i++)
#define RFOR(i,m,n) for(int i=(int)m;i>=(int)n;i--)
#define rep(i,n) FOR(i,0,n)
#define rrep(i,n) RFOR(i,n,0)
#define all(a) a.begin(),a.end()
const int vx[4] = {0,1,0,-1};
const int vy[4] = {1,0,-1,0};
#define F first
#define S second
#define PB push_back
#define EB emplace_back
#define int ll
#define vi vector<int>
#define IP pair<int,P>
#define PI pair<P,int>
#define PP pair<P,P>
#define Yes(f){cout<<(f?"Yes":"No")<<endl;}
#define YES(f){cout<<(f?"YES":"NO")<<endl;}
int Madd(int x,int y) {return (x+y)%M;}
int Msub(int x,int y) {return (x-y+M)%M;}
int Mmul(int x,int y) {return (x*y)%M;}
template< typename OperatorMonoid >
struct DuelSegTree {
using H = function< OperatorMonoid(OperatorMonoid, OperatorMonoid) >;
int sz, height;
vector< OperatorMonoid > lazy;
const H h;
const OperatorMonoid OM0;
DuelSegTree(int n, const H h, const OperatorMonoid OM0)
: h(h), OM0(OM0) {
sz = 1;
height = 0;
while(sz < n) sz <<= 1, height++;
lazy.assign(2 * sz, OM0);
}
inline void propagate(int k) {
if(lazy[k] != OM0) {
lazy[2 * k + 0] = h(lazy[2 * k + 0], lazy[k]);
lazy[2 * k + 1] = h(lazy[2 * k + 1], lazy[k]);
lazy[k] = OM0;
}
}
inline void thrust(int k) {
for(int i = height; i > 0; i--) propagate(k >> i);
}
void update(int a, int b, const OperatorMonoid &x) {
thrust(a += sz);
thrust(b += sz - 1);
for(int l = a, r = b + 1; l < r; l >>= 1, r >>= 1) {
if(l & 1) lazy[l] = h(lazy[l], x), ++l;
if(r & 1) --r, lazy[r] = h(lazy[r], x);
}
}
OperatorMonoid operator[](int k) {
thrust(k += sz);
return lazy[k];
}
};
signed main(){
cin.tie(0);
ios::sync_with_stdio(false);
cout<<fixed<<setprecision(20);
int n,q;
cin>>n>>q;
auto h=[](int a,int b){return b;};
DuelSegTree<int> seg(n,h,inf);
while(q--){
int t;
cin>>t;
if(t==0){
int l,r,x;
cin>>l>>r>>x;
seg.update(l,r+1,x);
}else{
int x;
cin>>x;
cout<<seg[x]<<endl;
}
}
}
| ### Prompt
Develop a solution in CPP to the problem described below:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
#define inf INT_MAX
#define INF LLONG_MAX
#define ll long long
#define ull unsigned long long
#define M (int)(1e9+7)
#define P pair<int,int>
#define FOR(i,m,n) for(int i=(int)m;i<(int)n;i++)
#define RFOR(i,m,n) for(int i=(int)m;i>=(int)n;i--)
#define rep(i,n) FOR(i,0,n)
#define rrep(i,n) RFOR(i,n,0)
#define all(a) a.begin(),a.end()
const int vx[4] = {0,1,0,-1};
const int vy[4] = {1,0,-1,0};
#define F first
#define S second
#define PB push_back
#define EB emplace_back
#define int ll
#define vi vector<int>
#define IP pair<int,P>
#define PI pair<P,int>
#define PP pair<P,P>
#define Yes(f){cout<<(f?"Yes":"No")<<endl;}
#define YES(f){cout<<(f?"YES":"NO")<<endl;}
int Madd(int x,int y) {return (x+y)%M;}
int Msub(int x,int y) {return (x-y+M)%M;}
int Mmul(int x,int y) {return (x*y)%M;}
template< typename OperatorMonoid >
struct DuelSegTree {
using H = function< OperatorMonoid(OperatorMonoid, OperatorMonoid) >;
int sz, height;
vector< OperatorMonoid > lazy;
const H h;
const OperatorMonoid OM0;
DuelSegTree(int n, const H h, const OperatorMonoid OM0)
: h(h), OM0(OM0) {
sz = 1;
height = 0;
while(sz < n) sz <<= 1, height++;
lazy.assign(2 * sz, OM0);
}
inline void propagate(int k) {
if(lazy[k] != OM0) {
lazy[2 * k + 0] = h(lazy[2 * k + 0], lazy[k]);
lazy[2 * k + 1] = h(lazy[2 * k + 1], lazy[k]);
lazy[k] = OM0;
}
}
inline void thrust(int k) {
for(int i = height; i > 0; i--) propagate(k >> i);
}
void update(int a, int b, const OperatorMonoid &x) {
thrust(a += sz);
thrust(b += sz - 1);
for(int l = a, r = b + 1; l < r; l >>= 1, r >>= 1) {
if(l & 1) lazy[l] = h(lazy[l], x), ++l;
if(r & 1) --r, lazy[r] = h(lazy[r], x);
}
}
OperatorMonoid operator[](int k) {
thrust(k += sz);
return lazy[k];
}
};
signed main(){
cin.tie(0);
ios::sync_with_stdio(false);
cout<<fixed<<setprecision(20);
int n,q;
cin>>n>>q;
auto h=[](int a,int b){return b;};
DuelSegTree<int> seg(n,h,inf);
while(q--){
int t;
cin>>t;
if(t==0){
int l,r,x;
cin>>l>>r>>x;
seg.update(l,r+1,x);
}else{
int x;
cin>>x;
cout<<seg[x]<<endl;
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int N,Q;
const int INF=INT_MAX;
struct LazySegmentTree{
private:
int n;
vector<int> node,lazy;
vector<bool> lazyFlag;
public:
LazySegmentTree(vector<int> v){
int sz=(int)v.size();
n=1; while(n<sz)n*=2;
node.resize(2*n-1);
lazy.resize(2*n-1,INF);
lazyFlag.resize(2*n-1,false);
for(int i=0;i<sz;i++)node[i+n-1]=v[i];
for(int i=n-2;i>=0;i--){node[i]=min(node[2*i+1],node[2*i+2]);}
}
void lazyEvaluate(int k,int l,int r){
if(lazyFlag[k]){
node[k]=lazy[k];
if(r-l>1){
lazy[2*k+1]=lazy[2*k+2]=lazy[k];
lazyFlag[2*k+1]=lazyFlag[2*k+2]=true;
}
lazyFlag[k]=false;
}
}
void update(int a,int b,int x,int k=0,int l=0,int r=-1){
if(r<0)r=n;
lazyEvaluate(k,l,r);
if(b<=l || r<=a)return;
if(a<=l && r<=b){
lazy[k]=x;
lazyFlag[k]=true;
lazyEvaluate(k,l,r);
}
else{
update(a,b,x,2*k+1,l,(l+r)/2);
update(a,b,x,2*k+2,(l+r)/2,r);
node[k]=min(node[2*k+1],node[2*k+2]);
}
}
int find(int a,int b,int k=0,int l=0,int r=-1){
if(r<0)r=n;
lazyEvaluate(k,l,r);
if(b<=l || r<=a)return INF;
if(a<=l && r<=b)return node[k];
int vl=find(a,b,2*k+1,l,(l+r)/2);
int vr=find(a,b,2*k+2,(l+r)/2,r);
return min(vl,vr);
}
};
int main(){
cin >> N >> Q;
LazySegmentTree seg(vector<int>(N,INF));
for(int i=0;i<Q;i++){
int q; cin >> q;
if(q==0){
int s,t,x; cin >> s >> t >> x;
seg.update(s,t+1,x);
}
else{
int i; cin >> i;
cout << seg.find(i,i+1) << endl;
}
}
}
| ### Prompt
Generate a Cpp solution to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int N,Q;
const int INF=INT_MAX;
struct LazySegmentTree{
private:
int n;
vector<int> node,lazy;
vector<bool> lazyFlag;
public:
LazySegmentTree(vector<int> v){
int sz=(int)v.size();
n=1; while(n<sz)n*=2;
node.resize(2*n-1);
lazy.resize(2*n-1,INF);
lazyFlag.resize(2*n-1,false);
for(int i=0;i<sz;i++)node[i+n-1]=v[i];
for(int i=n-2;i>=0;i--){node[i]=min(node[2*i+1],node[2*i+2]);}
}
void lazyEvaluate(int k,int l,int r){
if(lazyFlag[k]){
node[k]=lazy[k];
if(r-l>1){
lazy[2*k+1]=lazy[2*k+2]=lazy[k];
lazyFlag[2*k+1]=lazyFlag[2*k+2]=true;
}
lazyFlag[k]=false;
}
}
void update(int a,int b,int x,int k=0,int l=0,int r=-1){
if(r<0)r=n;
lazyEvaluate(k,l,r);
if(b<=l || r<=a)return;
if(a<=l && r<=b){
lazy[k]=x;
lazyFlag[k]=true;
lazyEvaluate(k,l,r);
}
else{
update(a,b,x,2*k+1,l,(l+r)/2);
update(a,b,x,2*k+2,(l+r)/2,r);
node[k]=min(node[2*k+1],node[2*k+2]);
}
}
int find(int a,int b,int k=0,int l=0,int r=-1){
if(r<0)r=n;
lazyEvaluate(k,l,r);
if(b<=l || r<=a)return INF;
if(a<=l && r<=b)return node[k];
int vl=find(a,b,2*k+1,l,(l+r)/2);
int vr=find(a,b,2*k+2,(l+r)/2,r);
return min(vl,vr);
}
};
int main(){
cin >> N >> Q;
LazySegmentTree seg(vector<int>(N,INF));
for(int i=0;i<Q;i++){
int q; cin >> q;
if(q==0){
int s,t,x; cin >> s >> t >> x;
seg.update(s,t+1,x);
}
else{
int i; cin >> i;
cout << seg.find(i,i+1) << endl;
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> P;
#define FOR(I,A,B) for(int I=int(A);I<int(B);++I)
struct RUQ{//平方分割 0-index
int n,rn,bn,Q=0;
vector<int> a,b,at,bt;
RUQ(int n_){
n = n_; rn = sqrt(n); bn = ceil((double)n/rn);
a.resize(n,2147483647);at.resize(n,0);
b.resize(bn,2147483647);bt.resize(n,0);
}
int value(int s){
return (at[s]<bt[s/rn]?b[s/rn]:a[s]);
}
void updete(int s,int t,int x){
Q++;
FOR(i,s,t+1){
if(i%rn==0 && (i+rn-1)<=t){
b[i/rn] = x;
bt[i/rn] = Q;
i += rn-1;
}else{
a[i] = x;
at[i] = Q;
}
}
}
};
int main(){
int n,q,s,t,x,com;
cin >> n >> q;
RUQ data(n);
FOR(Q,1,q+1){
cin>>com;
if(com){//find
cin >> s;
cout << data.value(s) << "\n";
}else{//updete
cin >> s >> t >> x;
data.updete(s,t,x);
}
}
}
| ### Prompt
Generate a cpp solution to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> P;
#define FOR(I,A,B) for(int I=int(A);I<int(B);++I)
struct RUQ{//平方分割 0-index
int n,rn,bn,Q=0;
vector<int> a,b,at,bt;
RUQ(int n_){
n = n_; rn = sqrt(n); bn = ceil((double)n/rn);
a.resize(n,2147483647);at.resize(n,0);
b.resize(bn,2147483647);bt.resize(n,0);
}
int value(int s){
return (at[s]<bt[s/rn]?b[s/rn]:a[s]);
}
void updete(int s,int t,int x){
Q++;
FOR(i,s,t+1){
if(i%rn==0 && (i+rn-1)<=t){
b[i/rn] = x;
bt[i/rn] = Q;
i += rn-1;
}else{
a[i] = x;
at[i] = Q;
}
}
}
};
int main(){
int n,q,s,t,x,com;
cin >> n >> q;
RUQ data(n);
FOR(Q,1,q+1){
cin>>com;
if(com){//find
cin >> s;
cout << data.value(s) << "\n";
}else{//updete
cin >> s >> t >> x;
data.updete(s,t,x);
}
}
}
``` |
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 2147483647
//static const double INF = (1 << 31)-1;
int n,q;
double d[400000],lazy[400000];
double x;
void lazy_evaluate(int k,int l,int r){
if(lazy[k] == INF)return;
else{
d[k] = lazy[k];
if(r - l > 1){
lazy[2*k+1] = lazy[k];
lazy[2*k+2] = lazy[k];
/*
lazy_evaluate(2*k+1);
lazy_evaluate(2*k+2);
*/
}
lazy[k] = INF;
}
}
/*
void update(int k,double a){
k += n - 1;
d[k] = a;
while(k > 0){
k = (k - 1)/2;
d[k] = min(d[k*2+1], d[k*2+2]);
}
}
*/
void query(int a, int b, int k=0, int l=0, int r=-1){
if(r < 0){
r = n;
}
//cout << "haittemasu" << endl;
lazy_evaluate(k,l,r);
if(b <= l || r <= a) return;
else if(a <= l && r <= b){
lazy[k] = x;
lazy_evaluate(k,l,r);
}
else{
query(a,b,2*k+1,l,(l+r)/2);
query(a,b,2*k+2,(l+r)/2,r);
d[k] = min(d[2*k+1], d[2*k+2]);
}
}
/*
double query(int a, int b, int k, int l, int r){
if(r <= a || b <= l){
return INF;
}
else if(a <= l && r <= b){
return d[k];
}
double vl = query(a, b, k*2+1, l, (l + r)/2);
double vr = query(a, b, k*2+2, (l + r)/2, r);
return min(vl, vr);
}
double findMin(int a, int b){
return query(a, b+1, 0, 0, n);
}
*/
void find(int a, int b, int k=0, int l=0, int r=-1){
if(r < 0){
r = n;
}
lazy_evaluate(k,l,r);
if(b <= l || r <= a){
return;
}
else if(a <= l && r <= b){
return;
}
//double vl =
find(a, b, k*2+1, l, (l + r)/2);
//double vr =
find(a, b, k*2+2, (l + r)/2, r);
//return min(vl, vr);
}
int main(){
int n_;
cin >> n_ >> q;
n=1;
while(n<n_){
n*=2;
}
for(int i = 0;i < n*2-1;i++){
d[i] = INF;
lazy[i] = INF;
}
int a, b;
double s,t;
for(int i = 0;i < q;i++){
cin >> a;
if(a == 0){
cin >> s >> t >> x;
query(s,t+1);
}
else{
cin >> b;
find(b,b+1);
printf("%.0f\n",d[b+n-1]);
}
}
/*
for(int i = 0;i < n*2-1;i++){
cout << d[i] << endl;
}
*/
return 0;
}
| ### Prompt
Please create a solution in Cpp to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 2147483647
//static const double INF = (1 << 31)-1;
int n,q;
double d[400000],lazy[400000];
double x;
void lazy_evaluate(int k,int l,int r){
if(lazy[k] == INF)return;
else{
d[k] = lazy[k];
if(r - l > 1){
lazy[2*k+1] = lazy[k];
lazy[2*k+2] = lazy[k];
/*
lazy_evaluate(2*k+1);
lazy_evaluate(2*k+2);
*/
}
lazy[k] = INF;
}
}
/*
void update(int k,double a){
k += n - 1;
d[k] = a;
while(k > 0){
k = (k - 1)/2;
d[k] = min(d[k*2+1], d[k*2+2]);
}
}
*/
void query(int a, int b, int k=0, int l=0, int r=-1){
if(r < 0){
r = n;
}
//cout << "haittemasu" << endl;
lazy_evaluate(k,l,r);
if(b <= l || r <= a) return;
else if(a <= l && r <= b){
lazy[k] = x;
lazy_evaluate(k,l,r);
}
else{
query(a,b,2*k+1,l,(l+r)/2);
query(a,b,2*k+2,(l+r)/2,r);
d[k] = min(d[2*k+1], d[2*k+2]);
}
}
/*
double query(int a, int b, int k, int l, int r){
if(r <= a || b <= l){
return INF;
}
else if(a <= l && r <= b){
return d[k];
}
double vl = query(a, b, k*2+1, l, (l + r)/2);
double vr = query(a, b, k*2+2, (l + r)/2, r);
return min(vl, vr);
}
double findMin(int a, int b){
return query(a, b+1, 0, 0, n);
}
*/
void find(int a, int b, int k=0, int l=0, int r=-1){
if(r < 0){
r = n;
}
lazy_evaluate(k,l,r);
if(b <= l || r <= a){
return;
}
else if(a <= l && r <= b){
return;
}
//double vl =
find(a, b, k*2+1, l, (l + r)/2);
//double vr =
find(a, b, k*2+2, (l + r)/2, r);
//return min(vl, vr);
}
int main(){
int n_;
cin >> n_ >> q;
n=1;
while(n<n_){
n*=2;
}
for(int i = 0;i < n*2-1;i++){
d[i] = INF;
lazy[i] = INF;
}
int a, b;
double s,t;
for(int i = 0;i < q;i++){
cin >> a;
if(a == 0){
cin >> s >> t >> x;
query(s,t+1);
}
else{
cin >> b;
find(b,b+1);
printf("%.0f\n",d[b+n-1]);
}
}
/*
for(int i = 0;i < n*2-1;i++){
cout << d[i] << endl;
}
*/
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
#define rep(i,x,y) for(int i=(x);i<(y);++i)
#define debug(x) #x << "=" << (x)
#ifdef DEBUG
#define _GLIBCXX_DEBUG
#define print(x) std::cerr << debug(x) << " (L:" << __LINE__ << ")" << std::endl
#else
#define print(x)
#endif
const int inf=1e9;
const int64_t inf64=1e18;
const double eps=1e-9;
template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec){
os << "[";
for (const auto &v : vec) {
os << v << ",";
}
os << "]";
return os;
}
using i64=int64_t;
class sqrt_decomposition{
public:
const int B=1000;
int n;
vector<int> data,lazy;
vector<bool> flag;
sqrt_decomposition(int n,int init):n(n),data(n,init),lazy((n+B-1)/B),flag((n+B-1)/B){}
void propagate(int i){
assert(flag[i]);
flag[i]=false;
rep(j,0,B){
if(i*B+j>=n) break;
data[i*B+j]=lazy[i];
}
}
//[s,t)
void update(int s,int t,int x){
while(s<t and s%B!=0){
if(flag[s/B]) propagate(s/B);
data[s]=x;
++s;
}
while(s<t and t%B!=0){
--t;
if(flag[t/B]) propagate(t/B);
data[t]=x;
}
while(s<t){
int i=s/B;
flag[i]=true;
lazy[i]=x;
s+=B;
}
}
int at(int i){
if(flag[i/B]) return lazy[i/B];
return data[i];
}
};
void solve(){
int n,q;
cin >> n >> q;
sqrt_decomposition sd(n,(1LL<<31)-1);
rep(i,0,q){
int k;
cin >> k;
if(k==0){
int s,t,x;
cin >> s >> t >> x;
++t;
sd.update(s,t,x);
}else{
int i;
cin >> i;
cout << sd.at(i) << endl;
}
}
}
int main(){
std::cin.tie(0);
std::ios::sync_with_stdio(false);
cout.setf(ios::fixed);
cout.precision(16);
solve();
return 0;
} | ### Prompt
Your task is to create a cpp solution to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
#define rep(i,x,y) for(int i=(x);i<(y);++i)
#define debug(x) #x << "=" << (x)
#ifdef DEBUG
#define _GLIBCXX_DEBUG
#define print(x) std::cerr << debug(x) << " (L:" << __LINE__ << ")" << std::endl
#else
#define print(x)
#endif
const int inf=1e9;
const int64_t inf64=1e18;
const double eps=1e-9;
template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec){
os << "[";
for (const auto &v : vec) {
os << v << ",";
}
os << "]";
return os;
}
using i64=int64_t;
class sqrt_decomposition{
public:
const int B=1000;
int n;
vector<int> data,lazy;
vector<bool> flag;
sqrt_decomposition(int n,int init):n(n),data(n,init),lazy((n+B-1)/B),flag((n+B-1)/B){}
void propagate(int i){
assert(flag[i]);
flag[i]=false;
rep(j,0,B){
if(i*B+j>=n) break;
data[i*B+j]=lazy[i];
}
}
//[s,t)
void update(int s,int t,int x){
while(s<t and s%B!=0){
if(flag[s/B]) propagate(s/B);
data[s]=x;
++s;
}
while(s<t and t%B!=0){
--t;
if(flag[t/B]) propagate(t/B);
data[t]=x;
}
while(s<t){
int i=s/B;
flag[i]=true;
lazy[i]=x;
s+=B;
}
}
int at(int i){
if(flag[i/B]) return lazy[i/B];
return data[i];
}
};
void solve(){
int n,q;
cin >> n >> q;
sqrt_decomposition sd(n,(1LL<<31)-1);
rep(i,0,q){
int k;
cin >> k;
if(k==0){
int s,t,x;
cin >> s >> t >> x;
++t;
sd.update(s,t,x);
}else{
int i;
cin >> i;
cout << sd.at(i) << endl;
}
}
}
int main(){
std::cin.tie(0);
std::ios::sync_with_stdio(false);
cout.setf(ios::fixed);
cout.precision(16);
solve();
return 0;
}
``` |
#include <iostream>
#include <vector>
using namespace std;
class SegTree{
private:
int n;
// 1-index (tree[0] is dummy), leaves are tree[n]-tree[2n-1]
vector<int> tree;
const int defVal = 2147483647;
public:
SegTree(int _n){
int t;
t = 1;
while (t<_n) t*=2;
n = t;
tree = vector<int>(2*n, defVal);
//cerr << "finish init" << endl;
//cerr << "tree size is " << 2*n << endl;
}
// stab function
void update(int l, int r, int x){ return _update(l,r,x,1,0,n);}
void _update(int l, int r, int x, int k, int kl, int kr){
// cerr << "update: "<<l<<" "<<r<<" "<<x<<" "<<k<<" "<<kl<<" "<<kr<<endl;
if (kr<=l || r<=kl) return;
else if (l<=kl && kr<=r){
tree[k] = x;
return;
} else {
if (tree[k]!=defVal){
tree[k*2]=tree[k]; tree[k*2+1]=tree[k];
tree[k]=defVal;
}
_update(l,r, x, k*2, kl,(kl+kr)/2);
_update(l,r, x, k*2+1, (kl+kr)/2,kr);
}
}
int find(int i){
return _find(i,1,0,n);
}
int _find(int i, int k, int kl, int kr){
// cerr << "find: "<<i<<" "<<k<<" "<<kl<<" "<<kr<<endl;
if (k>=n || tree[k]!=defVal) return tree[k];
if (i<(kl+kr)/2) return _find(i, k*2, kl,(kl+kr)/2);
else return _find(i, k*2+1, (kl+kr)/2,kr);
}
// debug
void print(){
cerr << "debug: ";
for (int i=1; i<2*n; i++){
if (i==n) cout << "| ";
if (tree[i]==defVal) cout << "DV ";
else cout << tree[i] << " ";
}
cout << endl;
}
};
int main(void){
int n,q;
SegTree *st;
cin >> n >> q;
st = new SegTree(n);
for (int i=0; i<q; i++){
int c,s,t,x,id;
cin >> c;
if (c==0){
cin >> s >> t >> x;
st->update(s,t+1,x);
}
else if (c==1){
cin >> id;
cout << st->find(id) << endl;
}
else cout << "illegal command" << endl;
//st->print();
}
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <iostream>
#include <vector>
using namespace std;
class SegTree{
private:
int n;
// 1-index (tree[0] is dummy), leaves are tree[n]-tree[2n-1]
vector<int> tree;
const int defVal = 2147483647;
public:
SegTree(int _n){
int t;
t = 1;
while (t<_n) t*=2;
n = t;
tree = vector<int>(2*n, defVal);
//cerr << "finish init" << endl;
//cerr << "tree size is " << 2*n << endl;
}
// stab function
void update(int l, int r, int x){ return _update(l,r,x,1,0,n);}
void _update(int l, int r, int x, int k, int kl, int kr){
// cerr << "update: "<<l<<" "<<r<<" "<<x<<" "<<k<<" "<<kl<<" "<<kr<<endl;
if (kr<=l || r<=kl) return;
else if (l<=kl && kr<=r){
tree[k] = x;
return;
} else {
if (tree[k]!=defVal){
tree[k*2]=tree[k]; tree[k*2+1]=tree[k];
tree[k]=defVal;
}
_update(l,r, x, k*2, kl,(kl+kr)/2);
_update(l,r, x, k*2+1, (kl+kr)/2,kr);
}
}
int find(int i){
return _find(i,1,0,n);
}
int _find(int i, int k, int kl, int kr){
// cerr << "find: "<<i<<" "<<k<<" "<<kl<<" "<<kr<<endl;
if (k>=n || tree[k]!=defVal) return tree[k];
if (i<(kl+kr)/2) return _find(i, k*2, kl,(kl+kr)/2);
else return _find(i, k*2+1, (kl+kr)/2,kr);
}
// debug
void print(){
cerr << "debug: ";
for (int i=1; i<2*n; i++){
if (i==n) cout << "| ";
if (tree[i]==defVal) cout << "DV ";
else cout << tree[i] << " ";
}
cout << endl;
}
};
int main(void){
int n,q;
SegTree *st;
cin >> n >> q;
st = new SegTree(n);
for (int i=0; i<q; i++){
int c,s,t,x,id;
cin >> c;
if (c==0){
cin >> s >> t >> x;
st->update(s,t+1,x);
}
else if (c==1){
cin >> id;
cout << st->find(id) << endl;
}
else cout << "illegal command" << endl;
//st->print();
}
return 0;
}
``` |
#include<cstdio>
#include<climits>
using namespace std;
static const int MAX_N = 100000;
int n, q;
int n_;
int laz[4 * MAX_N];
void init(){
n_ = 1;
while(n_ < n) n_ *= 2;
for(int i = 0; i < 2 * n_ - 1; i++){
laz[i] = -1;
}
laz[0] = INT_MAX;
}
void prop(int len, int k){
if(laz[k] == -1) return;
if(2 * k + 1 < 2 * n_ - 1){
laz[2 * k + 1] = laz[k];
laz[2 * k + 2] = laz[k];
}
}
void update(int a, int b, int x, int k, int l, int r){
if(r <= a || b <= l) return;
if(a <= l && r <= b){
laz[k] = x;
return;
}
prop(r - l, k);
laz[k] = -1;
update(a, b, x, 2 * k + 1, l, (l + r) / 2);
update(a, b, x, 2 * k + 2, (l + r) / 2, r);
}
int find(int index, int len, int k){
if(laz[k] != -1) return laz[k];
if(2 * k + 1 >= 2 * n_ - 1) return laz[k];
int diff = ((index % len < len / 2) ? 1 : 2);
return find(index, len / 2, 2 * k + diff);
}
int main(){
scanf("%d %d", &n, &q);
init();
for(int i = 0; i < q; i++){
int op;
scanf("%d", &op);
if(op == 0){
int s, t, x;
scanf("%d %d %d", &s, &t, &x);
update(s, t + 1, x, 0, 0, n_);
}else if(op == 1){
int index;
scanf("%d", &index);
printf("%d\n", find(index, n_, 0));
}
}
return 0;
}
| ### Prompt
In cpp, your task is to solve the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include<cstdio>
#include<climits>
using namespace std;
static const int MAX_N = 100000;
int n, q;
int n_;
int laz[4 * MAX_N];
void init(){
n_ = 1;
while(n_ < n) n_ *= 2;
for(int i = 0; i < 2 * n_ - 1; i++){
laz[i] = -1;
}
laz[0] = INT_MAX;
}
void prop(int len, int k){
if(laz[k] == -1) return;
if(2 * k + 1 < 2 * n_ - 1){
laz[2 * k + 1] = laz[k];
laz[2 * k + 2] = laz[k];
}
}
void update(int a, int b, int x, int k, int l, int r){
if(r <= a || b <= l) return;
if(a <= l && r <= b){
laz[k] = x;
return;
}
prop(r - l, k);
laz[k] = -1;
update(a, b, x, 2 * k + 1, l, (l + r) / 2);
update(a, b, x, 2 * k + 2, (l + r) / 2, r);
}
int find(int index, int len, int k){
if(laz[k] != -1) return laz[k];
if(2 * k + 1 >= 2 * n_ - 1) return laz[k];
int diff = ((index % len < len / 2) ? 1 : 2);
return find(index, len / 2, 2 * k + diff);
}
int main(){
scanf("%d %d", &n, &q);
init();
for(int i = 0; i < q; i++){
int op;
scanf("%d", &op);
if(op == 0){
int s, t, x;
scanf("%d %d %d", &s, &t, &x);
update(s, t + 1, x, 0, 0, n_);
}else if(op == 1){
int index;
scanf("%d", &index);
printf("%d\n", find(index, n_, 0));
}
}
return 0;
}
``` |
#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
typedef vector<int> vec;
class RUQ
{
public:
vec data;
vec lazy;
int n;
RUQ(int _n)
{
data.resize(4*_n);
lazy.resize(4*_n);
for(n=1;n<_n;n<<=1);
fill(data.begin(),data.end(),(1LL<<31)-1);
fill(lazy.begin(),lazy.end(),-1);
}
void Down(int p,int l,int r)
{
if(lazy[p]==-1) return;
if(l+1>=r)
{
data[p]=lazy[p];
}else
{
lazy[p*2+1]=lazy[p*2+2]=lazy[p];
}
lazy[p]=-1;
}
void update(int p,int l,int r,int x,int a,int b)
{
if(a>=r||b<=l) return;
if(a<=l&&r<=b)
{
lazy[p]=x;
return;
}
Down(p,l,r);
int mid=(l+r)/2;
update(p*2+1,l,mid,x,a,b);
update(p*2+2,mid,r,x,a,b);
}
int query(int p,int l,int r,int a,int b)
{
if(a>=r||b<=l) return -1;
Down(p,l,r);
if(a<=l&&r<=b) return data[p];
int mid=(l+r)/2;
if(a<mid)
{
return query(p*2+1,l,mid,a,b);
}else
{
return query(p*2+2,mid,r,a,b);
}
}
};
int main()
{
int n,q;
while(scanf("%d %d",&n,&q)!=EOF)
{
RUQ ruq(n);
while(q--)
{
int op,a,b,c;
scanf("%d",&op);
if(op)
{
scanf("%d",&a);
printf("%d\n",ruq.query(0,0,ruq.n,a,a+1));
}else
{
scanf("%d %d %d",&a,&b,&c);
ruq.update(0,0,ruq.n,c,a,b+1);
}
}
}
return 0;
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
typedef vector<int> vec;
class RUQ
{
public:
vec data;
vec lazy;
int n;
RUQ(int _n)
{
data.resize(4*_n);
lazy.resize(4*_n);
for(n=1;n<_n;n<<=1);
fill(data.begin(),data.end(),(1LL<<31)-1);
fill(lazy.begin(),lazy.end(),-1);
}
void Down(int p,int l,int r)
{
if(lazy[p]==-1) return;
if(l+1>=r)
{
data[p]=lazy[p];
}else
{
lazy[p*2+1]=lazy[p*2+2]=lazy[p];
}
lazy[p]=-1;
}
void update(int p,int l,int r,int x,int a,int b)
{
if(a>=r||b<=l) return;
if(a<=l&&r<=b)
{
lazy[p]=x;
return;
}
Down(p,l,r);
int mid=(l+r)/2;
update(p*2+1,l,mid,x,a,b);
update(p*2+2,mid,r,x,a,b);
}
int query(int p,int l,int r,int a,int b)
{
if(a>=r||b<=l) return -1;
Down(p,l,r);
if(a<=l&&r<=b) return data[p];
int mid=(l+r)/2;
if(a<mid)
{
return query(p*2+1,l,mid,a,b);
}else
{
return query(p*2+2,mid,r,a,b);
}
}
};
int main()
{
int n,q;
while(scanf("%d %d",&n,&q)!=EOF)
{
RUQ ruq(n);
while(q--)
{
int op,a,b,c;
scanf("%d",&op);
if(op)
{
scanf("%d",&a);
printf("%d\n",ruq.query(0,0,ruq.n,a,a+1));
}else
{
scanf("%d %d %d",&a,&b,&c);
ruq.update(0,0,ruq.n,c,a,b+1);
}
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
#define REP(i, n) for(int i = 0; i < n; i++)
#define REPR(i, n) for(int i = n - 1; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i <= n; i++)
#define FORR(i, m, n) for(int i = m; i >= n; i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using P = pair<ll, ll>;
int B = 320;
// バケット i は [i*B, (i+1)*B) の記録を持つ
// lazy[i] := バケットi の遅延を記録する
// 遅延が-1 ... 制約よりそんなことは発生しないことを利用して, これを遅延の単位元とできる
// i.e. lazy[i] == -1 def 消化時に何も伝播させないという遅延を与えるとみなす
vector<int> lazy(B, -1), a(100000, (1LL << 31) - 1);
// a_s, ..., a_t を x に変更する
void update(int s, int t, int x)
{
// s の属するバケット, t の属するバケット
int st = s / B, gt = t / B;
// st や gt に遅延が存在するとき更新前に消化しておく
// 遅延はパケットごとに考えるから, 遅延が存在⇔パケット内は全て等しい値で更新されている
// 区間に被覆されるパケットについてはこのあと新しい値へ更新されるので消化しなくてOK
if (lazy[st] != -1){
for (int i = st * B; i < (st + 1) * B; i++) a[i] = lazy[st];
lazy[st] = -1;
}
if (lazy[gt] != -1){
for (int i = gt * B; i < (gt + 1) * B; i++) a[i] = lazy[gt];
lazy[gt] = -1;
}
//... || ... s ... t ... || ...
//更新が同一パケット内で済むなら直接更新しても間に合う
if (st == gt){
for (int i = s; i <= t; i++) a[i] = x;
}
//... s ... || (被覆されるパケット) || ... t ...
else{
//端が属するパケット内は直接更新
for (int i = s; i < (st + 1) * B; i++) a[i] = x;
for (int i = gt * B; i <= t; i++) a[i] = x;
//区間に被覆されるパケットについては更新を遅延に残す
for (int i = st + 1; i < gt; i++){
lazy[i] = x;
}
}
}
// a_i を返す
int find(int s){
int st = s / B;
//遅延を処理する
if (lazy[st] != -1){
for (int i = st * B; i < (st + 1) * B; i++) a[i] = lazy[st];
lazy[st] = -1;
}
return a[s];
}
int main(){
cin.tie(0);
ios::sync_with_stdio(false);
int n, q;
cin >> n >> q;
REP(i, q){
int type;
cin >> type;
if(type == 0){
int s, t, x;
cin >> s >> t >> x;
update(s, t, x);
}
else{
int s;
cin >> s;
cout << find(s) << endl;
}
}
return 0;
}
| ### Prompt
In Cpp, your task is to solve the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
#define REP(i, n) for(int i = 0; i < n; i++)
#define REPR(i, n) for(int i = n - 1; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i <= n; i++)
#define FORR(i, m, n) for(int i = m; i >= n; i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using P = pair<ll, ll>;
int B = 320;
// バケット i は [i*B, (i+1)*B) の記録を持つ
// lazy[i] := バケットi の遅延を記録する
// 遅延が-1 ... 制約よりそんなことは発生しないことを利用して, これを遅延の単位元とできる
// i.e. lazy[i] == -1 def 消化時に何も伝播させないという遅延を与えるとみなす
vector<int> lazy(B, -1), a(100000, (1LL << 31) - 1);
// a_s, ..., a_t を x に変更する
void update(int s, int t, int x)
{
// s の属するバケット, t の属するバケット
int st = s / B, gt = t / B;
// st や gt に遅延が存在するとき更新前に消化しておく
// 遅延はパケットごとに考えるから, 遅延が存在⇔パケット内は全て等しい値で更新されている
// 区間に被覆されるパケットについてはこのあと新しい値へ更新されるので消化しなくてOK
if (lazy[st] != -1){
for (int i = st * B; i < (st + 1) * B; i++) a[i] = lazy[st];
lazy[st] = -1;
}
if (lazy[gt] != -1){
for (int i = gt * B; i < (gt + 1) * B; i++) a[i] = lazy[gt];
lazy[gt] = -1;
}
//... || ... s ... t ... || ...
//更新が同一パケット内で済むなら直接更新しても間に合う
if (st == gt){
for (int i = s; i <= t; i++) a[i] = x;
}
//... s ... || (被覆されるパケット) || ... t ...
else{
//端が属するパケット内は直接更新
for (int i = s; i < (st + 1) * B; i++) a[i] = x;
for (int i = gt * B; i <= t; i++) a[i] = x;
//区間に被覆されるパケットについては更新を遅延に残す
for (int i = st + 1; i < gt; i++){
lazy[i] = x;
}
}
}
// a_i を返す
int find(int s){
int st = s / B;
//遅延を処理する
if (lazy[st] != -1){
for (int i = st * B; i < (st + 1) * B; i++) a[i] = lazy[st];
lazy[st] = -1;
}
return a[s];
}
int main(){
cin.tie(0);
ios::sync_with_stdio(false);
int n, q;
cin >> n >> q;
REP(i, q){
int type;
cin >> type;
if(type == 0){
int s, t, x;
cin >> s >> t >> x;
update(s, t, x);
}
else{
int s;
cin >> s;
cout << find(s) << endl;
}
}
return 0;
}
``` |
#include <iostream>
#include <climits>
#include <vector>
#include <algorithm>
using namespace std;
int find(int x,int n,vector<int>& node) {
x += n - 1;
int ret = node[x];
while (x > 0) {
x = (x - 1) / 2;
ret = max(ret, node[x]);
}
return ret;
}
void update(int a, int b,int i, int k, int l, int r, vector<int>& node) {
if (r <= a || b <= l)return;
if (a <= l && r <= b) {
node[k] = i; return;
}
else {
update(a, b, i, 2 * k + 1, l, (l + r) / 2, node);
update(a, b, i, 2 * k + 2, (l + r) / 2, r, node);
}
}
int main() {
cin.tie(0); ios::sync_with_stdio(false);
int n_, q; cin >> n_ >> q;
int n = 1;
while (n < n_) n *= 2;
vector<int> node(2 * n - 1, 0);
vector<int> ans(q + 1, INT_MAX);
for (int i = 0; i < q; i++) {
int com; cin >> com;
if (com == 0) {
int s, t, x; cin >> s >> t >> x;
ans[i + 1] = x;
update(s, t + 1, i + 1, 0, 0, n, node);
}
else{
int x; cin >> x;
cout << ans[find(x,n,node)] << endl;
}
}
} | ### Prompt
Create a solution in Cpp for the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <iostream>
#include <climits>
#include <vector>
#include <algorithm>
using namespace std;
int find(int x,int n,vector<int>& node) {
x += n - 1;
int ret = node[x];
while (x > 0) {
x = (x - 1) / 2;
ret = max(ret, node[x]);
}
return ret;
}
void update(int a, int b,int i, int k, int l, int r, vector<int>& node) {
if (r <= a || b <= l)return;
if (a <= l && r <= b) {
node[k] = i; return;
}
else {
update(a, b, i, 2 * k + 1, l, (l + r) / 2, node);
update(a, b, i, 2 * k + 2, (l + r) / 2, r, node);
}
}
int main() {
cin.tie(0); ios::sync_with_stdio(false);
int n_, q; cin >> n_ >> q;
int n = 1;
while (n < n_) n *= 2;
vector<int> node(2 * n - 1, 0);
vector<int> ans(q + 1, INT_MAX);
for (int i = 0; i < q; i++) {
int com; cin >> com;
if (com == 0) {
int s, t, x; cin >> s >> t >> x;
ans[i + 1] = x;
update(s, t + 1, i + 1, 0, 0, n, node);
}
else{
int x; cin >> x;
cout << ans[find(x,n,node)] << endl;
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
const int sqrtN = 512;
const ll inf = (1LL<<31)-1;
int n, q;
cin >> n >> q;
int bucket_sz = (n+sqrtN-1)/sqrtN;
vector<ll> a(n, inf);
vector<ll> bucketUp(bucket_sz, 0);
vector<bool> bucketFlag(bucket_sz, false);
auto push = [&](int k) {
bucketFlag[k] = false;
int l = k*sqrtN, r = (k+1)*sqrtN;
for(int i = l; i < r; ++i) a[i] = bucketUp[k];
};
auto update = [&](int s, int t, int x) {
for(int i = 0; i < bucket_sz; ++i) {
int l = i*sqrtN, r = (i+1)*sqrtN;
if(r <= s || t <= l) continue;
if(s <= l && r <= t) {
bucketUp[i] = x;
bucketFlag[i] = true;
} else {
if(bucketFlag[i]) push(i);
for(int j = max(s, l); j < min(t, r); ++j) {
a[j] = x;
}
}
}
};
auto get = [&](int k)->ll{
if(bucketFlag[k/sqrtN]) push(k/sqrtN);
return a[k];
};
while(q--) {
int t;
cin >> t;
if(t == 0) {
ll s, t, x;
cin >> s >> t >> x;
update(s, t+1, x);
} else {
int k;
cin >> k;
cout << get(k) << endl;
}
}
return 0;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
const int sqrtN = 512;
const ll inf = (1LL<<31)-1;
int n, q;
cin >> n >> q;
int bucket_sz = (n+sqrtN-1)/sqrtN;
vector<ll> a(n, inf);
vector<ll> bucketUp(bucket_sz, 0);
vector<bool> bucketFlag(bucket_sz, false);
auto push = [&](int k) {
bucketFlag[k] = false;
int l = k*sqrtN, r = (k+1)*sqrtN;
for(int i = l; i < r; ++i) a[i] = bucketUp[k];
};
auto update = [&](int s, int t, int x) {
for(int i = 0; i < bucket_sz; ++i) {
int l = i*sqrtN, r = (i+1)*sqrtN;
if(r <= s || t <= l) continue;
if(s <= l && r <= t) {
bucketUp[i] = x;
bucketFlag[i] = true;
} else {
if(bucketFlag[i]) push(i);
for(int j = max(s, l); j < min(t, r); ++j) {
a[j] = x;
}
}
}
};
auto get = [&](int k)->ll{
if(bucketFlag[k/sqrtN]) push(k/sqrtN);
return a[k];
};
while(q--) {
int t;
cin >> t;
if(t == 0) {
ll s, t, x;
cin >> s >> t >> x;
update(s, t+1, x);
} else {
int k;
cin >> k;
cout << get(k) << endl;
}
}
return 0;
}
``` |
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
#define INF 2147483647
vector<pair<int,int> > a(300000,pair<int,int>(INF,0));
int n,si;
int find(int);
void update(int,int,int,int,int,int,int);
int main(){
int q,f;
si=1;
cin >> n >> q;
for(int i=0;;i++){
si*=2;
if(si>=n)break;
}
for(int i=0;i<q;i++){
cin >> f;
if(f==1){
int po,dat;
cin >> po;
dat=find(po);
cout << dat << endl;
}
else{
int st,go,d;
cin >> st >> go>>d;
update(st,go+1,0,0,si,d,i+1);
}
}
return 0;
}
int find(int p){
int ko=si-1+p;
pair<int,int> ma=a[ko];
while(ko>0){
ko=(ko-1)/2;
if(ma.second<a[ko].second){
ma=a[ko];
}
}
return ma.first;
}
void update (int s,int t,int k,int l,int r,int d,int ti){
int m=(l+r)/2;
if(s<=l&&r<=t){
a[k].first=d;
a[k].second=ti;
return;
}
else if(r<=s||t<=l)return;
else{
update(s,t,k*2+1,l,m,d,ti);
update(s,t,k*2+2,m,r,d,ti);
}
} | ### Prompt
In Cpp, your task is to solve the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
#define INF 2147483647
vector<pair<int,int> > a(300000,pair<int,int>(INF,0));
int n,si;
int find(int);
void update(int,int,int,int,int,int,int);
int main(){
int q,f;
si=1;
cin >> n >> q;
for(int i=0;;i++){
si*=2;
if(si>=n)break;
}
for(int i=0;i<q;i++){
cin >> f;
if(f==1){
int po,dat;
cin >> po;
dat=find(po);
cout << dat << endl;
}
else{
int st,go,d;
cin >> st >> go>>d;
update(st,go+1,0,0,si,d,i+1);
}
}
return 0;
}
int find(int p){
int ko=si-1+p;
pair<int,int> ma=a[ko];
while(ko>0){
ko=(ko-1)/2;
if(ma.second<a[ko].second){
ma=a[ko];
}
}
return ma.first;
}
void update (int s,int t,int k,int l,int r,int d,int ti){
int m=(l+r)/2;
if(s<=l&&r<=t){
a[k].first=d;
a[k].second=ti;
return;
}
else if(r<=s||t<=l)return;
else{
update(s,t,k*2+1,l,m,d,ti);
update(s,t,k*2+2,m,r,d,ti);
}
}
``` |
#include<iostream>
#include<algorithm>
#include<math.h>
#include<queue>
#include<vector>
#include<climits>
#include<map>
#include<string>
#include<functional>
#include<iomanip>
#include<deque>
#include<random>
#include<set>
using namespace std;
typedef long long ll;
typedef double lldo;
#define mp make_pair
#define pub push_back
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define sz(x) int(x.size())
ll gcd(ll a, ll b) { if (a % b == 0) { return b; } else return gcd(b, a % b); }
ll lcm(ll a, ll b) { if (a == 0) { return b; }return a / gcd(a, b) * b; }
template<class T>ll LBI(vector<T>& ar, T in) { return lower_bound(ar.begin(), ar.end(), in) - ar.begin(); }
template<class T>ll UBI(vector<T>& ar, T in) { return upper_bound(ar.begin(), ar.end(), in) - ar.begin(); }
ll n, q, com, s, t, x, i;
const int INF = (1LL << 31) - 1;
const int sqrtN = 512;
struct SqrtDecomposition {
int N, K;
vector<int> data;
vector<bool> lazyFlag;
vector<int> lazyUpdate;
SqrtDecomposition(int n) : N(n) {
K = (N + sqrtN - 1) / sqrtN;
data.assign(K * sqrtN, INF);
lazyFlag.assign(K, false);
lazyUpdate.assign(K, 0);
}
void eval(int k) {
if (lazyFlag[k]) {
lazyFlag[k] = false;
for (int i = k * sqrtN; i < (k + 1) * sqrtN; ++i) {
data[i] = lazyUpdate[k];
}
}
}
// [s, t)
void update(int s, int t, int x) {
for (int k = 0; k < K; ++k) {
int l = k * sqrtN, r = (k + 1) * sqrtN;
if (r <= s || t <= l)
continue;
if (s <= l && r <= t) {
lazyFlag[k] = true;
lazyUpdate[k] = x;
}
else {
eval(k);
for (int i = max(s, l); i < min(t, r); ++i) {
data[i] = x;
}
}
}
}
int find(int i) {
int k = i / sqrtN;
eval(k);
return data[i];
}
};
int main() {
cin >> n >> q;
SqrtDecomposition ruq(n);
rep(j, q) {
cin >> com;
if (com == 0) {
cin >> s >> t >> x;
ruq.update(s, t + 1, x);
}
else {
cin >> i;
cout << ruq.find(i) << endl;
}
}
return 0;
}
| ### Prompt
Generate a Cpp solution to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include<iostream>
#include<algorithm>
#include<math.h>
#include<queue>
#include<vector>
#include<climits>
#include<map>
#include<string>
#include<functional>
#include<iomanip>
#include<deque>
#include<random>
#include<set>
using namespace std;
typedef long long ll;
typedef double lldo;
#define mp make_pair
#define pub push_back
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define sz(x) int(x.size())
ll gcd(ll a, ll b) { if (a % b == 0) { return b; } else return gcd(b, a % b); }
ll lcm(ll a, ll b) { if (a == 0) { return b; }return a / gcd(a, b) * b; }
template<class T>ll LBI(vector<T>& ar, T in) { return lower_bound(ar.begin(), ar.end(), in) - ar.begin(); }
template<class T>ll UBI(vector<T>& ar, T in) { return upper_bound(ar.begin(), ar.end(), in) - ar.begin(); }
ll n, q, com, s, t, x, i;
const int INF = (1LL << 31) - 1;
const int sqrtN = 512;
struct SqrtDecomposition {
int N, K;
vector<int> data;
vector<bool> lazyFlag;
vector<int> lazyUpdate;
SqrtDecomposition(int n) : N(n) {
K = (N + sqrtN - 1) / sqrtN;
data.assign(K * sqrtN, INF);
lazyFlag.assign(K, false);
lazyUpdate.assign(K, 0);
}
void eval(int k) {
if (lazyFlag[k]) {
lazyFlag[k] = false;
for (int i = k * sqrtN; i < (k + 1) * sqrtN; ++i) {
data[i] = lazyUpdate[k];
}
}
}
// [s, t)
void update(int s, int t, int x) {
for (int k = 0; k < K; ++k) {
int l = k * sqrtN, r = (k + 1) * sqrtN;
if (r <= s || t <= l)
continue;
if (s <= l && r <= t) {
lazyFlag[k] = true;
lazyUpdate[k] = x;
}
else {
eval(k);
for (int i = max(s, l); i < min(t, r); ++i) {
data[i] = x;
}
}
}
}
int find(int i) {
int k = i / sqrtN;
eval(k);
return data[i];
}
};
int main() {
cin >> n >> q;
SqrtDecomposition ruq(n);
rep(j, q) {
cin >> com;
if (com == 0) {
cin >> s >> t >> x;
ruq.update(s, t + 1, x);
}
else {
cin >> i;
cout << ruq.find(i) << endl;
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
// Range Update Query
// ???[s, t) ???????´???? x ??????????????¨???
// ???i ?????????????´????????????????????????¨???
const int INF = (1LL << 31) - 1;
const int sqrtN = 512;
struct SqrtDecomposition {
int N, K;
vector<int> data;
vector<bool> lazyFlag;
vector<int> lazyUpdate;
SqrtDecomposition(int n) : N(n) {
// ?????±????????°??? N ??? sqrtN ??§?????£???????????????????????? ????????????
K = (N + sqrtN - 1) / sqrtN;
data.assign(K * sqrtN, INF);
lazyFlag.assign(K, false);
lazyUpdate.assign(K, 0);
}
// ?????§??????????????±????????? lazyUpdate ???????????\??£???????????´??????
// lazyUpdate ?????\??£?????????????????????????????? (?????¶??????)
// ?????±??????????????????????????? update ?????¨?????????????????? find ?????¨???????????????
// ????????????????????????????????????
void eval(int k) {
if (lazyFlag[k]) {
lazyFlag[k] = false;
for (int i = k * sqrtN; i < (k + 1) * sqrtN; ++i) {
data[i] = lazyUpdate[k];
}
}
}
// update [s, t)
void update(int s, int t, int x) {
for (int k = 0; k < K; ++k) {
// ?????±?????????????????¨??????
int l = k * sqrtN, r = (k + 1) * sqrtN;
// ?????????
if (r <= s || t <= l)
continue;
// ?????±??????????¢????????????????
// ??? ?????±????????¨????????????????????´????????§ lazeUpdate ????????¨??????
if (s <= l && r <= t) {
lazyFlag[k] = true;
lazyUpdate[k] = x;
}
// ?????±????????????????????§?????? ??? ??????????????´?????? O(\sqrt N)
else {
eval(k);
for (int i = max(s, l); i < min(t, r); ++i) {
data[i] = x;
}
}
}
}
// find index i
int find(int i) {
// k <- ??????????????±????????????????????????
int k = i / sqrtN;
eval(k);
return data[i];
}
};
int main() {
int n, q; cin >> n >> q;
SqrtDecomposition R(n);
for(int i=0; i<q; i++) {
int num; cin >> num;
if(num == 0) {
int s, t, x; cin >> s >> t >> x;
R.update(s, t+1, x);
}
else {
int idx; cin >> idx;
cout << R.find(idx) << endl;
}
}
return 0;
} | ### Prompt
Your challenge is to write a cpp solution to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
// Range Update Query
// ???[s, t) ???????´???? x ??????????????¨???
// ???i ?????????????´????????????????????????¨???
const int INF = (1LL << 31) - 1;
const int sqrtN = 512;
struct SqrtDecomposition {
int N, K;
vector<int> data;
vector<bool> lazyFlag;
vector<int> lazyUpdate;
SqrtDecomposition(int n) : N(n) {
// ?????±????????°??? N ??? sqrtN ??§?????£???????????????????????? ????????????
K = (N + sqrtN - 1) / sqrtN;
data.assign(K * sqrtN, INF);
lazyFlag.assign(K, false);
lazyUpdate.assign(K, 0);
}
// ?????§??????????????±????????? lazyUpdate ???????????\??£???????????´??????
// lazyUpdate ?????\??£?????????????????????????????? (?????¶??????)
// ?????±??????????????????????????? update ?????¨?????????????????? find ?????¨???????????????
// ????????????????????????????????????
void eval(int k) {
if (lazyFlag[k]) {
lazyFlag[k] = false;
for (int i = k * sqrtN; i < (k + 1) * sqrtN; ++i) {
data[i] = lazyUpdate[k];
}
}
}
// update [s, t)
void update(int s, int t, int x) {
for (int k = 0; k < K; ++k) {
// ?????±?????????????????¨??????
int l = k * sqrtN, r = (k + 1) * sqrtN;
// ?????????
if (r <= s || t <= l)
continue;
// ?????±??????????¢????????????????
// ??? ?????±????????¨????????????????????´????????§ lazeUpdate ????????¨??????
if (s <= l && r <= t) {
lazyFlag[k] = true;
lazyUpdate[k] = x;
}
// ?????±????????????????????§?????? ??? ??????????????´?????? O(\sqrt N)
else {
eval(k);
for (int i = max(s, l); i < min(t, r); ++i) {
data[i] = x;
}
}
}
}
// find index i
int find(int i) {
// k <- ??????????????±????????????????????????
int k = i / sqrtN;
eval(k);
return data[i];
}
};
int main() {
int n, q; cin >> n >> q;
SqrtDecomposition R(n);
for(int i=0; i<q; i++) {
int num; cin >> num;
if(num == 0) {
int s, t, x; cin >> s >> t >> x;
R.update(s, t+1, x);
}
else {
int idx; cin >> idx;
cout << R.find(idx) << endl;
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int arr[100005 << 2], value[100005], n, q;
void updata(int a, int b, int x, int k = 0, int l = 0, int r = (1 << 17)){
if (b <= l || r <= a)return;
if (a <= l && r <= b){
arr[k] = max(arr[k], x);
return;
}
int m = (l + r) / 2;
updata(a, b, x, k * 2 + 1, l, m);
updata(a, b, x, k * 2 + 2, m, r);
}
int query(int x){
x += (1 << 17) - 1;
int res = arr[x];
while (x){
x = (x - 1) / 2;
res = max(res, arr[x]);
}
return res;
}
int main()
{
value[0] = 2147483647;
scanf("%d%d", &n, &q);
for (int i = 1; i <= q; ++i){
int ch, a, b;
scanf("%d", &ch);
if (ch){
scanf("%d", &a);
printf("%d\n", value[query(a)]);
}else{
scanf("%d%d%d", &a, &b, &value[i]);
++b;
updata(a, b, i);
}
}
return 0;
}
| ### Prompt
In Cpp, your task is to solve the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int arr[100005 << 2], value[100005], n, q;
void updata(int a, int b, int x, int k = 0, int l = 0, int r = (1 << 17)){
if (b <= l || r <= a)return;
if (a <= l && r <= b){
arr[k] = max(arr[k], x);
return;
}
int m = (l + r) / 2;
updata(a, b, x, k * 2 + 1, l, m);
updata(a, b, x, k * 2 + 2, m, r);
}
int query(int x){
x += (1 << 17) - 1;
int res = arr[x];
while (x){
x = (x - 1) / 2;
res = max(res, arr[x]);
}
return res;
}
int main()
{
value[0] = 2147483647;
scanf("%d%d", &n, &q);
for (int i = 1; i <= q; ++i){
int ch, a, b;
scanf("%d", &ch);
if (ch){
scanf("%d", &a);
printf("%d\n", value[query(a)]);
}else{
scanf("%d%d%d", &a, &b, &value[i]);
++b;
updata(a, b, i);
}
}
return 0;
}
``` |
#include <algorithm>
#include <cstdio>
#include <iostream>
#include <map>
#include <cmath>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
#include <stdlib.h>
#include <stdio.h>
#include <bitset>
#include <cstring>
#include <deque>
#include <iomanip>
#include <limits>
#include <fstream>
using namespace std;
#define FOR(I,A,B) for(int I = (A); I < (B); ++I)
#define CLR(mat) memset(mat, 0, sizeof(mat))
typedef long long ll;
const ll INF = (1LL << 31) - 1;
struct RUQ {
int N, K;
vector<ll> data;
vector<ll> lazyUpdate; // -1ならupdateしない
RUQ(int N): N(N) {
K = sqrt(N) + 1;
data.resize(K * K + 1, INF);
lazyUpdate.resize(K + 1, -1);
}
// kは分割した配列のindex
void evil(int k) {
if(lazyUpdate[k] != -1) {
for(int i = k * K; i < (k + 1) * K; i++) {
data[i] = lazyUpdate[k];
}
lazyUpdate[k] = -1;
}
}
// [s, t)
void update(int s, int t, int x) {
for(int k = 0; k < K; k++) {
int l = k * K, r = (k + 1) * K;
// 区間でない
if(r <= s || t <= l) continue;
// 完全に含まれている
if(s <= l && r <= t) {
lazyUpdate[k] = x;
} else {
evil(k);
for(int i = max(s, l); i < min(t, r); i++) {
data[i] = x;
}
}
}
}
int find(int i) {
int k = i / K;
evil(k);
return data[i];
}
};
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int n, q;
cin >> n >> q;
RUQ ruq(n);
FOR(i,0,q) {
ll a,s,t,x;
cin>>a;
if(a){
cin>>s;
cout<<ruq.find(s)<<endl;
} else {
cin>>s>>t>>x;
ruq.update(s,t+1,x);
}
}
}
| ### Prompt
Generate a cpp solution to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <algorithm>
#include <cstdio>
#include <iostream>
#include <map>
#include <cmath>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
#include <stdlib.h>
#include <stdio.h>
#include <bitset>
#include <cstring>
#include <deque>
#include <iomanip>
#include <limits>
#include <fstream>
using namespace std;
#define FOR(I,A,B) for(int I = (A); I < (B); ++I)
#define CLR(mat) memset(mat, 0, sizeof(mat))
typedef long long ll;
const ll INF = (1LL << 31) - 1;
struct RUQ {
int N, K;
vector<ll> data;
vector<ll> lazyUpdate; // -1ならupdateしない
RUQ(int N): N(N) {
K = sqrt(N) + 1;
data.resize(K * K + 1, INF);
lazyUpdate.resize(K + 1, -1);
}
// kは分割した配列のindex
void evil(int k) {
if(lazyUpdate[k] != -1) {
for(int i = k * K; i < (k + 1) * K; i++) {
data[i] = lazyUpdate[k];
}
lazyUpdate[k] = -1;
}
}
// [s, t)
void update(int s, int t, int x) {
for(int k = 0; k < K; k++) {
int l = k * K, r = (k + 1) * K;
// 区間でない
if(r <= s || t <= l) continue;
// 完全に含まれている
if(s <= l && r <= t) {
lazyUpdate[k] = x;
} else {
evil(k);
for(int i = max(s, l); i < min(t, r); i++) {
data[i] = x;
}
}
}
}
int find(int i) {
int k = i / K;
evil(k);
return data[i];
}
};
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int n, q;
cin >> n >> q;
RUQ ruq(n);
FOR(i,0,q) {
ll a,s,t,x;
cin>>a;
if(a){
cin>>s;
cout<<ruq.find(s)<<endl;
} else {
cin>>s>>t>>x;
ruq.update(s,t+1,x);
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
#define rep(i,j,k) for(int i = (int)j;i <= (int)k;i ++)
#define debug(x) cerr<<#x<<":"<<x<<endl
const int maxn=(int)1e5+5;
const int inf=(1ll<<31)-1;
#define lson (rt<<1)
#define rson (rt<<1|1)
int tag[maxn*4];//标记打成-1，用~-1==0简化代码
void upd(int rt,int l,int r,int x,int y,int v){
if(x<=l&&r<=y)tag[rt]=v;
else{
if(~tag[rt])tag[lson]=tag[rson]=tag[rt],tag[rt]=-1;
int mid=(l+r)/2;
if(y<=mid)upd(lson,l,mid,x,y,v);
else if(x>mid) upd(rson,mid+1,r,x,y,v);
else upd(lson,l,mid,x,mid,v),upd(rson,mid+1,r,mid+1,y,v);
}
}
int find(int rt,int l,int r,int x){
if(~tag[rt])return tag[rt];
int mid=(l+r)>>1;
if(x<=mid)return find(lson,l,mid,x);
else return find(rson,mid+1,r,x);
}
int main(){
ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int n,q;
cin>>n>>q;
rep(i,1,(n<<2))tag[i]=inf;
rep(i,1,q){
int op,x,y,v;
cin>>op;
if(op){
cin>>x;
cout<<find(1,1,n,x+1)<<endl;
}else{
int v;
cin>>x>>y>>v;
upd(1,1,n,x+1,y+1,v);
}
}
}
/*
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
1≤n≤100000
1≤q≤100000
0≤s≤t<n
0≤i<n
0≤x<2^31−1
*/
| ### Prompt
Please provide a CPP coded solution to the problem described below:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
#define rep(i,j,k) for(int i = (int)j;i <= (int)k;i ++)
#define debug(x) cerr<<#x<<":"<<x<<endl
const int maxn=(int)1e5+5;
const int inf=(1ll<<31)-1;
#define lson (rt<<1)
#define rson (rt<<1|1)
int tag[maxn*4];//标记打成-1，用~-1==0简化代码
void upd(int rt,int l,int r,int x,int y,int v){
if(x<=l&&r<=y)tag[rt]=v;
else{
if(~tag[rt])tag[lson]=tag[rson]=tag[rt],tag[rt]=-1;
int mid=(l+r)/2;
if(y<=mid)upd(lson,l,mid,x,y,v);
else if(x>mid) upd(rson,mid+1,r,x,y,v);
else upd(lson,l,mid,x,mid,v),upd(rson,mid+1,r,mid+1,y,v);
}
}
int find(int rt,int l,int r,int x){
if(~tag[rt])return tag[rt];
int mid=(l+r)>>1;
if(x<=mid)return find(lson,l,mid,x);
else return find(rson,mid+1,r,x);
}
int main(){
ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int n,q;
cin>>n>>q;
rep(i,1,(n<<2))tag[i]=inf;
rep(i,1,q){
int op,x,y,v;
cin>>op;
if(op){
cin>>x;
cout<<find(1,1,n,x+1)<<endl;
}else{
int v;
cin>>x>>y>>v;
upd(1,1,n,x+1,y+1,v);
}
}
}
/*
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
1≤n≤100000
1≤q≤100000
0≤s≤t<n
0≤i<n
0≤x<2^31−1
*/
``` |
#include<iostream>
#include<climits>
using namespace std;
// 完成版。
int tree[(1 << 18)];
void Update(int s, int t, int m, int left = 0, int right = (1 << 17), int key = 0){
if(t < left || right <= s) return;
if(s <= left && right - 1 <= t){ tree[key] = max(tree[key], m); return; }
Update(s, t, m, left, (left + right) / 2, 2 * key + 1);
Update(s, t, m, (left + right) / 2, right, 2 * key + 2);
}
int Find(int i){
i += (1 << 17) - 1;
int m = tree[i];
while(i){
i = (i - 1) / 2;
m = max(tree[i], m);
};
return m;
}
int main(){
int i;
for(i = 0; i < (1 << 18); i++) tree[i] = 0;
int n, q;
int x[100000];
for(i = 0; i < 100000; i++) x[i] = INT_MAX;
int s, t, value, m;
m = 1;
cin >> n >> q;
int query;
while(q){
q--;
cin >> query;
if(query){
cin >> i;
cout << x[Find(i)] << endl;
}else{
cin >> s >> t >> value;
x[m] = value;
Update(s, t, m);
m++;
}
};
return 0;
}
| ### Prompt
Your challenge is to write a cpp solution to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include<iostream>
#include<climits>
using namespace std;
// 完成版。
int tree[(1 << 18)];
void Update(int s, int t, int m, int left = 0, int right = (1 << 17), int key = 0){
if(t < left || right <= s) return;
if(s <= left && right - 1 <= t){ tree[key] = max(tree[key], m); return; }
Update(s, t, m, left, (left + right) / 2, 2 * key + 1);
Update(s, t, m, (left + right) / 2, right, 2 * key + 2);
}
int Find(int i){
i += (1 << 17) - 1;
int m = tree[i];
while(i){
i = (i - 1) / 2;
m = max(tree[i], m);
};
return m;
}
int main(){
int i;
for(i = 0; i < (1 << 18); i++) tree[i] = 0;
int n, q;
int x[100000];
for(i = 0; i < 100000; i++) x[i] = INT_MAX;
int s, t, value, m;
m = 1;
cin >> n >> q;
int query;
while(q){
q--;
cin >> query;
if(query){
cin >> i;
cout << x[Find(i)] << endl;
}else{
cin >> s >> t >> value;
x[m] = value;
Update(s, t, m);
m++;
}
};
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define N (1<<18)
ll ai[N*2];
ll value[N*2];
ll max_num = 1;
int find(int k){
k += N-1;
int ans = ai[k];
int v = value[k];
while(k > 0){
k = (k-1)/2;
if(value[k] > v){
ans = ai[k];
v = value[k];
}
}
return ans;
}
void update(int s, int t, int x, int i, int k = 0, int l = 0, int r = N){
if(r <= s || t <= l) return;
if(s <= l && r <= t){
value[k] = i;
ai[k] = x;
}else{
update(s, t, x, i, k*2+1, l, (l+r)/2);
update(s, t, x, i, k*2+2, (l+r)/2, r);
}
}
int main(){
for(int i = 0; i < 31; i++) max_num *= 2;
max_num -= 1;
for(int i = 0; i < N*2; i++){
ai[i] = max_num;
value[i] = -1;
}
int n, q;
cin >> n >> q;
for(int i = 0; i < q; i++){
int com, s, t, x;
cin >> com;
if(com == 1){
cin >> s;
cout << find(s) << endl;
}else{
cin >> s >> t >> x;
update(s, t+1, x, i);
}
}
}
| ### Prompt
Create a solution in Cpp for the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define N (1<<18)
ll ai[N*2];
ll value[N*2];
ll max_num = 1;
int find(int k){
k += N-1;
int ans = ai[k];
int v = value[k];
while(k > 0){
k = (k-1)/2;
if(value[k] > v){
ans = ai[k];
v = value[k];
}
}
return ans;
}
void update(int s, int t, int x, int i, int k = 0, int l = 0, int r = N){
if(r <= s || t <= l) return;
if(s <= l && r <= t){
value[k] = i;
ai[k] = x;
}else{
update(s, t, x, i, k*2+1, l, (l+r)/2);
update(s, t, x, i, k*2+2, (l+r)/2, r);
}
}
int main(){
for(int i = 0; i < 31; i++) max_num *= 2;
max_num -= 1;
for(int i = 0; i < N*2; i++){
ai[i] = max_num;
value[i] = -1;
}
int n, q;
cin >> n >> q;
for(int i = 0; i < q; i++){
int com, s, t, x;
cin >> com;
if(com == 1){
cin >> s;
cout << find(s) << endl;
}else{
cin >> s >> t >> x;
update(s, t+1, x, i);
}
}
}
``` |
#include<bits/stdc++.h>
using namespace std;
#define MAX_N 1<<17
int n,dat[2*(MAX_N)-1];
int valu[2*(MAX_N)-1];//更新順
//INT＿MAXで初期化
void init(int n_){
n=1;
while(n<n_)n*=2;
for(int i=0;i<2*n-1;i++){
dat[i]=INT_MAX;
valu[i]=-1;
}
}
//k番目の要素を返す
int find(int k){
k+=n-1;
int ans=dat[k];
int va=valu[k];
while(k>0){
k=(k-1)/2;
if(valu[k]>va){
ans=dat[k];
va=valu[k];
}
}
return ans;
}
//[a,b)の範囲を更新
void update(int a,int b,int x,int i,int k=0,int l=0,int r=n){
if(r<=a||b<=l)return;
if(a<=l&&r<=b){
valu[k]=i;
dat[k]=x;
}else{
update(a,b,x,i,k*2+1,l,(l+r)/2);
update(a,b,x,i,k*2+2,(l+r)/2,r);
}
}
int main(){
int q;
cin>>n>>q;
init(n);
for(int i=0;i<q;i++){
int com,s,t,x;
cin>>com; //com=1でfind,elseでupdate
if(com){
cin>>s;
cout<<find(s)<<endl;
}else{
cin>>s>>t>>x;
update(s,t+1,x,i);
}
}
return 0;
}
| ### Prompt
Generate a cpp solution to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
#define MAX_N 1<<17
int n,dat[2*(MAX_N)-1];
int valu[2*(MAX_N)-1];//更新順
//INT＿MAXで初期化
void init(int n_){
n=1;
while(n<n_)n*=2;
for(int i=0;i<2*n-1;i++){
dat[i]=INT_MAX;
valu[i]=-1;
}
}
//k番目の要素を返す
int find(int k){
k+=n-1;
int ans=dat[k];
int va=valu[k];
while(k>0){
k=(k-1)/2;
if(valu[k]>va){
ans=dat[k];
va=valu[k];
}
}
return ans;
}
//[a,b)の範囲を更新
void update(int a,int b,int x,int i,int k=0,int l=0,int r=n){
if(r<=a||b<=l)return;
if(a<=l&&r<=b){
valu[k]=i;
dat[k]=x;
}else{
update(a,b,x,i,k*2+1,l,(l+r)/2);
update(a,b,x,i,k*2+2,(l+r)/2,r);
}
}
int main(){
int q;
cin>>n>>q;
init(n);
for(int i=0;i<q;i++){
int com,s,t,x;
cin>>com; //com=1でfind,elseでupdate
if(com){
cin>>s;
cout<<find(s)<<endl;
}else{
cin>>s>>t>>x;
update(s,t+1,x,i);
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int INF = (1LL << 31) - 1;
const int sqrtN = 512;
struct SqrtDecomposition {
int N, K;
vector<int> data;
vector<bool> lazyFlag;
vector<int> lazyUpdate;
SqrtDecomposition(int n) : N(n) {
K = (N + sqrtN - 1) / sqrtN;
data.assign(K * sqrtN, INF);
lazyFlag.assign(K, false);
lazyUpdate.assign(K, 0);
}
void eval(int k) {
if (lazyFlag[k]) {
lazyFlag[k] = false;
for (int i = k * sqrtN; i < (k + 1) * sqrtN; ++i) {
data[i] = lazyUpdate[k];
}
}
}
// [s, t)
void update(int s, int t, int x) {
int p = max(0LL, s / sqrtN - 1);
int q = min(K, t / sqrtN + 1);
for (int k = p; k < q; ++k) {
int l = k * sqrtN, r = (k + 1) * sqrtN;
if (r <= s || t <= l)
continue;
if (s <= l && r <= t) {
lazyFlag[k] = true;
lazyUpdate[k] = x;
} else {
eval(k);
for (int i = max(s, l); i < min(t, r); ++i) {
data[i] = x;
}
}
}
}
int find(int i) {
int k = i / sqrtN;
eval(k);
return data[i];
}
};
signed main() {
ios::sync_with_stdio(false);
int N, Q;
cin >> N >> Q;
SqrtDecomposition decomp(N);
while (Q--) {
int c;
cin >> c;
if (c == 0) {
int s, t, x;
cin >> s >> t >> x;
decomp.update(s, t + 1, x);
} else {
int i;
cin >> i;
cout << decomp.find(i) << endl;
}
}
} | ### Prompt
In CPP, your task is to solve the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int INF = (1LL << 31) - 1;
const int sqrtN = 512;
struct SqrtDecomposition {
int N, K;
vector<int> data;
vector<bool> lazyFlag;
vector<int> lazyUpdate;
SqrtDecomposition(int n) : N(n) {
K = (N + sqrtN - 1) / sqrtN;
data.assign(K * sqrtN, INF);
lazyFlag.assign(K, false);
lazyUpdate.assign(K, 0);
}
void eval(int k) {
if (lazyFlag[k]) {
lazyFlag[k] = false;
for (int i = k * sqrtN; i < (k + 1) * sqrtN; ++i) {
data[i] = lazyUpdate[k];
}
}
}
// [s, t)
void update(int s, int t, int x) {
int p = max(0LL, s / sqrtN - 1);
int q = min(K, t / sqrtN + 1);
for (int k = p; k < q; ++k) {
int l = k * sqrtN, r = (k + 1) * sqrtN;
if (r <= s || t <= l)
continue;
if (s <= l && r <= t) {
lazyFlag[k] = true;
lazyUpdate[k] = x;
} else {
eval(k);
for (int i = max(s, l); i < min(t, r); ++i) {
data[i] = x;
}
}
}
}
int find(int i) {
int k = i / sqrtN;
eval(k);
return data[i];
}
};
signed main() {
ios::sync_with_stdio(false);
int N, Q;
cin >> N >> Q;
SqrtDecomposition decomp(N);
while (Q--) {
int c;
cin >> c;
if (c == 0) {
int s, t, x;
cin >> s >> t >> x;
decomp.update(s, t + 1, x);
} else {
int i;
cin >> i;
cout << decomp.find(i) << endl;
}
}
}
``` |
#include <iostream>
#include <vector>
using namespace std;
const int INF = 2147483647;
struct SegmentTree {
int n;
vector<int> heap;
SegmentTree(int n): n(n) {
heap.assign(4 * n, INF);
}
void update(int nodeId, int l, int r, int ql, int qr, int v) {
if (qr < l || ql > r) {
return;
}
else if (ql <= l && r <= qr) {
heap[nodeId] = v;
}
else {
pushdown(nodeId);
int m = l + (r - l)/2;
update(nodeId * 2, l, m, ql, qr, v);
update(nodeId * 2 + 1, m + 1, r, ql, qr, v);
}
}
int query(int nodeId, int l, int r, int q) {
if (q < l || q > r) {
return -1;
}
else if (l == r) {
return heap[nodeId];
}
else {
pushdown(nodeId);
int m = l + (r - l)/2;
return max(query(nodeId * 2, l, m, q), query(nodeId * 2 + 1, m + 1, r, q));
}
}
void pushdown(int nodeId) {
//cout << "push: " << nodeId << endl;
if (heap[nodeId] != -1) {
heap[nodeId * 2] = heap[nodeId];
heap[nodeId * 2 + 1] = heap[nodeId];
heap[nodeId] = -1;
}
}
};
int main() {
int n, q;
cin >> n >> q;
SegmentTree st(n);
for (int i = 0; i < q; i++) {
int cmd;
cin >> cmd;
if (cmd == 0) {
int ql, qr, v;
cin >> ql >> qr >> v;
st.update(1, 0, n -1, ql, qr, v);
// for (int i = 0; i < 4 * n; i++) {
// cout << i << ":" << st.heap[i] << endl;
// }
}
else {
int pos;
cin >> pos;
cout << st.query(1, 0, n-1, pos) << endl;
}
}
return 0;
} | ### Prompt
Please provide a Cpp coded solution to the problem described below:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <iostream>
#include <vector>
using namespace std;
const int INF = 2147483647;
struct SegmentTree {
int n;
vector<int> heap;
SegmentTree(int n): n(n) {
heap.assign(4 * n, INF);
}
void update(int nodeId, int l, int r, int ql, int qr, int v) {
if (qr < l || ql > r) {
return;
}
else if (ql <= l && r <= qr) {
heap[nodeId] = v;
}
else {
pushdown(nodeId);
int m = l + (r - l)/2;
update(nodeId * 2, l, m, ql, qr, v);
update(nodeId * 2 + 1, m + 1, r, ql, qr, v);
}
}
int query(int nodeId, int l, int r, int q) {
if (q < l || q > r) {
return -1;
}
else if (l == r) {
return heap[nodeId];
}
else {
pushdown(nodeId);
int m = l + (r - l)/2;
return max(query(nodeId * 2, l, m, q), query(nodeId * 2 + 1, m + 1, r, q));
}
}
void pushdown(int nodeId) {
//cout << "push: " << nodeId << endl;
if (heap[nodeId] != -1) {
heap[nodeId * 2] = heap[nodeId];
heap[nodeId * 2 + 1] = heap[nodeId];
heap[nodeId] = -1;
}
}
};
int main() {
int n, q;
cin >> n >> q;
SegmentTree st(n);
for (int i = 0; i < q; i++) {
int cmd;
cin >> cmd;
if (cmd == 0) {
int ql, qr, v;
cin >> ql >> qr >> v;
st.update(1, 0, n -1, ql, qr, v);
// for (int i = 0; i < 4 * n; i++) {
// cout << i << ":" << st.heap[i] << endl;
// }
}
else {
int pos;
cin >> pos;
cout << st.query(1, 0, n-1, pos) << endl;
}
}
return 0;
}
``` |
#line 1 "test/aoj/DSL2D.test.cpp"
#define PROBLEM "http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=DSL_2_D"
#include <iostream>
#line 1 "segment_tree/dual_segment_tree.hpp"
#include <vector>
#include <cstdint>
//===
template<class Monoid>
struct DualSegmentTree {
using T = typename Monoid::value_type;
std::vector<T> lazy;
DualSegmentTree() = default;
explicit DualSegmentTree(uint32_t n):
lazy(n << 1, Monoid::identity()) {};
inline int size() {
return (int)(lazy.size() >> 1);
};
inline void propagate(uint32_t k) {
if (k >= size()) return;
lazy[(k << 1) | 0] = Monoid::operation(lazy[(k << 1) | 0], lazy[k]);
lazy[(k << 1) | 1] = Monoid::operation(lazy[(k << 1) | 1], lazy[k]);
lazy[k] = Monoid::identity();
};
inline void push_down(uint32_t k) {
for (uint32_t i = 31; i > 0; i--) propagate(k >> i);
};
// [l, r)
void update(uint32_t l, uint32_t r, T op) {
l += size();
r += size();
push_down(l);
push_down(r);
while (l < r) {
if (l & 1) lazy[l] = Monoid::operation(lazy[l], op), l++;
if (r & 1) --r, lazy[r] = Monoid::operation(lazy[r], op);
l >>= 1;
r >>= 1;
}
};
T operator [] (uint32_t k) {
k += size();
push_down(k);
return lazy[k];
};
};
//===
#line 4 "test/aoj/DSL2D.test.cpp"
using namespace std;
using llong = long long;
struct RUQ {
using value_type = long long;
using T = value_type;
inline static T identity() {
return (1ll << 31) - 1;
};
inline static T operation(T a, T b) {
if (a == identity()) return b;
else if (b == identity()) return a;
else return b;
};
};
int main() {
llong n, q;
llong com;
llong s, t, x;
llong idx;
cin >> n >> q;
DualSegmentTree<RUQ> seg(n);
for (int i = 0; i < q; i++) {
cin >> com;
if (com == 0) {
cin >> s >> t >> x;
seg.update(s, t + 1, x);
}
else if (com == 1) {
cin >> idx;
cout << seg[idx] << '\n';
}
}
};
| ### Prompt
Your task is to create a Cpp solution to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#line 1 "test/aoj/DSL2D.test.cpp"
#define PROBLEM "http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=DSL_2_D"
#include <iostream>
#line 1 "segment_tree/dual_segment_tree.hpp"
#include <vector>
#include <cstdint>
//===
template<class Monoid>
struct DualSegmentTree {
using T = typename Monoid::value_type;
std::vector<T> lazy;
DualSegmentTree() = default;
explicit DualSegmentTree(uint32_t n):
lazy(n << 1, Monoid::identity()) {};
inline int size() {
return (int)(lazy.size() >> 1);
};
inline void propagate(uint32_t k) {
if (k >= size()) return;
lazy[(k << 1) | 0] = Monoid::operation(lazy[(k << 1) | 0], lazy[k]);
lazy[(k << 1) | 1] = Monoid::operation(lazy[(k << 1) | 1], lazy[k]);
lazy[k] = Monoid::identity();
};
inline void push_down(uint32_t k) {
for (uint32_t i = 31; i > 0; i--) propagate(k >> i);
};
// [l, r)
void update(uint32_t l, uint32_t r, T op) {
l += size();
r += size();
push_down(l);
push_down(r);
while (l < r) {
if (l & 1) lazy[l] = Monoid::operation(lazy[l], op), l++;
if (r & 1) --r, lazy[r] = Monoid::operation(lazy[r], op);
l >>= 1;
r >>= 1;
}
};
T operator [] (uint32_t k) {
k += size();
push_down(k);
return lazy[k];
};
};
//===
#line 4 "test/aoj/DSL2D.test.cpp"
using namespace std;
using llong = long long;
struct RUQ {
using value_type = long long;
using T = value_type;
inline static T identity() {
return (1ll << 31) - 1;
};
inline static T operation(T a, T b) {
if (a == identity()) return b;
else if (b == identity()) return a;
else return b;
};
};
int main() {
llong n, q;
llong com;
llong s, t, x;
llong idx;
cin >> n >> q;
DualSegmentTree<RUQ> seg(n);
for (int i = 0; i < q; i++) {
cin >> com;
if (com == 0) {
cin >> s >> t >> x;
seg.update(s, t + 1, x);
}
else if (com == 1) {
cin >> idx;
cout << seg[idx] << '\n';
}
}
};
``` |
#include<bits/stdc++.h>
using namespace std;
#define int long long
typedef pair<int,int> P;
struct RUP{
int n;
vector<P> dat;
RUP(){}
RUP(int n_){init(n_);}
void init(int n_){
n=1;
while(n<n_) n*=2;
dat.clear();
dat.resize(2*n-1);
for(int i=0;i<2*n-1;i++) dat[i].first=-1,dat[i].second=INT_MAX;
}
int query(int k){
k+=n-1;
P p=dat[k];
while(k>0){
k=(k-1)/2;
p=max(p,dat[k]);
}
return p.second;
}
void update(int a,int b,P p,int k,int l,int r){
if(r<=a||b<=l) return;
if(a<=l&&r<=b) {
dat[k]=p;
}else{
update(a,b,p,k*2+1,l,(l+r)/2);
update(a,b,p,k*2+2,(l+r)/2,r);
}
}
void update(int a,int b,P p){
update(a,b,p,0,0,n);
}
};
signed main(){
int n,q;
cin>>n>>q;
RUP rup(n);
for(int i=0;i<q;i++){
int f;
cin>>f;
if(!f){
int s,t,x;
cin>>s>>t>>x;
rup.update(s,t+1,P(i,x));
}else{
int u;
cin>>u;
cout<<rup.query(u)<<endl;
}
}
return 0;
}
/*
verified on 2017/02/24
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=DSL_2_D
*/ | ### Prompt
Develop a solution in Cpp to the problem described below:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
#define int long long
typedef pair<int,int> P;
struct RUP{
int n;
vector<P> dat;
RUP(){}
RUP(int n_){init(n_);}
void init(int n_){
n=1;
while(n<n_) n*=2;
dat.clear();
dat.resize(2*n-1);
for(int i=0;i<2*n-1;i++) dat[i].first=-1,dat[i].second=INT_MAX;
}
int query(int k){
k+=n-1;
P p=dat[k];
while(k>0){
k=(k-1)/2;
p=max(p,dat[k]);
}
return p.second;
}
void update(int a,int b,P p,int k,int l,int r){
if(r<=a||b<=l) return;
if(a<=l&&r<=b) {
dat[k]=p;
}else{
update(a,b,p,k*2+1,l,(l+r)/2);
update(a,b,p,k*2+2,(l+r)/2,r);
}
}
void update(int a,int b,P p){
update(a,b,p,0,0,n);
}
};
signed main(){
int n,q;
cin>>n>>q;
RUP rup(n);
for(int i=0;i<q;i++){
int f;
cin>>f;
if(!f){
int s,t,x;
cin>>s>>t>>x;
rup.update(s,t+1,P(i,x));
}else{
int u;
cin>>u;
cout<<rup.query(u)<<endl;
}
}
return 0;
}
/*
verified on 2017/02/24
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=DSL_2_D
*/
``` |
#include <iostream>
#include <string>
#include <cstring>
#include <fstream>
#include <random>
#include <cmath>
#include <iomanip>
#include <climits>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <deque>
#include <map>
#include <set>
#include <unordered_set>
#include <queue>
#include <stack>
#include <list>
#include <unordered_map>
#include <bitset>
#include <sstream>
#include <new>
#include <typeinfo>
#include <iterator>
typedef long long ll;
typedef unsigned long long ull;
constexpr ll mop = 1000000007;
constexpr ll mop2 = 998244353;
using namespace std;
constexpr ll DEF_VAL = (1LL << 31) - 1;
constexpr ll SEG_LEN = (1LL << 18);
pair<ll,ll> segTree[SEG_LEN * 2]; // segTree[i] = {times, val}
void initialize() {
for (ll i = 0; i < SEG_LEN * 2; i++) {
segTree[i] = make_pair(0, DEF_VAL);
}
}
void add(ll l, ll r, ll val, ll times) {
l += SEG_LEN;
r += SEG_LEN;
while (l < r) {
if (l % 2 == 1) {
segTree[l++] = make_pair(times, val);
}
l /= 2;
if (r % 2 == 1) {
segTree[--r] = make_pair(times, val);
}
r /= 2;
}
}
ll get(ll idx) {
idx += SEG_LEN;
pair<ll,ll> ans = make_pair(0, DEF_VAL);
while (idx > 0) {
if (ans.first < segTree[idx].first) {
ans = segTree[idx];
}
idx /= 2;
}
return ans.second;
}
int main() {
ll aLen, Q;
cin >> aLen >> Q;
ll times = 0;
for (ll i = 0; i < Q; i++) {
ll com;
cin >> com;
if (com == 0) {
ll l, r, val;
cin >> l >> r >> val;
add(l, r+1, val, ++times);
}
else {
ll idx;
cin >> idx;
cout << get(idx) << endl;
}
}
}
| ### Prompt
Please create a solution in Cpp to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <iostream>
#include <string>
#include <cstring>
#include <fstream>
#include <random>
#include <cmath>
#include <iomanip>
#include <climits>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <deque>
#include <map>
#include <set>
#include <unordered_set>
#include <queue>
#include <stack>
#include <list>
#include <unordered_map>
#include <bitset>
#include <sstream>
#include <new>
#include <typeinfo>
#include <iterator>
typedef long long ll;
typedef unsigned long long ull;
constexpr ll mop = 1000000007;
constexpr ll mop2 = 998244353;
using namespace std;
constexpr ll DEF_VAL = (1LL << 31) - 1;
constexpr ll SEG_LEN = (1LL << 18);
pair<ll,ll> segTree[SEG_LEN * 2]; // segTree[i] = {times, val}
void initialize() {
for (ll i = 0; i < SEG_LEN * 2; i++) {
segTree[i] = make_pair(0, DEF_VAL);
}
}
void add(ll l, ll r, ll val, ll times) {
l += SEG_LEN;
r += SEG_LEN;
while (l < r) {
if (l % 2 == 1) {
segTree[l++] = make_pair(times, val);
}
l /= 2;
if (r % 2 == 1) {
segTree[--r] = make_pair(times, val);
}
r /= 2;
}
}
ll get(ll idx) {
idx += SEG_LEN;
pair<ll,ll> ans = make_pair(0, DEF_VAL);
while (idx > 0) {
if (ans.first < segTree[idx].first) {
ans = segTree[idx];
}
idx /= 2;
}
return ans.second;
}
int main() {
ll aLen, Q;
cin >> aLen >> Q;
ll times = 0;
for (ll i = 0; i < Q; i++) {
ll com;
cin >> com;
if (com == 0) {
ll l, r, val;
cin >> l >> r >> val;
add(l, r+1, val, ++times);
}
else {
ll idx;
cin >> idx;
cout << get(idx) << endl;
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int N, Q;
int const INF = INT_MAX;
struct LazySegmentTree {
private:
int n;
vector<int> node, lazy;
vector<bool> lazyFlag;
public:
LazySegmentTree(vector<int> v) {
int sz = (int)v.size();
n = 1; while(n < sz) n *= 2;
node.resize(2*n-1);
lazy.resize(2*n-1, INF);
lazyFlag.resize(2*n-1, false);
for(int i=0; i<sz; i++) node[i+n-1] = v[i];
for(int i=n-2; i>=0; i--) node[i] = min(node[i*2+1], node[i*2+2]);
}
void lazy_evaluate(int k, int l, int r) {
if(lazyFlag[k]) {
node[k] = lazy[k];
if(r - l > 1) {
lazy[2*k+1] = lazy[k];
lazy[2*k+2] = lazy[k];
lazyFlag[2*k+1] = true;
lazyFlag[2*k+2] = true;
}
lazyFlag[k] = false;
}
}
void update(int a, int b, int x, int k=0, int l=0, int r=-1) {
if(r < 0) r = n;
lazy_evaluate(k, l, r);
if(r <= a || b <= l) return;
if(a <= l && r <= b) {
lazy[k] = x;
lazyFlag[k] = true;
lazy_evaluate(k, l, r);
}
else {
update(a, b, x, 2*k+1, l, (l+r)/2);
update(a, b, x, 2*k+2, (l+r)/2, r);
node[k] = min(node[k*2+1], node[k*2+2]);
}
}
int find(int a, int b, int k=0, int l=0, int r=-1) {
if(r < 0) r = n;
lazy_evaluate(k, l, r);
if(r <= a || b <= l) return INF;
if(a <= l && r <= b) return node[k];
int vl = find(a, b, 2*k+1, l, (l+r)/2);
int vr = find(a, b, 2*k+2, (l+r)/2, r);
return min(vl, vr);
}
};
int main() {
cin >> N >> Q;
LazySegmentTree seg( vector<int>(N, INF) );
for(int i=0; i<Q; i++) {
int query; cin >> query;
if(query == 0) {
int s, t, x; cin >> s >> t >> x;
seg.update(s, t+1, x);
}
else {
int i; cin >> i;
cout << seg.find(i, i+1) << endl;
}
}
return 0;
} | ### Prompt
Please provide a cpp coded solution to the problem described below:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int N, Q;
int const INF = INT_MAX;
struct LazySegmentTree {
private:
int n;
vector<int> node, lazy;
vector<bool> lazyFlag;
public:
LazySegmentTree(vector<int> v) {
int sz = (int)v.size();
n = 1; while(n < sz) n *= 2;
node.resize(2*n-1);
lazy.resize(2*n-1, INF);
lazyFlag.resize(2*n-1, false);
for(int i=0; i<sz; i++) node[i+n-1] = v[i];
for(int i=n-2; i>=0; i--) node[i] = min(node[i*2+1], node[i*2+2]);
}
void lazy_evaluate(int k, int l, int r) {
if(lazyFlag[k]) {
node[k] = lazy[k];
if(r - l > 1) {
lazy[2*k+1] = lazy[k];
lazy[2*k+2] = lazy[k];
lazyFlag[2*k+1] = true;
lazyFlag[2*k+2] = true;
}
lazyFlag[k] = false;
}
}
void update(int a, int b, int x, int k=0, int l=0, int r=-1) {
if(r < 0) r = n;
lazy_evaluate(k, l, r);
if(r <= a || b <= l) return;
if(a <= l && r <= b) {
lazy[k] = x;
lazyFlag[k] = true;
lazy_evaluate(k, l, r);
}
else {
update(a, b, x, 2*k+1, l, (l+r)/2);
update(a, b, x, 2*k+2, (l+r)/2, r);
node[k] = min(node[k*2+1], node[k*2+2]);
}
}
int find(int a, int b, int k=0, int l=0, int r=-1) {
if(r < 0) r = n;
lazy_evaluate(k, l, r);
if(r <= a || b <= l) return INF;
if(a <= l && r <= b) return node[k];
int vl = find(a, b, 2*k+1, l, (l+r)/2);
int vr = find(a, b, 2*k+2, (l+r)/2, r);
return min(vl, vr);
}
};
int main() {
cin >> N >> Q;
LazySegmentTree seg( vector<int>(N, INF) );
for(int i=0; i<Q; i++) {
int query; cin >> query;
if(query == 0) {
int s, t, x; cin >> s >> t >> x;
seg.update(s, t+1, x);
}
else {
int i; cin >> i;
cout << seg.find(i, i+1) << endl;
}
}
return 0;
}
``` |
#include "bits/stdc++.h"
using namespace std;
#ifdef _DEBUG
#include "dump.hpp"
#else
#define dump(...)
#endif
//#define int long long
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(b)-1;i>=(a);i--)
#define all(c) begin(c),end(c)
const int INF = (1ll << 31) - 1;
const int MOD = (int)(1e9) + 7;
const double PI = acos(-1);
const double EPS = 1e-9;
template<class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return true; } return false; }
template<class T> bool chmin(T &a, const T &b) { if (a > b) { a = b; return true; } return false; }
struct SegmentTree {
int n;
vector<int>d;
SegmentTree(int m) {
for (n = 1; n < m; n <<= 1);
d.assign(2 * n, INF);
}
void update(int a, int b, int x = -1, int k = 1, int l = 0, int r = -1) {
if (r == -1)r = n;
if (r <= a || b <= l)return;
else if (a <= l&&r <= b) {
if (x >= 0)d[k] = x;
}
else {
if (d[k] != INF)d[2 * k] = d[2 * k + 1] = d[k], d[k] = INF;
update(a, b, x, 2 * k, l, (l + r) / 2);
update(a, b, x, 2 * k + 1, (l + r) / 2, r);
}
}
int find(int i) {
update(i, i + 1);
return d[n + i];
}
};
signed main() {
cin.tie(0);
ios::sync_with_stdio(false);
int n, q; cin >> n >> q;
SegmentTree seg(n);
rep(i, 0, q) {
int com; cin >> com;
if (com) {
int x; cin >> x;
cout << seg.find(x) << endl;
}
else {
int s, t, x; cin >> s >> t >> x;
seg.update(s, t + 1, x);
}
}
return 0;
} | ### Prompt
Please create a solution in cpp to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include "bits/stdc++.h"
using namespace std;
#ifdef _DEBUG
#include "dump.hpp"
#else
#define dump(...)
#endif
//#define int long long
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(b)-1;i>=(a);i--)
#define all(c) begin(c),end(c)
const int INF = (1ll << 31) - 1;
const int MOD = (int)(1e9) + 7;
const double PI = acos(-1);
const double EPS = 1e-9;
template<class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return true; } return false; }
template<class T> bool chmin(T &a, const T &b) { if (a > b) { a = b; return true; } return false; }
struct SegmentTree {
int n;
vector<int>d;
SegmentTree(int m) {
for (n = 1; n < m; n <<= 1);
d.assign(2 * n, INF);
}
void update(int a, int b, int x = -1, int k = 1, int l = 0, int r = -1) {
if (r == -1)r = n;
if (r <= a || b <= l)return;
else if (a <= l&&r <= b) {
if (x >= 0)d[k] = x;
}
else {
if (d[k] != INF)d[2 * k] = d[2 * k + 1] = d[k], d[k] = INF;
update(a, b, x, 2 * k, l, (l + r) / 2);
update(a, b, x, 2 * k + 1, (l + r) / 2, r);
}
}
int find(int i) {
update(i, i + 1);
return d[n + i];
}
};
signed main() {
cin.tie(0);
ios::sync_with_stdio(false);
int n, q; cin >> n >> q;
SegmentTree seg(n);
rep(i, 0, q) {
int com; cin >> com;
if (com) {
int x; cin >> x;
cout << seg.find(x) << endl;
}
else {
int s, t, x; cin >> s >> t >> x;
seg.update(s, t + 1, x);
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
class lazy_segtree {
public:
int size;
vector<int> node,lazy;
lazy_segtree(int n_) {
size=1;
while (size<n_) size*=2;
node.resize(2 * size, INT_MAX);
lazy.resize(2 * size, -1);
}
void setlazy(int k, int v) {
if (k>=2*size-1) return ;
lazy[k] = v;
node[k] = v;
}
void push(int k) {
if (k>=2*size-1) return ;
if (lazy[k] == -1) return ;
setlazy(k * 2 + 1, lazy[k]);
setlazy(k * 2 + 2, lazy[k]);
lazy[k] = -1;
}
void range_update(int queryL,int queryR,int val,int k=0,int nodeL=0,int nodeR=-1) {
if (nodeR==-1) nodeR=size;
if (nodeR<=queryL || queryR<=nodeL) return ;
if (queryL<=nodeL && nodeR<=queryR) {
setlazy(k,val);
return ;
}
else {
push(k);
int nodeM=(nodeL+nodeR)/2;
range_update(queryL,queryR,val,k*2+1,nodeL,nodeM);
range_update(queryL,queryR,val,k*2+2,nodeM,nodeR);
return ;
}
}
int get_node(int queryL,int queryR,int k=0,int nodeL=0,int nodeR=-1) {
if (nodeR==-1) nodeR=size;
if (nodeR<=queryL || queryR<=nodeL) return -1;
if (queryL<=nodeL && nodeR<=queryR) return node[k];
else {
push(k);
int nodeM=(nodeL+nodeR)/2;
int tmp1=get_node(queryL,queryR,k*2+1,nodeL,nodeM);
int tmp2=get_node(queryL,queryR,k*2+2,nodeM,nodeR);
if (tmp1!=-1) return tmp1;
else if(tmp2!=-1) return tmp2;
}
}
};
int n,q;
int main(){
cin >> n >> q;
lazy_segtree ls(n);
for (int i=0; i<q; i++) {
int que,s,t,x,idx;
cin >> que;
if (que==0) {
cin >> s >> t >> x;
ls.range_update(s,t+1,x);
}
else {
cin >> idx;
cout << ls.get_node(idx,idx+1) << endl;
}
}
return 0;
} | ### Prompt
Generate a Cpp solution to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
class lazy_segtree {
public:
int size;
vector<int> node,lazy;
lazy_segtree(int n_) {
size=1;
while (size<n_) size*=2;
node.resize(2 * size, INT_MAX);
lazy.resize(2 * size, -1);
}
void setlazy(int k, int v) {
if (k>=2*size-1) return ;
lazy[k] = v;
node[k] = v;
}
void push(int k) {
if (k>=2*size-1) return ;
if (lazy[k] == -1) return ;
setlazy(k * 2 + 1, lazy[k]);
setlazy(k * 2 + 2, lazy[k]);
lazy[k] = -1;
}
void range_update(int queryL,int queryR,int val,int k=0,int nodeL=0,int nodeR=-1) {
if (nodeR==-1) nodeR=size;
if (nodeR<=queryL || queryR<=nodeL) return ;
if (queryL<=nodeL && nodeR<=queryR) {
setlazy(k,val);
return ;
}
else {
push(k);
int nodeM=(nodeL+nodeR)/2;
range_update(queryL,queryR,val,k*2+1,nodeL,nodeM);
range_update(queryL,queryR,val,k*2+2,nodeM,nodeR);
return ;
}
}
int get_node(int queryL,int queryR,int k=0,int nodeL=0,int nodeR=-1) {
if (nodeR==-1) nodeR=size;
if (nodeR<=queryL || queryR<=nodeL) return -1;
if (queryL<=nodeL && nodeR<=queryR) return node[k];
else {
push(k);
int nodeM=(nodeL+nodeR)/2;
int tmp1=get_node(queryL,queryR,k*2+1,nodeL,nodeM);
int tmp2=get_node(queryL,queryR,k*2+2,nodeM,nodeR);
if (tmp1!=-1) return tmp1;
else if(tmp2!=-1) return tmp2;
}
}
};
int n,q;
int main(){
cin >> n >> q;
lazy_segtree ls(n);
for (int i=0; i<q; i++) {
int que,s,t,x,idx;
cin >> que;
if (que==0) {
cin >> s >> t >> x;
ls.range_update(s,t+1,x);
}
else {
cin >> idx;
cout << ls.get_node(idx,idx+1) << endl;
}
}
return 0;
}
``` |
#include <iostream>
#include <climits>
#include <algorithm>
#define lson l, m, o << 1
#define rson m + 1, r, o << 1 | 1
const int N = 1e5 + 1;
int mn[N << 2];
int lazy[N << 2];
void up(int o) {
mn[o] = std::min(mn[o << 1], mn[o << 1 | 1]);
}
void down(int o) {
if (lazy[o] != -1) {
mn[o << 1] = lazy[o];
mn[o << 1 | 1] = lazy[o];
lazy[o << 1] = lazy[o];
lazy[o << 1 | 1] = lazy[o];
lazy[o] = -1;
}
}
void build(int l, int r, int o) {
lazy[o] = -1;
if (l == r) {
mn[o] = INT_MAX;
return;
}
int m = (l + r) >> 1;
build(lson);
build(rson);
up(o);
}
void modify(int v, int L, int R, int l, int r, int o) {
if (L <= l && R >= r) {
lazy[o] = v;
mn[o] = v;
return;
}
down(o);
int m = (l + r) >> 1;
if (L <= m) modify(v, L, R, lson);
if (R > m) modify(v, L, R, rson);
up(o);
}
int query(int L, int R, int l, int r, int o) {
if (L <= l && R >= r) {
return mn[o];
}
down(o);
int m = (l + r) >> 1;
int res = INT_MAX;
if (L <= m) res = std::min(res, query(L, R, lson));
if (R > m) res = std::min(res, query(L, R, rson));
return res;
}
int main() {
int n, q;
while (std::cin >> n >> q) {
build(1, n, 1);
while (q --) {
int op;
std::cin >> op;
if (op == 0) {
int l, r, v;
std::cin >> l >> r >> v;
l ++;
r ++;
modify(v, l, r, 1, n, 1);
} else {
int l;
std::cin >> l;
l ++;
std::cout << query(l, l, 1, n, 1) << std::endl;
}
}
}
return 0;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <iostream>
#include <climits>
#include <algorithm>
#define lson l, m, o << 1
#define rson m + 1, r, o << 1 | 1
const int N = 1e5 + 1;
int mn[N << 2];
int lazy[N << 2];
void up(int o) {
mn[o] = std::min(mn[o << 1], mn[o << 1 | 1]);
}
void down(int o) {
if (lazy[o] != -1) {
mn[o << 1] = lazy[o];
mn[o << 1 | 1] = lazy[o];
lazy[o << 1] = lazy[o];
lazy[o << 1 | 1] = lazy[o];
lazy[o] = -1;
}
}
void build(int l, int r, int o) {
lazy[o] = -1;
if (l == r) {
mn[o] = INT_MAX;
return;
}
int m = (l + r) >> 1;
build(lson);
build(rson);
up(o);
}
void modify(int v, int L, int R, int l, int r, int o) {
if (L <= l && R >= r) {
lazy[o] = v;
mn[o] = v;
return;
}
down(o);
int m = (l + r) >> 1;
if (L <= m) modify(v, L, R, lson);
if (R > m) modify(v, L, R, rson);
up(o);
}
int query(int L, int R, int l, int r, int o) {
if (L <= l && R >= r) {
return mn[o];
}
down(o);
int m = (l + r) >> 1;
int res = INT_MAX;
if (L <= m) res = std::min(res, query(L, R, lson));
if (R > m) res = std::min(res, query(L, R, rson));
return res;
}
int main() {
int n, q;
while (std::cin >> n >> q) {
build(1, n, 1);
while (q --) {
int op;
std::cin >> op;
if (op == 0) {
int l, r, v;
std::cin >> l >> r >> v;
l ++;
r ++;
modify(v, l, r, 1, n, 1);
} else {
int l;
std::cin >> l;
l ++;
std::cout << query(l, l, 1, n, 1) << std::endl;
}
}
}
return 0;
}
``` |
#include<iostream>
#include<algorithm>
#include <stdio.h>
using namespace std;
typedef pair<int,int> p;
#define TREE_N (1<<17)
int c,n;
p tree[ TREE_N * 2 -1],ans;
void update(int i,int x,int t){
i+=TREE_N-1;
if(t==-1)tree[i]=p(-1,x);
else ans=tree[i];
while(i>0){
i=(i-1)/2;
if(t==-1)tree[i]=p(t,x);
else ans=max(ans,tree[i]);
}
}
void rec(int t,int a,int b,int k=0,int l=0,int r=TREE_N){
if((l<a&&a<r)||(l<b&&b<r)){
int m=(l+r)/2;
rec(t,a,b,k*2+1,l,m);
rec(t,a,b,k*2+2,m,r);
}else if(a<=l&&r<=b){
tree[k]=p(t,c);
}
}
/*int Rec(int a,int b,int k=0,int l=0,int r=TREE_N){
if((l<a&&a<r)||(l<b&&b<r)){
int m=(l+r)/2;
int left = Rec(a,b,k*2+1,l,m);
int right = Rec(a,b,k*2+2,m,r);
return min(left,right);
}else if(a<=l&&r<=b){
return tree[k];
}else{
return 2147483647;
}
}*/
int main(){
int q,com,a,b;
cin>>n>>q;
for(int i=0;i<n;i++)update(i,2147483647,-1);
for(int i=0;i<q;i++){
cin>>com;
if(com){
cin>>a;
update(a,0,0);
cout<<ans.second<<endl;
}
else {
cin>>a>>b>>c;
rec(i,a,b+1);
}
}
return 0;
} | ### Prompt
Generate a CPP solution to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include<iostream>
#include<algorithm>
#include <stdio.h>
using namespace std;
typedef pair<int,int> p;
#define TREE_N (1<<17)
int c,n;
p tree[ TREE_N * 2 -1],ans;
void update(int i,int x,int t){
i+=TREE_N-1;
if(t==-1)tree[i]=p(-1,x);
else ans=tree[i];
while(i>0){
i=(i-1)/2;
if(t==-1)tree[i]=p(t,x);
else ans=max(ans,tree[i]);
}
}
void rec(int t,int a,int b,int k=0,int l=0,int r=TREE_N){
if((l<a&&a<r)||(l<b&&b<r)){
int m=(l+r)/2;
rec(t,a,b,k*2+1,l,m);
rec(t,a,b,k*2+2,m,r);
}else if(a<=l&&r<=b){
tree[k]=p(t,c);
}
}
/*int Rec(int a,int b,int k=0,int l=0,int r=TREE_N){
if((l<a&&a<r)||(l<b&&b<r)){
int m=(l+r)/2;
int left = Rec(a,b,k*2+1,l,m);
int right = Rec(a,b,k*2+2,m,r);
return min(left,right);
}else if(a<=l&&r<=b){
return tree[k];
}else{
return 2147483647;
}
}*/
int main(){
int q,com,a,b;
cin>>n>>q;
for(int i=0;i<n;i++)update(i,2147483647,-1);
for(int i=0;i<q;i++){
cin>>com;
if(com){
cin>>a;
update(a,0,0);
cout<<ans.second<<endl;
}
else {
cin>>a>>b>>c;
rec(i,a,b+1);
}
}
return 0;
}
``` |
#include "bits/stdc++.h"
using namespace std;
#ifdef _DEBUG
#include "dump.hpp"
#else
#define dump(...)
#endif
//#define int long long
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(b)-1;i>=(a);i--)
#define all(c) begin(c),end(c)
const int INF = sizeof(int) == sizeof(long long) ? 0x3f3f3f3f3f3f3f3fLL : 0x3f3f3f3f;
const int MOD = (int)(1e9) + 7;
template<class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return true; } return false; }
template<class T> bool chmin(T &a, const T &b) { if (b < a) { a = b; return true; } return false; }
const int MAX_N = 1e5;
const int SQRT_N = 317;
const int BUCKET_SIZE = 1 + (MAX_N - 1) / SQRT_N;
int dat[MAX_N] = {};
int lazy_bucket[SQRT_N] = {};
bool lazy_flag[SQRT_N] = {};
void init(int n) {
rep(i, 0, n) {
dat[i] = INT_MAX;
}
}
int get(int x) {
int k = x / SQRT_N;
return lazy_flag[k] ? lazy_bucket[k] : dat[x];
}
void update(int l, int r, int val) {
for (int k = 0; k < BUCKET_SIZE; k++) {
int bl = k * SQRT_N, br = (k + 1) * SQRT_N;
if (r <= bl || br <= l)
continue;
if (l <= bl && br <= r) {
lazy_bucket[k] = val;
lazy_flag[k] = true;
}
else {
if (lazy_flag[k]) {
for (int i = bl; i < br; i++) {
dat[i] = lazy_bucket[k];
}
lazy_flag[k] = false;
}
for (int i = max(l, bl); i < min(r, br); i++) {
dat[i] = val;
}
}
}
}
signed main() {
cin.tie(0);
ios::sync_with_stdio(false);
int n, q; cin >> n >> q;
init(n);
rep(i, 0, q) {
int c; cin >> c;
if (c == 0) {
int s, t, x; cin >> s >> t >> x;
update(s, t + 1, x);
}
else {
int i; cin >> i;
cout << get(i) << endl;
}
}
return 0;
} | ### Prompt
Develop a solution in CPP to the problem described below:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include "bits/stdc++.h"
using namespace std;
#ifdef _DEBUG
#include "dump.hpp"
#else
#define dump(...)
#endif
//#define int long long
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(b)-1;i>=(a);i--)
#define all(c) begin(c),end(c)
const int INF = sizeof(int) == sizeof(long long) ? 0x3f3f3f3f3f3f3f3fLL : 0x3f3f3f3f;
const int MOD = (int)(1e9) + 7;
template<class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return true; } return false; }
template<class T> bool chmin(T &a, const T &b) { if (b < a) { a = b; return true; } return false; }
const int MAX_N = 1e5;
const int SQRT_N = 317;
const int BUCKET_SIZE = 1 + (MAX_N - 1) / SQRT_N;
int dat[MAX_N] = {};
int lazy_bucket[SQRT_N] = {};
bool lazy_flag[SQRT_N] = {};
void init(int n) {
rep(i, 0, n) {
dat[i] = INT_MAX;
}
}
int get(int x) {
int k = x / SQRT_N;
return lazy_flag[k] ? lazy_bucket[k] : dat[x];
}
void update(int l, int r, int val) {
for (int k = 0; k < BUCKET_SIZE; k++) {
int bl = k * SQRT_N, br = (k + 1) * SQRT_N;
if (r <= bl || br <= l)
continue;
if (l <= bl && br <= r) {
lazy_bucket[k] = val;
lazy_flag[k] = true;
}
else {
if (lazy_flag[k]) {
for (int i = bl; i < br; i++) {
dat[i] = lazy_bucket[k];
}
lazy_flag[k] = false;
}
for (int i = max(l, bl); i < min(r, br); i++) {
dat[i] = val;
}
}
}
}
signed main() {
cin.tie(0);
ios::sync_with_stdio(false);
int n, q; cin >> n >> q;
init(n);
rep(i, 0, q) {
int c; cin >> c;
if (c == 0) {
int s, t, x; cin >> s >> t >> x;
update(s, t + 1, x);
}
else {
int i; cin >> i;
cout << get(i) << endl;
}
}
return 0;
}
``` |
#include<bits/stdc++.h>
#define int long long
using namespace std;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }
class SegmentTree
{
private:
int n;
vector<int> node;
public:
SegmentTree(int sz){
n=1;
while(n<sz) n*=2;
node.resize(2*n);
for(int i=1;i<n;i++) node[i]=-1;
for(int i=n;i<2*n;i++) node[i]=INT_MAX;
}
void update(int a,int b,int val,int k=1,int l=0,int r=-1){
if(r<0) r=n;
if(r<=a or b<=l) return;
if(a<=l and r<=b){
node[k]=val;
return;
}
if(node[k]!=-1){
node[2*k]=node[2*k+1]=node[k];
node[k]=-1;
}
update(a,b,val,2*k,l,(l+r)/2);
update(a,b,val,2*k+1,(l+r)/2,r);
}
int get(int idx){
idx+=n;
int res=node[idx];
while(idx>>=1){
if(node[idx]!=-1) res=node[idx];
}
return res;
}
};
signed main()
{
cin.tie(0);
ios::sync_with_stdio(false);
int N,Q; cin>>N>>Q;
SegmentTree seg(N);
while(Q--){
int q; cin>>q;
if(q==0){
int s,t,x; cin>>s>>t>>x;
seg.update(s,t+1,x);
}else{
int i; cin>>i;
cout<<seg.get(i)<<endl;
}
}
return 0;
}
| ### Prompt
Generate a Cpp solution to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include<bits/stdc++.h>
#define int long long
using namespace std;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }
class SegmentTree
{
private:
int n;
vector<int> node;
public:
SegmentTree(int sz){
n=1;
while(n<sz) n*=2;
node.resize(2*n);
for(int i=1;i<n;i++) node[i]=-1;
for(int i=n;i<2*n;i++) node[i]=INT_MAX;
}
void update(int a,int b,int val,int k=1,int l=0,int r=-1){
if(r<0) r=n;
if(r<=a or b<=l) return;
if(a<=l and r<=b){
node[k]=val;
return;
}
if(node[k]!=-1){
node[2*k]=node[2*k+1]=node[k];
node[k]=-1;
}
update(a,b,val,2*k,l,(l+r)/2);
update(a,b,val,2*k+1,(l+r)/2,r);
}
int get(int idx){
idx+=n;
int res=node[idx];
while(idx>>=1){
if(node[idx]!=-1) res=node[idx];
}
return res;
}
};
signed main()
{
cin.tie(0);
ios::sync_with_stdio(false);
int N,Q; cin>>N>>Q;
SegmentTree seg(N);
while(Q--){
int q; cin>>q;
if(q==0){
int s,t,x; cin>>s>>t>>x;
seg.update(s,t+1,x);
}else{
int i; cin>>i;
cout<<seg.get(i)<<endl;
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
#define pb push_back
using namespace std;
typedef long long ll;
const int maxn=400005;
struct {
ll mn=0x7fffffff,lazy=-1;
} S[maxn];
int n,q;
void update(int id,int ql,int qr,int v,int l,int r) {
if (ql==l && qr==r) {
S[id].lazy=v;
return;
}
if (S[id].lazy!=-1) {
S[id].mn=S[id].lazy;
if (l!=r) S[id<<1].lazy=S[id<<1|1].lazy=S[id].lazy;
S[id].lazy=-1;
}
int m=l+r>>1;
if (qr<=m) update(id<<1,ql,qr,v,l,m);
else if (ql>m) update(id<<1|1,ql,qr,v,m+1,r);
else update(id<<1,ql,m,v,l,m), update(id<<1|1,m+1,qr,v,m+1,r);
int lm=S[id<<1].lazy==-1?S[id<<1].mn:S[id<<1].lazy;
int rm=S[id<1|1].lazy==-1?S[id<<1|1].mn:S[id<<1|1].lazy;
S[id].mn=min(lm,rm);
}
int query(int id,int i,int l,int r) {
if (S[id].lazy!=-1) {
S[id].mn=S[id].lazy;
if (l!=r) S[id<<1].lazy=S[id<<1|1].lazy=S[id].lazy;
S[id].lazy=-1;
}
if (l==r) return S[id].mn;
int m=l+r>>1;
if (i<=m) return query(id<<1,i,l,m);
else return query(id<<1|1,i,m+1,r);
}
int main() {
cin>>n>>q;
for (int i=0;i<q;i++) {
bool op; cin>>op;
if (op) {
int k; cin>>k;
cout<<query(1,k,0,n-1)<<'\n';
}
else {
int s,t,x;
cin>>s>>t>>x;
update(1,s,t,x,0,n-1);
}
}
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <bits/stdc++.h>
#define pb push_back
using namespace std;
typedef long long ll;
const int maxn=400005;
struct {
ll mn=0x7fffffff,lazy=-1;
} S[maxn];
int n,q;
void update(int id,int ql,int qr,int v,int l,int r) {
if (ql==l && qr==r) {
S[id].lazy=v;
return;
}
if (S[id].lazy!=-1) {
S[id].mn=S[id].lazy;
if (l!=r) S[id<<1].lazy=S[id<<1|1].lazy=S[id].lazy;
S[id].lazy=-1;
}
int m=l+r>>1;
if (qr<=m) update(id<<1,ql,qr,v,l,m);
else if (ql>m) update(id<<1|1,ql,qr,v,m+1,r);
else update(id<<1,ql,m,v,l,m), update(id<<1|1,m+1,qr,v,m+1,r);
int lm=S[id<<1].lazy==-1?S[id<<1].mn:S[id<<1].lazy;
int rm=S[id<1|1].lazy==-1?S[id<<1|1].mn:S[id<<1|1].lazy;
S[id].mn=min(lm,rm);
}
int query(int id,int i,int l,int r) {
if (S[id].lazy!=-1) {
S[id].mn=S[id].lazy;
if (l!=r) S[id<<1].lazy=S[id<<1|1].lazy=S[id].lazy;
S[id].lazy=-1;
}
if (l==r) return S[id].mn;
int m=l+r>>1;
if (i<=m) return query(id<<1,i,l,m);
else return query(id<<1|1,i,m+1,r);
}
int main() {
cin>>n>>q;
for (int i=0;i<q;i++) {
bool op; cin>>op;
if (op) {
int k; cin>>k;
cout<<query(1,k,0,n-1)<<'\n';
}
else {
int s,t,x;
cin>>s>>t>>x;
update(1,s,t,x,0,n-1);
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
struct RU {
using t1 = int;
using t2 = int;
static t2 id2() { return -1; }
static t1 op2(const t1& l, const t2& r) { return r == id2() ? l : r; }
static t2 op3(const t2& l, const t2& r) { return r == id2() ? l : r; }
};
template <typename M>
class lazy_segment_tree {
using T1 = typename M::t1;
using T2 = typename M::t2;
const int h, n;
vector<T1> data;
vector<T2> lazy;
void push(int node) {
if (lazy[node] == M::id2()) return;
lazy[node << 1] = M::op3(lazy[node << 1], lazy[node]);
lazy[(node << 1) | 1] = M::op3(lazy[(node << 1) | 1], lazy[node]);
lazy[node] = M::id2();
}
public:
lazy_segment_tree(int n_, T1 v1)
: h(ceil(log2(n_))), n(1 << h), data(n_, v1), lazy(n << 1, M::id2()) {}
lazy_segment_tree(const vector<T1>& data_)
: h(ceil(log2(data_.size()))), n(1 << h), data(data_), lazy(n << 1, M::id2()) {}
void update(int l, int r, T2 val) {
l += n, r += n;
for (int i = h; i > 0; i--) push(l >> i), push(r >> i);
r++;
while (l < r) {
if (l & 1) lazy[l] = M::op3(lazy[l], val), l++;
if (r & 1) r--, lazy[r] = M::op3(lazy[r], val);
l >>= 1; r >>= 1;
}
}
T1 find(int p) {
T1 res = data[p]; p += n;
while (p) res = M::op2(res, lazy[p]), p >>= 1;
return res;
}
};
int main()
{
ios::sync_with_stdio(false), cin.tie(0);
int n, q;
cin >> n >> q;
lazy_segment_tree<RU> lst(n, INT_MAX);
while (q--) {
int com;
cin >> com;
if (com == 0) {
int s, t, x;
cin >> s >> t >> x;
lst.update(s, t, x);
}
else {
int p;
cin >> p;
printf("%d\n", lst.find(p));
}
}
return 0;
}
| ### Prompt
Please formulate a CPP solution to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
struct RU {
using t1 = int;
using t2 = int;
static t2 id2() { return -1; }
static t1 op2(const t1& l, const t2& r) { return r == id2() ? l : r; }
static t2 op3(const t2& l, const t2& r) { return r == id2() ? l : r; }
};
template <typename M>
class lazy_segment_tree {
using T1 = typename M::t1;
using T2 = typename M::t2;
const int h, n;
vector<T1> data;
vector<T2> lazy;
void push(int node) {
if (lazy[node] == M::id2()) return;
lazy[node << 1] = M::op3(lazy[node << 1], lazy[node]);
lazy[(node << 1) | 1] = M::op3(lazy[(node << 1) | 1], lazy[node]);
lazy[node] = M::id2();
}
public:
lazy_segment_tree(int n_, T1 v1)
: h(ceil(log2(n_))), n(1 << h), data(n_, v1), lazy(n << 1, M::id2()) {}
lazy_segment_tree(const vector<T1>& data_)
: h(ceil(log2(data_.size()))), n(1 << h), data(data_), lazy(n << 1, M::id2()) {}
void update(int l, int r, T2 val) {
l += n, r += n;
for (int i = h; i > 0; i--) push(l >> i), push(r >> i);
r++;
while (l < r) {
if (l & 1) lazy[l] = M::op3(lazy[l], val), l++;
if (r & 1) r--, lazy[r] = M::op3(lazy[r], val);
l >>= 1; r >>= 1;
}
}
T1 find(int p) {
T1 res = data[p]; p += n;
while (p) res = M::op2(res, lazy[p]), p >>= 1;
return res;
}
};
int main()
{
ios::sync_with_stdio(false), cin.tie(0);
int n, q;
cin >> n >> q;
lazy_segment_tree<RU> lst(n, INT_MAX);
while (q--) {
int com;
cin >> com;
if (com == 0) {
int s, t, x;
cin >> s >> t >> x;
lst.update(s, t, x);
}
else {
int p;
cin >> p;
printf("%d\n", lst.find(p));
}
}
return 0;
}
``` |
#include<bits/stdc++.h>
using namespace std;
const int MAX_N = 1<<18;
typedef pair<int,int> P;
int n_;
P dat[2*MAX_N-1];
void init(int n) {
n_=1;
while (n>n_) {
n_*=2;
}
for (int i=0; i<2*n_-1; i++) {
dat[i].first=-1;
dat[i].second=INT_MAX;
}
}
int find(int i) {
i+=n_-1;
P p=dat[i];
while (i>0) {
i=(i-1)/2;
p=max(p,dat[i]);
}
return p.second;
}
void update(int a,int b,P p,int k=0,int l=0,int r=n_) {
if (r<=a || b<=l) {
return;
}
if (a<=l && r<=b) {
dat[k]=p;
} else {
update(a,b,p,k*2+1,l,(l+r)/2);
update(a,b,p,k*2+2,(l+r)/2,r);
}
}
int main() {
int n, q;
cin >> n >> q;
init(n);
for (int i=0;i<q;i++) {
int f;
cin >> f;
if (!f) {
int s, t, x;
cin >> s >> t >> x;
update(s,t+1,P(i,x));
} else {
int u;
cin >> u;
printf("%d\n",find(u));
}
}
return 0;
} | ### Prompt
Please provide a Cpp coded solution to the problem described below:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
const int MAX_N = 1<<18;
typedef pair<int,int> P;
int n_;
P dat[2*MAX_N-1];
void init(int n) {
n_=1;
while (n>n_) {
n_*=2;
}
for (int i=0; i<2*n_-1; i++) {
dat[i].first=-1;
dat[i].second=INT_MAX;
}
}
int find(int i) {
i+=n_-1;
P p=dat[i];
while (i>0) {
i=(i-1)/2;
p=max(p,dat[i]);
}
return p.second;
}
void update(int a,int b,P p,int k=0,int l=0,int r=n_) {
if (r<=a || b<=l) {
return;
}
if (a<=l && r<=b) {
dat[k]=p;
} else {
update(a,b,p,k*2+1,l,(l+r)/2);
update(a,b,p,k*2+2,(l+r)/2,r);
}
}
int main() {
int n, q;
cin >> n >> q;
init(n);
for (int i=0;i<q;i++) {
int f;
cin >> f;
if (!f) {
int s, t, x;
cin >> s >> t >> x;
update(s,t+1,P(i,x));
} else {
int u;
cin >> u;
printf("%d\n",find(u));
}
}
return 0;
}
``` |
#include<bits/stdc++.h>
using namespace std;
int A[300000],B[300000];
int flag[300000];
int n,q;
void evaluate(int k){
if(flag[k]==0) return ;
A[k*2+1]=A[k*2+2]=B[k];
B[k*2+1]=B[k*2+2]=B[k];
flag[k*2+1]=flag[k*2+2]=1;
flag[k]=0;
return ;
}
int update(int s,int t,int x,int k,int l,int r){
if(s>=r || l>=t) return A[k];
if(s<=l && r<=t){
A[k]=x;
B[k]=x;
flag[k]=1;
return A[k];
}
evaluate(k);
int al=update(s,t,x,k*2+1,l,(l+r)/2);
int ar=update(s,t,x,k*2+2,(l+r)/2,r);
return A[k]=min(al,ar);
}
int find(int i,int k){
if(flag[k]) A[i]=B[k];
if(k==0) return A[i];
return find(i,(k-1)/2);
}
int main(){
int com,s,t,x,k;
for(int i=0;i<300000;i++) A[i]=B[i]=INT_MAX;
cin>>n>>q;
int ra=1;
while(ra<=n) ra*=2;
n=ra;
for(int i=0;i<q;i++){
cin>>com;
if(com==0){
cin>>s>>t>>x;
update(s,t+1,x,0,0,n);
}
else if(com==1){
cin>>k;
cout<<find(k+n-1,k+n-1)<<endl;
}
}
//for(int i=0;i<10;i++) cout<<A[i]<<endl;
return 0;
}
| ### Prompt
Create a solution in Cpp for the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
int A[300000],B[300000];
int flag[300000];
int n,q;
void evaluate(int k){
if(flag[k]==0) return ;
A[k*2+1]=A[k*2+2]=B[k];
B[k*2+1]=B[k*2+2]=B[k];
flag[k*2+1]=flag[k*2+2]=1;
flag[k]=0;
return ;
}
int update(int s,int t,int x,int k,int l,int r){
if(s>=r || l>=t) return A[k];
if(s<=l && r<=t){
A[k]=x;
B[k]=x;
flag[k]=1;
return A[k];
}
evaluate(k);
int al=update(s,t,x,k*2+1,l,(l+r)/2);
int ar=update(s,t,x,k*2+2,(l+r)/2,r);
return A[k]=min(al,ar);
}
int find(int i,int k){
if(flag[k]) A[i]=B[k];
if(k==0) return A[i];
return find(i,(k-1)/2);
}
int main(){
int com,s,t,x,k;
for(int i=0;i<300000;i++) A[i]=B[i]=INT_MAX;
cin>>n>>q;
int ra=1;
while(ra<=n) ra*=2;
n=ra;
for(int i=0;i<q;i++){
cin>>com;
if(com==0){
cin>>s>>t>>x;
update(s,t+1,x,0,0,n);
}
else if(com==1){
cin>>k;
cout<<find(k+n-1,k+n-1)<<endl;
}
}
//for(int i=0;i<10;i++) cout<<A[i]<<endl;
return 0;
}
``` |
#include<bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
#ifndef MAX
#define MAX 400000
#endif
//Range Update Query+Range Sum Query
template<class T>
class RUQ_RSQ {
public:
int n;
T dat[MAX], lazy[MAX], ZERO, DEFAULT;
void init(int n_, T d = INT_MAX, T t = 0) {
DEFAULT = d;
ZERO = T();
n = 1; while (n < n_)n <<= 1;
for (int i = 0; i < 2 * n - 1; i++) {
dat[i] = t; lazy[i] = DEFAULT;
}
}
inline void push(int k, int s) {
if (lazy[k] == DEFAULT)return;
dat[k] = lazy[k] * s;
if (k < n - 1) {
lazy[k * 2 + 1] = lazy[k];
lazy[k * 2 + 2] = lazy[k];
}
lazy[k] = DEFAULT;
}
inline void update_node(int k) {
dat[k] = dat[k * 2 + 1] + dat[k * 2 + 2];
}
inline void update(int a, int b, T x, int k, int l, int r) {
push(k, r - l);
if (r <= a || b <= l)return;
if (a <= l&&r <= b) {
lazy[k] = x; push(k, r - l); return;
}
update(a, b, x, k * 2 + 1, l, (l + r) / 2);
update(a, b, x, k * 2 + 2, (l + r) / 2, r);
update_node(k);
}
inline T query(int a, int b, int k, int l, int r) {
push(k, r - l);
if (r <= a || b <= l)return ZERO;
if (a <= l&&r <= b)return dat[k];
T lb = query(a, b, k * 2 + 1, l, (l + r) / 2);
T rb = query(a, b, k * 2 + 2, (l + r) / 2, r);
update_node(k);
return lb + rb;
}
inline void update(int a, int b, T x) {
update(a, b, x, 0, 0, n);
}
inline void update(int a, T x) {
update(a, a + 1, x);
}
inline T query(int a, int b) {
return query(a, b, 0, 0, n);
}
inline T query(int a) {
return query(a, a + 1);
}
};
RUQ_RSQ<int>seg;
int main() {
int n, q; scanf("%d%d", &n, &q);
seg.init(n, INT_MAX, INT_MAX);
rep(i, q) {
int a; scanf("%d", &a);
if (a == 0) {
int s, t, x; scanf("%d%d%d", &s, &t, &x); t++;
seg.update(s, t, x);
}
else {
int t; scanf("%d", &t);
printf("%d\n", seg.query(t));
}
}
} | ### Prompt
Your task is to create a Cpp solution to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include<bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
#ifndef MAX
#define MAX 400000
#endif
//Range Update Query+Range Sum Query
template<class T>
class RUQ_RSQ {
public:
int n;
T dat[MAX], lazy[MAX], ZERO, DEFAULT;
void init(int n_, T d = INT_MAX, T t = 0) {
DEFAULT = d;
ZERO = T();
n = 1; while (n < n_)n <<= 1;
for (int i = 0; i < 2 * n - 1; i++) {
dat[i] = t; lazy[i] = DEFAULT;
}
}
inline void push(int k, int s) {
if (lazy[k] == DEFAULT)return;
dat[k] = lazy[k] * s;
if (k < n - 1) {
lazy[k * 2 + 1] = lazy[k];
lazy[k * 2 + 2] = lazy[k];
}
lazy[k] = DEFAULT;
}
inline void update_node(int k) {
dat[k] = dat[k * 2 + 1] + dat[k * 2 + 2];
}
inline void update(int a, int b, T x, int k, int l, int r) {
push(k, r - l);
if (r <= a || b <= l)return;
if (a <= l&&r <= b) {
lazy[k] = x; push(k, r - l); return;
}
update(a, b, x, k * 2 + 1, l, (l + r) / 2);
update(a, b, x, k * 2 + 2, (l + r) / 2, r);
update_node(k);
}
inline T query(int a, int b, int k, int l, int r) {
push(k, r - l);
if (r <= a || b <= l)return ZERO;
if (a <= l&&r <= b)return dat[k];
T lb = query(a, b, k * 2 + 1, l, (l + r) / 2);
T rb = query(a, b, k * 2 + 2, (l + r) / 2, r);
update_node(k);
return lb + rb;
}
inline void update(int a, int b, T x) {
update(a, b, x, 0, 0, n);
}
inline void update(int a, T x) {
update(a, a + 1, x);
}
inline T query(int a, int b) {
return query(a, b, 0, 0, n);
}
inline T query(int a) {
return query(a, a + 1);
}
};
RUQ_RSQ<int>seg;
int main() {
int n, q; scanf("%d%d", &n, &q);
seg.init(n, INT_MAX, INT_MAX);
rep(i, q) {
int a; scanf("%d", &a);
if (a == 0) {
int s, t, x; scanf("%d%d%d", &s, &t, &x); t++;
seg.update(s, t, x);
}
else {
int t; scanf("%d", &t);
printf("%d\n", seg.query(t));
}
}
}
``` |
#include<cstdio>
#include<vector>
#include<algorithm>
#include<utility>
#include<numeric>
#include<iostream>
#include<array>
#include<string>
#include<sstream>
#include<stack>
#include<queue>
#include<list>
#include<functional>
#define _USE_MATH_DEFINES
#include<math.h>
#include<map>
#define SENTINEL 1000000001
#define min(a,b) (a)>(b)?(b):(a)
#define max(a,b) (a)>(b)?(a):(b)
using namespace std;
typedef pair<int, int> pii;
long long seg[262149], n, v, a, b, c, d, INF = (1LL << 31) - 1, size_ = 1;
void update(long long p, long long q, long long r, long long s, long long t, long long u, long long v) {
if (s <= p || q <= r) { if (v != INF)seg[u] = v; return; }
if (p <= r && s <= q) { seg[u] = t; return; }
long long w = v; if (w == INF)w = seg[u]; seg[u] = INF;
update(p, q, r, (r + s) / 2, t, u * 2, w);
update(p, q, (r + s) / 2, s, t, u * 2 + 1, w);
}
long long find(long long p, long long q, long long r, long long s, long long u) {
if (s <= p || q <= r)return INF;
if ((r <= p && q <= s) && (seg[u] != INF || s - r == 1)) { return seg[u]; }
long long a1 = find(p, q, r, (r + s) / 2, u * 2);
long long a2 = find(p, q, (r + s) / 2, s, u * 2 + 1);
return min(a1, a2);
}
int main() {
cin >> n >> v; while (size_ < n)size_ *= 2;
for (int i = 1; i <= size_ * 2; i++)seg[i] = INF;
for (int i = 1; i <= v; i++) {
scanf("%d", &a);
if (a == 0) { scanf("%lld%lld%lld", &b, &c, &d); c++; update(b, c, 0, size_, d, 1, INF); }
if (a == 1) { scanf("%lld", &b); printf("%lld\n", find(b, b + 1, 0, size_, 1)); }
}
return 0;
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include<cstdio>
#include<vector>
#include<algorithm>
#include<utility>
#include<numeric>
#include<iostream>
#include<array>
#include<string>
#include<sstream>
#include<stack>
#include<queue>
#include<list>
#include<functional>
#define _USE_MATH_DEFINES
#include<math.h>
#include<map>
#define SENTINEL 1000000001
#define min(a,b) (a)>(b)?(b):(a)
#define max(a,b) (a)>(b)?(a):(b)
using namespace std;
typedef pair<int, int> pii;
long long seg[262149], n, v, a, b, c, d, INF = (1LL << 31) - 1, size_ = 1;
void update(long long p, long long q, long long r, long long s, long long t, long long u, long long v) {
if (s <= p || q <= r) { if (v != INF)seg[u] = v; return; }
if (p <= r && s <= q) { seg[u] = t; return; }
long long w = v; if (w == INF)w = seg[u]; seg[u] = INF;
update(p, q, r, (r + s) / 2, t, u * 2, w);
update(p, q, (r + s) / 2, s, t, u * 2 + 1, w);
}
long long find(long long p, long long q, long long r, long long s, long long u) {
if (s <= p || q <= r)return INF;
if ((r <= p && q <= s) && (seg[u] != INF || s - r == 1)) { return seg[u]; }
long long a1 = find(p, q, r, (r + s) / 2, u * 2);
long long a2 = find(p, q, (r + s) / 2, s, u * 2 + 1);
return min(a1, a2);
}
int main() {
cin >> n >> v; while (size_ < n)size_ *= 2;
for (int i = 1; i <= size_ * 2; i++)seg[i] = INF;
for (int i = 1; i <= v; i++) {
scanf("%d", &a);
if (a == 0) { scanf("%lld%lld%lld", &b, &c, &d); c++; update(b, c, 0, size_, d, 1, INF); }
if (a == 1) { scanf("%lld", &b); printf("%lld\n", find(b, b + 1, 0, size_, 1)); }
}
return 0;
}
``` |
#include "bits/stdc++.h"
using namespace std;
#ifdef _DEBUG
#include "dump.hpp"
#else
#define dump(...)
#endif
//#define int long long
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(b)-1;i>=(a);i--)
#define all(c) begin(c),end(c)
const int INF = (1ll << 31) - 1;
const int MOD = (int)(1e9) + 7;
const double PI = acos(-1);
const double EPS = 1e-9;
template<class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return true; } return false; }
template<class T> bool chmin(T &a, const T &b) { if (a > b) { a = b; return true; } return false; }
struct RangeUpdateQuery {
int n;
vector<int>d;
RangeUpdateQuery(int m) {
for (n = 1; n < m; n <<= 1);
d.assign(2 * n, INF);
}
void update(int a, int b, int x = -1, int k = 1, int l = 0, int r = -1) {
if (r == -1)r = n;
if (r <= a || b <= l)return;
if (a <= l&&r <= b) {
if (x >= 0)d[k] = x;
return;
}
else {
if (d[k] != INF)d[2 * k] = d[2 * k + 1] = d[k], d[k] = INF;
update(a, b, x, 2 * k, l, (l + r) / 2);
update(a, b, x, 2 * k + 1, (l + r) / 2, r);
}
}
int find(int i) {
update(i, i + 1);
return d[n + i];
}
};
signed main() {
cin.tie(0);
ios::sync_with_stdio(false);
int n, q; cin >> n >> q;
RangeUpdateQuery ruq(n);
rep(z, 0, q) {
int com; cin >> com;
if (com) {
int i; cin >> i;
cout << ruq.find(i) << endl;
}
else {
int s, t, x; cin >> s >> t >> x;
ruq.update(s, t+1, x);
}
}
return 0;
} | ### Prompt
Please formulate a Cpp solution to the following problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(i): output the value of ai.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ i < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 i
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(i).
Output
For each find operation, print the value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0
1 1
Output
1
3
Input
1 3
1 0
0 0 0 5
1 0
Output
2147483647
5
### Response
```cpp
#include "bits/stdc++.h"
using namespace std;
#ifdef _DEBUG
#include "dump.hpp"
#else
#define dump(...)
#endif
//#define int long long
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(b)-1;i>=(a);i--)
#define all(c) begin(c),end(c)
const int INF = (1ll << 31) - 1;
const int MOD = (int)(1e9) + 7;
const double PI = acos(-1);
const double EPS = 1e-9;
template<class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return true; } return false; }
template<class T> bool chmin(T &a, const T &b) { if (a > b) { a = b; return true; } return false; }
struct RangeUpdateQuery {
int n;
vector<int>d;
RangeUpdateQuery(int m) {
for (n = 1; n < m; n <<= 1);
d.assign(2 * n, INF);
}
void update(int a, int b, int x = -1, int k = 1, int l = 0, int r = -1) {
if (r == -1)r = n;
if (r <= a || b <= l)return;
if (a <= l&&r <= b) {
if (x >= 0)d[k] = x;
return;
}
else {
if (d[k] != INF)d[2 * k] = d[2 * k + 1] = d[k], d[k] = INF;
update(a, b, x, 2 * k, l, (l + r) / 2);
update(a, b, x, 2 * k + 1, (l + r) / 2, r);
}
}
int find(int i) {
update(i, i + 1);
return d[n + i];
}
};
signed main() {
cin.tie(0);
ios::sync_with_stdio(false);
int n, q; cin >> n >> q;
RangeUpdateQuery ruq(n);
rep(z, 0, q) {
int com; cin >> com;
if (com) {
int i; cin >> i;
cout << ruq.find(i) << endl;
}
else {
int s, t, x; cin >> s >> t >> x;
ruq.update(s, t+1, x);
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
pair<int, int> operator-(pair<int, int> a, pair<int, int> b) {
return pair<int, int>(a.first - b.first, a.second - b.second);
}
long long operator%(pair<int, int> a, pair<int, int> b) {
return 1LL * a.first * b.second - 1LL * a.second * b.first;
}
long long operator*(pair<int, int> a, pair<int, int> b) {
return 1LL * a.first * b.first + 1LL * a.second * b.second;
}
long long dist(pair<int, int> a, pair<int, int> b) { return (a - b) * (a - b); }
int CCW(pair<int, int> a, pair<int, int> b, pair<int, int> c) {
pair<int, int> p1 = b - a, p2 = c - a;
if (p1 % p2 > 0) return 1;
if (p1 % p2 < 0) return -1;
return 0;
}
pair<int, int> cpiv;
bool comp(pair<int, int> a, pair<int, int> b) {
int type = CCW(cpiv, a, b);
if (type == 0) return dist(cpiv, a) < dist(cpiv, b);
return CCW(cpiv, a, b) == 1;
}
struct polygon {
vector<pair<int, int> > v;
vector<pair<int, int> > ans;
void add_point(const pair<int, int>& a) { v.push_back(a); }
void clear() { v.clear(); }
void normalize() {
ans.clear();
for (auto& i : v)
if (i < v.back()) swap(i, v.back());
ans.push_back(v.back());
v.pop_back();
cpiv = ans[0];
sort(v.begin(), v.end(), comp);
for (auto i : v) {
bool keep = 1;
while (ans.size() > 1) {
int n = ans.size();
int type = CCW(ans[n - 2], ans[n - 1], i);
if (type == 1) break;
if (type == -1) {
ans.pop_back();
continue;
}
keep = 0;
if (dist(ans[n - 2], i) > dist(ans[n - 2], ans[n - 1])) ans[n - 1] = i;
break;
}
if (keep) ans.push_back(i);
}
if (ans.size() > 2) {
int n = ans.size();
if (CCW(ans[0], ans[n - 1], ans[n - 2]) == 0) ans.pop_back();
}
}
vector<long long> gethash() {
int n = ans.size();
vector<long long> samp;
for (int i = 1; i < n - 1; i++) {
samp.push_back(dist(ans[i], ans[i - 1]));
samp.push_back((ans[i - 1] - ans[i]) * (ans[i + 1] - ans[i]));
}
samp.push_back(dist(ans[n - 1], ans[n - 2]));
return samp;
}
} pol[2];
pair<int, int> a[2][N];
string solve(vector<long long>& a, vector<long long>& b) {
vector<int> tr(a.size(), 0);
for (int i = 1; i < a.size(); i++) {
tr[i] = tr[i - 1];
while (tr[i] != 0 && a[tr[i]] != a[i]) tr[i] = tr[tr[i] - 1];
if (a[i] == a[tr[i]]) tr[i]++;
}
int len = 0;
for (int i = 0; i < b.size(); i++) {
long long val = b[i];
while (len != 0 && a[len] != val) len = tr[len - 1];
if (a[len] == val) len++;
if (len == a.size()) return "YES";
}
return "NO";
}
int main() {
int n1, n2;
cin >> n1 >> n2;
for (int i = 0; i < n1; i++) cin >> a[0][i].first >> a[0][i].second;
for (int i = 0; i < n2; i++) cin >> a[1][i].first >> a[1][i].second;
pol[0].clear();
pol[1].clear();
for (int i = 0; i < n1; i++) pol[0].add_point(a[0][i]);
for (int i = 0; i < n2; i++) pol[1].add_point(a[1][i]);
pol[0].normalize();
pol[1].normalize();
if (pol[0].ans.size() != pol[1].ans.size()) {
cout << "NO";
return 0;
}
int nt = pol[0].ans.size();
for (int i = 0; i < nt; i++) pol[0].ans.push_back(pol[0].ans[i]);
vector<long long> a = pol[0].gethash();
vector<long long> b = pol[1].gethash();
cout << solve(b, a);
}
| ### Prompt
Please formulate a cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
pair<int, int> operator-(pair<int, int> a, pair<int, int> b) {
return pair<int, int>(a.first - b.first, a.second - b.second);
}
long long operator%(pair<int, int> a, pair<int, int> b) {
return 1LL * a.first * b.second - 1LL * a.second * b.first;
}
long long operator*(pair<int, int> a, pair<int, int> b) {
return 1LL * a.first * b.first + 1LL * a.second * b.second;
}
long long dist(pair<int, int> a, pair<int, int> b) { return (a - b) * (a - b); }
int CCW(pair<int, int> a, pair<int, int> b, pair<int, int> c) {
pair<int, int> p1 = b - a, p2 = c - a;
if (p1 % p2 > 0) return 1;
if (p1 % p2 < 0) return -1;
return 0;
}
pair<int, int> cpiv;
bool comp(pair<int, int> a, pair<int, int> b) {
int type = CCW(cpiv, a, b);
if (type == 0) return dist(cpiv, a) < dist(cpiv, b);
return CCW(cpiv, a, b) == 1;
}
struct polygon {
vector<pair<int, int> > v;
vector<pair<int, int> > ans;
void add_point(const pair<int, int>& a) { v.push_back(a); }
void clear() { v.clear(); }
void normalize() {
ans.clear();
for (auto& i : v)
if (i < v.back()) swap(i, v.back());
ans.push_back(v.back());
v.pop_back();
cpiv = ans[0];
sort(v.begin(), v.end(), comp);
for (auto i : v) {
bool keep = 1;
while (ans.size() > 1) {
int n = ans.size();
int type = CCW(ans[n - 2], ans[n - 1], i);
if (type == 1) break;
if (type == -1) {
ans.pop_back();
continue;
}
keep = 0;
if (dist(ans[n - 2], i) > dist(ans[n - 2], ans[n - 1])) ans[n - 1] = i;
break;
}
if (keep) ans.push_back(i);
}
if (ans.size() > 2) {
int n = ans.size();
if (CCW(ans[0], ans[n - 1], ans[n - 2]) == 0) ans.pop_back();
}
}
vector<long long> gethash() {
int n = ans.size();
vector<long long> samp;
for (int i = 1; i < n - 1; i++) {
samp.push_back(dist(ans[i], ans[i - 1]));
samp.push_back((ans[i - 1] - ans[i]) * (ans[i + 1] - ans[i]));
}
samp.push_back(dist(ans[n - 1], ans[n - 2]));
return samp;
}
} pol[2];
pair<int, int> a[2][N];
string solve(vector<long long>& a, vector<long long>& b) {
vector<int> tr(a.size(), 0);
for (int i = 1; i < a.size(); i++) {
tr[i] = tr[i - 1];
while (tr[i] != 0 && a[tr[i]] != a[i]) tr[i] = tr[tr[i] - 1];
if (a[i] == a[tr[i]]) tr[i]++;
}
int len = 0;
for (int i = 0; i < b.size(); i++) {
long long val = b[i];
while (len != 0 && a[len] != val) len = tr[len - 1];
if (a[len] == val) len++;
if (len == a.size()) return "YES";
}
return "NO";
}
int main() {
int n1, n2;
cin >> n1 >> n2;
for (int i = 0; i < n1; i++) cin >> a[0][i].first >> a[0][i].second;
for (int i = 0; i < n2; i++) cin >> a[1][i].first >> a[1][i].second;
pol[0].clear();
pol[1].clear();
for (int i = 0; i < n1; i++) pol[0].add_point(a[0][i]);
for (int i = 0; i < n2; i++) pol[1].add_point(a[1][i]);
pol[0].normalize();
pol[1].normalize();
if (pol[0].ans.size() != pol[1].ans.size()) {
cout << "NO";
return 0;
}
int nt = pol[0].ans.size();
for (int i = 0; i < nt; i++) pol[0].ans.push_back(pol[0].ans[i]);
vector<long long> a = pol[0].gethash();
vector<long long> b = pol[1].gethash();
cout << solve(b, a);
}
``` |
#include <bits/stdc++.h>
using namespace std;
struct vec2l {
long x, y;
};
string vec_to_str(const vec2l& val) {
return "(" + to_string(val.x) + ", " + to_string(val.y) + ")";
}
vector<vec2l> read_points(int n) {
vector<vec2l> result;
result.reserve(n);
for (int i = 0; i < n; ++i) {
vec2l p;
scanf("%ld %ld", &p.x, &p.y);
result.push_back(p);
}
return result;
}
long long cross_product(const vec2l& a, const vec2l& b) {
long long ax = static_cast<long long>(a.x);
long long ay = static_cast<long long>(a.y);
long long bx = static_cast<long long>(b.x);
long long by = static_cast<long long>(b.y);
return (ax * by - bx * ay);
}
long long size_squared(const vec2l& a) {
long long ax = static_cast<long long>(a.x);
long long ay = static_cast<long long>(a.y);
return (ax * ax + ay * ay);
}
long long dot_product(const vec2l& a, const vec2l& b) {
long long ax = static_cast<long long>(a.x);
long long ay = static_cast<long long>(a.y);
long long bx = static_cast<long long>(b.x);
long long by = static_cast<long long>(b.y);
return (ax * bx + ay * by);
}
void quick_hull(vector<vec2l>& hull, const vector<vec2l>& points,
const vec2l& p1, const vec2l& p2) {
vec2l p{p2.x - p1.x, p2.y - p1.y};
vector<vec2l> filtered;
filtered.reserve(points.size());
long long max_height = -1;
int highest = -1;
vec2l max_b;
for (const auto& point : points) {
vec2l b{point.x - p1.x, point.y - p1.y};
long long height = cross_product(p, b);
if (height > 0) {
filtered.push_back(point);
if (height > max_height ||
(height == max_height && dot_product(p, b) < dot_product(p, max_b))) {
max_height = height;
highest = filtered.size() - 1;
max_b = b;
}
}
}
if (filtered.size() == 0) {
hull.push_back(p2);
} else if (filtered.size() == 1) {
hull.push_back(filtered[0]);
hull.push_back(p2);
} else if (filtered.size() > 1) {
quick_hull(hull, filtered, p1, filtered[highest]);
quick_hull(hull, filtered, filtered[highest], p2);
}
}
vector<vec2l> hull(const vector<vec2l>& points) {
vector<vec2l> hull;
hull.reserve(points.size());
vec2l left = points[0];
vec2l right = points[0];
for (const vec2l& point : points) {
if (point.x < left.x || (point.x == left.x && point.y < left.y)) {
left = point;
}
if (point.x > right.x || (point.x == right.x && point.y > right.y)) {
right = point;
}
}
if (left.x == right.x) {
hull.push_back(left);
if (left.y != right.y) {
hull.push_back(right);
}
return hull;
}
quick_hull(hull, points, left, right);
quick_hull(hull, points, right, left);
return hull;
}
struct edge {
long long length_squared;
long long dot;
};
string edge_to_str(const edge& val) {
return "(" + to_string(val.length_squared) + ", " + to_string(val.dot) + ")";
}
bool edges_equal(const edge& a, const edge& b) {
return a.length_squared == b.length_squared && a.dot == b.dot;
}
vector<edge> edge_list(const vector<vec2l>& h) {
vector<edge> result;
result.reserve(h.size());
for (int i = 0; i < h.size(); ++i) {
vec2l a = h[i];
vec2l b = h[(i + 1) % h.size()];
vec2l c = h[(i + 2) % h.size()];
vec2l ab = {b.x - a.x, b.y - a.y};
vec2l bc = {c.x - b.x, c.y - b.y};
result.push_back({size_squared(ab), dot_product(ab, bc)});
}
return result;
}
bool list_match(const vector<edge>& a, const vector<edge>& b) {
vector<int> prefix(b.size(), 0);
for (int i = 1; i < prefix.size(); ++i) {
int p = prefix[i - 1];
while (p >= 0 && !edges_equal(b[i], b[p])) {
p = (p == 0) ? -1 : prefix[p - 1];
}
prefix[i] = p + 1;
}
int i = 0;
int j = 0;
while (i < a.size()) {
if (edges_equal(a[i], b[j])) {
++i;
++j;
if (j >= b.size()) {
return true;
}
} else {
if (j > 0) {
j = prefix[j];
} else {
++i;
}
}
}
return false;
}
int main() {
int n, m;
scanf("%d %d", &n, &m);
vector<vec2l> a = read_points(n);
vector<vec2l> b = read_points(m);
vector<vec2l> a_hull = hull(a);
vector<vec2l> b_hull = hull(b);
if (a_hull.size() == 1 && b_hull.size() == 1) {
printf("YES\n");
return 0;
} else if (a_hull.size() != b_hull.size()) {
printf("NO\n");
return 0;
}
vector<edge> a_edges = edge_list(a_hull);
vector<edge> b_edges = edge_list(b_hull);
int a_edges_size = a_edges.size();
a_edges.reserve(a_edges_size * 2);
for (int i = 0; i < a_edges_size; ++i) {
a_edges.push_back(a_edges[i]);
}
printf(list_match(a_edges, b_edges) ? "YES\n" : "NO\n");
}
| ### Prompt
Please create a solution in cpp to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
struct vec2l {
long x, y;
};
string vec_to_str(const vec2l& val) {
return "(" + to_string(val.x) + ", " + to_string(val.y) + ")";
}
vector<vec2l> read_points(int n) {
vector<vec2l> result;
result.reserve(n);
for (int i = 0; i < n; ++i) {
vec2l p;
scanf("%ld %ld", &p.x, &p.y);
result.push_back(p);
}
return result;
}
long long cross_product(const vec2l& a, const vec2l& b) {
long long ax = static_cast<long long>(a.x);
long long ay = static_cast<long long>(a.y);
long long bx = static_cast<long long>(b.x);
long long by = static_cast<long long>(b.y);
return (ax * by - bx * ay);
}
long long size_squared(const vec2l& a) {
long long ax = static_cast<long long>(a.x);
long long ay = static_cast<long long>(a.y);
return (ax * ax + ay * ay);
}
long long dot_product(const vec2l& a, const vec2l& b) {
long long ax = static_cast<long long>(a.x);
long long ay = static_cast<long long>(a.y);
long long bx = static_cast<long long>(b.x);
long long by = static_cast<long long>(b.y);
return (ax * bx + ay * by);
}
void quick_hull(vector<vec2l>& hull, const vector<vec2l>& points,
const vec2l& p1, const vec2l& p2) {
vec2l p{p2.x - p1.x, p2.y - p1.y};
vector<vec2l> filtered;
filtered.reserve(points.size());
long long max_height = -1;
int highest = -1;
vec2l max_b;
for (const auto& point : points) {
vec2l b{point.x - p1.x, point.y - p1.y};
long long height = cross_product(p, b);
if (height > 0) {
filtered.push_back(point);
if (height > max_height ||
(height == max_height && dot_product(p, b) < dot_product(p, max_b))) {
max_height = height;
highest = filtered.size() - 1;
max_b = b;
}
}
}
if (filtered.size() == 0) {
hull.push_back(p2);
} else if (filtered.size() == 1) {
hull.push_back(filtered[0]);
hull.push_back(p2);
} else if (filtered.size() > 1) {
quick_hull(hull, filtered, p1, filtered[highest]);
quick_hull(hull, filtered, filtered[highest], p2);
}
}
vector<vec2l> hull(const vector<vec2l>& points) {
vector<vec2l> hull;
hull.reserve(points.size());
vec2l left = points[0];
vec2l right = points[0];
for (const vec2l& point : points) {
if (point.x < left.x || (point.x == left.x && point.y < left.y)) {
left = point;
}
if (point.x > right.x || (point.x == right.x && point.y > right.y)) {
right = point;
}
}
if (left.x == right.x) {
hull.push_back(left);
if (left.y != right.y) {
hull.push_back(right);
}
return hull;
}
quick_hull(hull, points, left, right);
quick_hull(hull, points, right, left);
return hull;
}
struct edge {
long long length_squared;
long long dot;
};
string edge_to_str(const edge& val) {
return "(" + to_string(val.length_squared) + ", " + to_string(val.dot) + ")";
}
bool edges_equal(const edge& a, const edge& b) {
return a.length_squared == b.length_squared && a.dot == b.dot;
}
vector<edge> edge_list(const vector<vec2l>& h) {
vector<edge> result;
result.reserve(h.size());
for (int i = 0; i < h.size(); ++i) {
vec2l a = h[i];
vec2l b = h[(i + 1) % h.size()];
vec2l c = h[(i + 2) % h.size()];
vec2l ab = {b.x - a.x, b.y - a.y};
vec2l bc = {c.x - b.x, c.y - b.y};
result.push_back({size_squared(ab), dot_product(ab, bc)});
}
return result;
}
bool list_match(const vector<edge>& a, const vector<edge>& b) {
vector<int> prefix(b.size(), 0);
for (int i = 1; i < prefix.size(); ++i) {
int p = prefix[i - 1];
while (p >= 0 && !edges_equal(b[i], b[p])) {
p = (p == 0) ? -1 : prefix[p - 1];
}
prefix[i] = p + 1;
}
int i = 0;
int j = 0;
while (i < a.size()) {
if (edges_equal(a[i], b[j])) {
++i;
++j;
if (j >= b.size()) {
return true;
}
} else {
if (j > 0) {
j = prefix[j];
} else {
++i;
}
}
}
return false;
}
int main() {
int n, m;
scanf("%d %d", &n, &m);
vector<vec2l> a = read_points(n);
vector<vec2l> b = read_points(m);
vector<vec2l> a_hull = hull(a);
vector<vec2l> b_hull = hull(b);
if (a_hull.size() == 1 && b_hull.size() == 1) {
printf("YES\n");
return 0;
} else if (a_hull.size() != b_hull.size()) {
printf("NO\n");
return 0;
}
vector<edge> a_edges = edge_list(a_hull);
vector<edge> b_edges = edge_list(b_hull);
int a_edges_size = a_edges.size();
a_edges.reserve(a_edges_size * 2);
for (int i = 0; i < a_edges_size; ++i) {
a_edges.push_back(a_edges[i]);
}
printf(list_match(a_edges, b_edges) ? "YES\n" : "NO\n");
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 2;
const int INF = 1e9 + 7;
const double EPS = 1e-10;
struct Point {
int x, y;
Point() {}
Point(int x, int y) : x(x), y(y) {}
friend Point operator-(Point p1, Point p2) {
return Point(p1.x - p2.x, p1.y - p2.y);
}
friend bool operator<(Point p1, Point p2) {
return (p1.y == p2.y ? p1.x < p2.x : p1.y < p2.y);
}
};
double norma(Point v) { return sqrt(1LL * v.x * v.x + 1LL * v.y * v.y); }
double cross(Point v1, Point v2) {
return 1LL * v1.x * v2.y - 1LL * v2.x * v1.y;
}
vector<Point> ConvexHull(vector<Point> P) {
if (P.size() < 3) return P;
Point p = *(min_element(P.begin(), P.end()));
sort(P.begin(), P.end(), [p](Point p1, Point p2) {
double area = cross(p1 - p, p2 - p);
if (area == 0) return norma(p1 - p) < norma(p2 - p);
return area > 0;
});
vector<Point> C;
C.push_back(P[0]);
C.push_back(P[1]);
int it = 2, szC = 2;
while (it < P.size()) {
if (szC > 1 && cross(P[it] - C[szC - 1], C[szC - 2] - C[szC - 1]) <= 0) {
C.pop_back();
szC--;
} else {
C.push_back(P[it]);
szC++;
it++;
}
}
return C;
}
struct data {
double a, b, c;
friend bool operator==(data d1, data d2) {
bool eq1 = abs(d1.a - d2.a) <= EPS;
bool eq2 = abs(d1.b - d2.b) <= EPS;
bool eq3 = abs(d1.c - d2.c) <= EPS;
return eq1 && eq2 && eq3;
}
friend bool operator!=(data d1, data d2) { return !(d1 == d2); }
};
vector<int> F;
void buildF(vector<data> &P) {
F = vector<int>(P.size(), 0);
for (int itP = 1, len = 0; itP < P.size(); itP++) {
for (; len != -1 && P[itP] != P[len]; len = len ? F[len - 1] : -1)
;
F[itP] = ++len;
}
}
bool KMP(vector<data> &P, vector<data> &T) {
buildF(P);
for (int itT = 0, itP = 0; itT < T.size(); itT++) {
for (; itP != -1 && T[itT] != P[itP]; itP = itP ? F[itP - 1] : -1)
;
if (P.size() == ++itP) {
return true;
}
}
return false;
}
vector<data> getData(vector<Point> &P) {
int sz = P.size();
vector<data> Data;
for (int i = 0; i < sz; i++) {
Data.push_back({norma(P[i] - P[(i + 1) % sz]),
norma(P[(i + 1) % sz] - P[(i + 2) % sz]),
norma(P[(i + 2) % sz] - P[i])});
}
return Data;
}
bool equals(vector<Point> v1, vector<Point> v2) {
if (v1.size() != v2.size()) return false;
if (v1.size() == 2) {
return norma(v1[0] - v1[1]) == norma(v2[0] - v2[1]);
}
for (int i = 0; i < v1.size(); i++) {
v2.push_back(v2[i]);
}
vector<data> P = getData(v1);
vector<data> T = getData(v2);
return KMP(P, T);
}
int main() {
int n, m;
scanf("%d %d", &n, &m);
vector<Point> P1(n);
for (int i = 0; i < n; i++) {
scanf("%d %d", &P1[i].x, &P1[i].y);
}
vector<Point> P2(m);
for (int i = 0; i < m; i++) {
scanf("%d %d", &P2[i].x, &P2[i].y);
}
vector<Point> v1 = ConvexHull(P1);
vector<Point> v2 = ConvexHull(P2);
if (equals(v1, v2))
cout << "YES";
else
cout << "NO";
return 0;
}
| ### Prompt
Generate a Cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 2;
const int INF = 1e9 + 7;
const double EPS = 1e-10;
struct Point {
int x, y;
Point() {}
Point(int x, int y) : x(x), y(y) {}
friend Point operator-(Point p1, Point p2) {
return Point(p1.x - p2.x, p1.y - p2.y);
}
friend bool operator<(Point p1, Point p2) {
return (p1.y == p2.y ? p1.x < p2.x : p1.y < p2.y);
}
};
double norma(Point v) { return sqrt(1LL * v.x * v.x + 1LL * v.y * v.y); }
double cross(Point v1, Point v2) {
return 1LL * v1.x * v2.y - 1LL * v2.x * v1.y;
}
vector<Point> ConvexHull(vector<Point> P) {
if (P.size() < 3) return P;
Point p = *(min_element(P.begin(), P.end()));
sort(P.begin(), P.end(), [p](Point p1, Point p2) {
double area = cross(p1 - p, p2 - p);
if (area == 0) return norma(p1 - p) < norma(p2 - p);
return area > 0;
});
vector<Point> C;
C.push_back(P[0]);
C.push_back(P[1]);
int it = 2, szC = 2;
while (it < P.size()) {
if (szC > 1 && cross(P[it] - C[szC - 1], C[szC - 2] - C[szC - 1]) <= 0) {
C.pop_back();
szC--;
} else {
C.push_back(P[it]);
szC++;
it++;
}
}
return C;
}
struct data {
double a, b, c;
friend bool operator==(data d1, data d2) {
bool eq1 = abs(d1.a - d2.a) <= EPS;
bool eq2 = abs(d1.b - d2.b) <= EPS;
bool eq3 = abs(d1.c - d2.c) <= EPS;
return eq1 && eq2 && eq3;
}
friend bool operator!=(data d1, data d2) { return !(d1 == d2); }
};
vector<int> F;
void buildF(vector<data> &P) {
F = vector<int>(P.size(), 0);
for (int itP = 1, len = 0; itP < P.size(); itP++) {
for (; len != -1 && P[itP] != P[len]; len = len ? F[len - 1] : -1)
;
F[itP] = ++len;
}
}
bool KMP(vector<data> &P, vector<data> &T) {
buildF(P);
for (int itT = 0, itP = 0; itT < T.size(); itT++) {
for (; itP != -1 && T[itT] != P[itP]; itP = itP ? F[itP - 1] : -1)
;
if (P.size() == ++itP) {
return true;
}
}
return false;
}
vector<data> getData(vector<Point> &P) {
int sz = P.size();
vector<data> Data;
for (int i = 0; i < sz; i++) {
Data.push_back({norma(P[i] - P[(i + 1) % sz]),
norma(P[(i + 1) % sz] - P[(i + 2) % sz]),
norma(P[(i + 2) % sz] - P[i])});
}
return Data;
}
bool equals(vector<Point> v1, vector<Point> v2) {
if (v1.size() != v2.size()) return false;
if (v1.size() == 2) {
return norma(v1[0] - v1[1]) == norma(v2[0] - v2[1]);
}
for (int i = 0; i < v1.size(); i++) {
v2.push_back(v2[i]);
}
vector<data> P = getData(v1);
vector<data> T = getData(v2);
return KMP(P, T);
}
int main() {
int n, m;
scanf("%d %d", &n, &m);
vector<Point> P1(n);
for (int i = 0; i < n; i++) {
scanf("%d %d", &P1[i].x, &P1[i].y);
}
vector<Point> P2(m);
for (int i = 0; i < m; i++) {
scanf("%d %d", &P2[i].x, &P2[i].y);
}
vector<Point> v1 = ConvexHull(P1);
vector<Point> v2 = ConvexHull(P2);
if (equals(v1, v2))
cout << "YES";
else
cout << "NO";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
struct tpoint {
long long x, y;
tpoint() {}
tpoint(long long _x, long long _y) : x(_x), y(_y) {}
} p1[N], hu1[N], p2[N], hu2[N];
tpoint operator-(tpoint a, tpoint b) { return tpoint(a.x - b.x, a.y - b.y); }
long long cross(tpoint a, tpoint b) { return a.x * b.y - b.x * a.y; }
bool operator<(const tpoint &a, const tpoint &b) {
return a.x < b.x || (a.x == b.x && a.y < b.y);
}
int andrew(int n, tpoint hull[], tpoint p[]) {
sort(p, p + n);
int m = 0, k;
for (int i = 0; i < (n); ++i) {
while (m > 1 && cross(hull[m - 1] - hull[m - 2], p[i] - hull[m - 2]) <= 0)
m--;
hull[m++] = p[i];
}
k = m;
for (int i = n - 1; i >= 0; --i) {
while (m > k && cross(hull[m - 1] - hull[m - 2], p[i] - hull[m - 2]) <= 0)
m--;
hull[m++] = p[i];
}
if (n > 1) m--;
return m;
}
long long dot(tpoint a, tpoint b, tpoint c) {
return (b.x - a.x) * (c.x - a.x) + (b.y - a.y) * (c.y - a.y);
}
long long dis(tpoint a, tpoint b) {
long long xx = (a.x - b.x);
long long yy = (a.y - b.y);
return xx * xx + yy * yy;
}
long long a1[N * 6], a2[N * 6];
long long di1[N * 2], di2[N * 2];
int nex[N * 2];
void cal_next(long long s[], int len) {
nex[0] = -1;
int k = -1;
for (int q = 1; q < (len); ++q) {
while (k > -1 && s[k + 1] != s[q]) k = nex[k];
if (s[k + 1] == s[q]) k++;
nex[q] = k;
}
}
int kmp(long long *str, int slen, long long *ptr, int plen) {
memset(nex, 0, sizeof(nex));
cal_next(ptr, plen);
int k = -1;
for (int i = 0; i < (slen); ++i) {
while (k > -1 && ptr[k + 1] != str[i]) k = nex[k];
if (ptr[k + 1] == str[i]) k++;
if (k == plen - 1) {
return 1;
}
}
return 0;
}
int main() {
int n, m, k1, k2;
scanf("%d%d", &n, &m);
for (int i = 0; i < (n); ++i) {
scanf("%lld%lld", &p1[i].x, &p1[i].y);
}
k1 = andrew(n, hu1, p1);
for (int i = 0; i < (m); ++i) {
scanf("%lld%lld", &p2[i].x, &p2[i].y);
}
k2 = andrew(m, hu2, p2);
if (k1 != k2) {
cout << "NO" << endl;
return 0;
}
int top1 = 0, top2 = 0;
for (int i = 0; i < (k1); ++i) {
a1[top1++] = dis(hu1[(i - 1 + k1) % k1], hu1[i]);
a1[top1++] = dot(hu1[i], hu1[(i - 1 + k1) % k1], hu1[(i + 1) % k1]);
a1[top1++] = dis(hu1[(i + 1) % k1], hu1[i]);
}
for (int i = 0; i < (k2); ++i) {
a2[top2++] = dis(hu2[(i - 1 + k2) % k2], hu2[i]);
a2[top2++] = dot(hu2[i], hu2[(i - 1 + k2) % k2], hu2[(i + 1) % k2]);
a2[top2++] = dis(hu2[(i + 1) % k2], hu2[i]);
}
for (int i = top1; i < (top1 * 2); ++i) a1[i] = a1[i - top1];
int fg1 = kmp(a1, 2 * top1, a2, top2);
if (fg1) {
cout << "YES" << endl;
} else
cout << "NO" << endl;
return 0;
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
struct tpoint {
long long x, y;
tpoint() {}
tpoint(long long _x, long long _y) : x(_x), y(_y) {}
} p1[N], hu1[N], p2[N], hu2[N];
tpoint operator-(tpoint a, tpoint b) { return tpoint(a.x - b.x, a.y - b.y); }
long long cross(tpoint a, tpoint b) { return a.x * b.y - b.x * a.y; }
bool operator<(const tpoint &a, const tpoint &b) {
return a.x < b.x || (a.x == b.x && a.y < b.y);
}
int andrew(int n, tpoint hull[], tpoint p[]) {
sort(p, p + n);
int m = 0, k;
for (int i = 0; i < (n); ++i) {
while (m > 1 && cross(hull[m - 1] - hull[m - 2], p[i] - hull[m - 2]) <= 0)
m--;
hull[m++] = p[i];
}
k = m;
for (int i = n - 1; i >= 0; --i) {
while (m > k && cross(hull[m - 1] - hull[m - 2], p[i] - hull[m - 2]) <= 0)
m--;
hull[m++] = p[i];
}
if (n > 1) m--;
return m;
}
long long dot(tpoint a, tpoint b, tpoint c) {
return (b.x - a.x) * (c.x - a.x) + (b.y - a.y) * (c.y - a.y);
}
long long dis(tpoint a, tpoint b) {
long long xx = (a.x - b.x);
long long yy = (a.y - b.y);
return xx * xx + yy * yy;
}
long long a1[N * 6], a2[N * 6];
long long di1[N * 2], di2[N * 2];
int nex[N * 2];
void cal_next(long long s[], int len) {
nex[0] = -1;
int k = -1;
for (int q = 1; q < (len); ++q) {
while (k > -1 && s[k + 1] != s[q]) k = nex[k];
if (s[k + 1] == s[q]) k++;
nex[q] = k;
}
}
int kmp(long long *str, int slen, long long *ptr, int plen) {
memset(nex, 0, sizeof(nex));
cal_next(ptr, plen);
int k = -1;
for (int i = 0; i < (slen); ++i) {
while (k > -1 && ptr[k + 1] != str[i]) k = nex[k];
if (ptr[k + 1] == str[i]) k++;
if (k == plen - 1) {
return 1;
}
}
return 0;
}
int main() {
int n, m, k1, k2;
scanf("%d%d", &n, &m);
for (int i = 0; i < (n); ++i) {
scanf("%lld%lld", &p1[i].x, &p1[i].y);
}
k1 = andrew(n, hu1, p1);
for (int i = 0; i < (m); ++i) {
scanf("%lld%lld", &p2[i].x, &p2[i].y);
}
k2 = andrew(m, hu2, p2);
if (k1 != k2) {
cout << "NO" << endl;
return 0;
}
int top1 = 0, top2 = 0;
for (int i = 0; i < (k1); ++i) {
a1[top1++] = dis(hu1[(i - 1 + k1) % k1], hu1[i]);
a1[top1++] = dot(hu1[i], hu1[(i - 1 + k1) % k1], hu1[(i + 1) % k1]);
a1[top1++] = dis(hu1[(i + 1) % k1], hu1[i]);
}
for (int i = 0; i < (k2); ++i) {
a2[top2++] = dis(hu2[(i - 1 + k2) % k2], hu2[i]);
a2[top2++] = dot(hu2[i], hu2[(i - 1 + k2) % k2], hu2[(i + 1) % k2]);
a2[top2++] = dis(hu2[(i + 1) % k2], hu2[i]);
}
for (int i = top1; i < (top1 * 2); ++i) a1[i] = a1[i - top1];
int fg1 = kmp(a1, 2 * top1, a2, top2);
if (fg1) {
cout << "YES" << endl;
} else
cout << "NO" << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int max_n = 100111, inf = 1000111222;
struct point {
int x, y;
point() { x = y = 0; }
point(int x, int y) : x(x), y(y) {}
void read() { scanf("%d%d", &x, &y); }
void write(string s = "") { printf("%d %d%s", x, y, s.c_str()); }
void write_point(string s = "") { printf("(%d %d)%s", x, y, s.c_str()); }
bool operator<(const point &a) const {
return x < a.x || x == a.x && y < a.y;
}
point operator-(const point &a) const { return point(x - a.x, y - a.y); }
};
long long vect_pr(point a, point b) {
return 1LL * a.x * b.y - 1LL * a.y * b.x;
}
bool cv(point a, point b, point c) { return vect_pr(b - a, c - b) < 0; }
bool ccv(point a, point b, point c) { return vect_pr(b - a, c - b) > 0; }
vector<point> convex_hull(int n, point p[]) {
sort(p, p + n);
vector<point> res, up, down;
if (n == 1) {
res.push_back(p[0]);
res.push_back(p[0]);
return res;
}
for (int i = 0; i < n; ++i) {
if (cv(p[0], p[i], p[n - 1]) || i == 0 || i + 1 == n) {
while (up.size() > 1 && !cv(up[up.size() - 2], up[up.size() - 1], p[i])) {
up.pop_back();
}
up.push_back(p[i]);
}
}
for (int i = 0; i < n; ++i) {
if (ccv(p[0], p[i], p[n - 1]) || i == 0 || i + 1 == n) {
while (down.size() > 1 &&
!ccv(down[down.size() - 2], down[down.size() - 1], p[i])) {
down.pop_back();
}
down.push_back(p[i]);
}
}
down.pop_back();
reverse(down.begin(), down.end());
down.pop_back();
res = up;
for (int i = 0; i < down.size(); ++i) {
res.push_back(down[i]);
}
return res;
}
int n[2];
point p[max_n];
vector<point> v[2];
vector<pair<long long, long long>> a, b;
long long dist(const point &p) { return 1LL * p.x * p.x + 1LL * p.y * p.y; }
vector<pair<long long, long long>> get(const vector<point> &v) {
vector<pair<long long, long long>> res;
for (int i = 0; i < v.size(); ++i) {
res.push_back({dist(v[i] - v[(i + 1) % v.size()]),
vect_pr(v[(i + 2) % v.size()] - v[(i + 1) % v.size()],
v[(i + 1) % v.size()] - v[i])});
}
return res;
}
int pr[3 * max_n];
void get_prefix_function(const vector<pair<long long, long long>> &s, int p[]) {
p[0] = 0;
for (int i = 1; i < s.size(); ++i) {
p[i] = p[i - 1];
while (p[i] > 0 && s[i] != s[p[i]]) {
p[i] = p[p[i] - 1];
}
if (s[i] == s[p[i]]) {
++p[i];
}
}
}
int main() {
scanf("%d%d", &n[0], &n[1]);
for (int i = 0; i < 2; ++i) {
for (int j = 0; j < n[i]; ++j) {
scanf("%d%d", &p[j].x, &p[j].y);
}
v[i] = convex_hull(n[i], p);
}
if (v[0].size() != v[1].size()) {
puts("NO");
return 0;
}
a = get(v[0]);
b = get(v[1]);
a.push_back({-1, -1});
for (int i = 0; i < 2 * b.size(); ++i) {
a.push_back(b[i % b.size()]);
}
get_prefix_function(a, pr);
for (int i = b.size() + 1; i < a.size(); ++i) {
if (pr[i] == b.size()) {
puts("YES");
return 0;
}
}
puts("NO");
return 0;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int max_n = 100111, inf = 1000111222;
struct point {
int x, y;
point() { x = y = 0; }
point(int x, int y) : x(x), y(y) {}
void read() { scanf("%d%d", &x, &y); }
void write(string s = "") { printf("%d %d%s", x, y, s.c_str()); }
void write_point(string s = "") { printf("(%d %d)%s", x, y, s.c_str()); }
bool operator<(const point &a) const {
return x < a.x || x == a.x && y < a.y;
}
point operator-(const point &a) const { return point(x - a.x, y - a.y); }
};
long long vect_pr(point a, point b) {
return 1LL * a.x * b.y - 1LL * a.y * b.x;
}
bool cv(point a, point b, point c) { return vect_pr(b - a, c - b) < 0; }
bool ccv(point a, point b, point c) { return vect_pr(b - a, c - b) > 0; }
vector<point> convex_hull(int n, point p[]) {
sort(p, p + n);
vector<point> res, up, down;
if (n == 1) {
res.push_back(p[0]);
res.push_back(p[0]);
return res;
}
for (int i = 0; i < n; ++i) {
if (cv(p[0], p[i], p[n - 1]) || i == 0 || i + 1 == n) {
while (up.size() > 1 && !cv(up[up.size() - 2], up[up.size() - 1], p[i])) {
up.pop_back();
}
up.push_back(p[i]);
}
}
for (int i = 0; i < n; ++i) {
if (ccv(p[0], p[i], p[n - 1]) || i == 0 || i + 1 == n) {
while (down.size() > 1 &&
!ccv(down[down.size() - 2], down[down.size() - 1], p[i])) {
down.pop_back();
}
down.push_back(p[i]);
}
}
down.pop_back();
reverse(down.begin(), down.end());
down.pop_back();
res = up;
for (int i = 0; i < down.size(); ++i) {
res.push_back(down[i]);
}
return res;
}
int n[2];
point p[max_n];
vector<point> v[2];
vector<pair<long long, long long>> a, b;
long long dist(const point &p) { return 1LL * p.x * p.x + 1LL * p.y * p.y; }
vector<pair<long long, long long>> get(const vector<point> &v) {
vector<pair<long long, long long>> res;
for (int i = 0; i < v.size(); ++i) {
res.push_back({dist(v[i] - v[(i + 1) % v.size()]),
vect_pr(v[(i + 2) % v.size()] - v[(i + 1) % v.size()],
v[(i + 1) % v.size()] - v[i])});
}
return res;
}
int pr[3 * max_n];
void get_prefix_function(const vector<pair<long long, long long>> &s, int p[]) {
p[0] = 0;
for (int i = 1; i < s.size(); ++i) {
p[i] = p[i - 1];
while (p[i] > 0 && s[i] != s[p[i]]) {
p[i] = p[p[i] - 1];
}
if (s[i] == s[p[i]]) {
++p[i];
}
}
}
int main() {
scanf("%d%d", &n[0], &n[1]);
for (int i = 0; i < 2; ++i) {
for (int j = 0; j < n[i]; ++j) {
scanf("%d%d", &p[j].x, &p[j].y);
}
v[i] = convex_hull(n[i], p);
}
if (v[0].size() != v[1].size()) {
puts("NO");
return 0;
}
a = get(v[0]);
b = get(v[1]);
a.push_back({-1, -1});
for (int i = 0; i < 2 * b.size(); ++i) {
a.push_back(b[i % b.size()]);
}
get_prefix_function(a, pr);
for (int i = b.size() + 1; i < a.size(); ++i) {
if (pr[i] == b.size()) {
puts("YES");
return 0;
}
}
puts("NO");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 600005;
long long read() {
long long x = 0;
char ch = getchar();
while (!isdigit(ch)) ch = getchar();
while (isdigit(ch)) x = (x << 1) + (x << 3) + ch - 48, ch = getchar();
return x;
}
int n, m;
struct Point {
int x, y;
} p1[N], p2[N], O;
long long sqr(int x) { return 1LL * x * x; }
long long dis(Point a, Point b) { return sqr(a.x - b.x) + sqr(a.y - b.y); }
long long cross(Point a, Point b, Point c) {
return 1LL * (b.x - a.x) * (c.y - a.y) - 1LL * (c.x - a.x) * (b.y - a.y);
}
bool cmp_O(Point a, Point b) {
if (a.y == b.y) return a.x < b.x;
return a.y < b.y;
}
bool cmp_Angle(Point a, Point b) {
long long c = cross(O, a, b);
if (c == 0) return dis(O, a) < dis(O, b);
return c > 0;
}
int st[N], top;
int Make_Convex(Point P[], int n) {
for (int i = 2; i <= n; i++)
if (!cmp_O(P[1], P[i])) swap(P[1], P[i]);
O = P[1];
sort(P + 2, P + n + 1, cmp_Angle);
top = 0;
st[++top] = 1, st[++top] = 2;
for (int i = 3; i <= n; i++) {
while (top >= 2 && cross(P[st[top - 1]], P[st[top]], P[i]) <= 0) top--;
st[++top] = i;
}
for (int i = 1; i <= top; i++) P[i] = P[st[i]];
return top;
}
long long s1[N], s2[N];
int Fail[N];
void KMP(long long s[], int n) {
Fail[0] = Fail[1] = 0;
for (int i = 2; i <= n; i++) {
int k = Fail[i - 1];
while (k > 0 && s[i] != s[k + 1]) k = Fail[k];
Fail[i] = k + (s[k + 1] == s[i] ? 1 : 0);
}
}
bool check() {
int k = 0;
for (int i = 1; i <= n * 6; i++) {
while (k > 0 && s2[k + 1] != s1[i]) k = Fail[k];
if (s2[k + 1] == s1[i]) k++;
if (k >= m * 3) {
if (i % 3 == 0)
return 1;
else
k = Fail[k];
}
}
return 0;
}
int main() {
n = read(), m = read();
for (int i = 1; i <= n; i++) p1[i].x = read(), p1[i].y = read();
for (int i = 1; i <= m; i++) p2[i].x = read(), p2[i].y = read();
n = Make_Convex(p1, n);
m = Make_Convex(p2, m);
for (int i = 1; i <= n; i++) {
s1[i * 3 - 2] = dis(p1[i], p1[i % n + 1]);
s1[i * 3 - 1] = cross(p1[i], p1[i % n + 1], p1[(i + 1) % n + 1]);
s1[i * 3] = dis(p1[i], p1[(i + 1) % n + 1]);
}
for (int i = 1; i <= m; i++) {
s2[i * 3 - 2] = dis(p2[i], p2[i % m + 1]);
s2[i * 3 - 1] = cross(p2[i], p2[i % m + 1], p2[(i + 1) % m + 1]);
s2[i * 3] = dis(p2[i], p2[(i + 1) % m + 1]);
}
for (int i = 1; i <= n * 3; i++) s1[i + n * 3] = s1[i];
KMP(s2, m * 3);
puts(n == m && check() ? "YES" : "NO");
return 0;
}
| ### Prompt
Generate a cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 600005;
long long read() {
long long x = 0;
char ch = getchar();
while (!isdigit(ch)) ch = getchar();
while (isdigit(ch)) x = (x << 1) + (x << 3) + ch - 48, ch = getchar();
return x;
}
int n, m;
struct Point {
int x, y;
} p1[N], p2[N], O;
long long sqr(int x) { return 1LL * x * x; }
long long dis(Point a, Point b) { return sqr(a.x - b.x) + sqr(a.y - b.y); }
long long cross(Point a, Point b, Point c) {
return 1LL * (b.x - a.x) * (c.y - a.y) - 1LL * (c.x - a.x) * (b.y - a.y);
}
bool cmp_O(Point a, Point b) {
if (a.y == b.y) return a.x < b.x;
return a.y < b.y;
}
bool cmp_Angle(Point a, Point b) {
long long c = cross(O, a, b);
if (c == 0) return dis(O, a) < dis(O, b);
return c > 0;
}
int st[N], top;
int Make_Convex(Point P[], int n) {
for (int i = 2; i <= n; i++)
if (!cmp_O(P[1], P[i])) swap(P[1], P[i]);
O = P[1];
sort(P + 2, P + n + 1, cmp_Angle);
top = 0;
st[++top] = 1, st[++top] = 2;
for (int i = 3; i <= n; i++) {
while (top >= 2 && cross(P[st[top - 1]], P[st[top]], P[i]) <= 0) top--;
st[++top] = i;
}
for (int i = 1; i <= top; i++) P[i] = P[st[i]];
return top;
}
long long s1[N], s2[N];
int Fail[N];
void KMP(long long s[], int n) {
Fail[0] = Fail[1] = 0;
for (int i = 2; i <= n; i++) {
int k = Fail[i - 1];
while (k > 0 && s[i] != s[k + 1]) k = Fail[k];
Fail[i] = k + (s[k + 1] == s[i] ? 1 : 0);
}
}
bool check() {
int k = 0;
for (int i = 1; i <= n * 6; i++) {
while (k > 0 && s2[k + 1] != s1[i]) k = Fail[k];
if (s2[k + 1] == s1[i]) k++;
if (k >= m * 3) {
if (i % 3 == 0)
return 1;
else
k = Fail[k];
}
}
return 0;
}
int main() {
n = read(), m = read();
for (int i = 1; i <= n; i++) p1[i].x = read(), p1[i].y = read();
for (int i = 1; i <= m; i++) p2[i].x = read(), p2[i].y = read();
n = Make_Convex(p1, n);
m = Make_Convex(p2, m);
for (int i = 1; i <= n; i++) {
s1[i * 3 - 2] = dis(p1[i], p1[i % n + 1]);
s1[i * 3 - 1] = cross(p1[i], p1[i % n + 1], p1[(i + 1) % n + 1]);
s1[i * 3] = dis(p1[i], p1[(i + 1) % n + 1]);
}
for (int i = 1; i <= m; i++) {
s2[i * 3 - 2] = dis(p2[i], p2[i % m + 1]);
s2[i * 3 - 1] = cross(p2[i], p2[i % m + 1], p2[(i + 1) % m + 1]);
s2[i * 3] = dis(p2[i], p2[(i + 1) % m + 1]);
}
for (int i = 1; i <= n * 3; i++) s1[i + n * 3] = s1[i];
KMP(s2, m * 3);
puts(n == m && check() ? "YES" : "NO");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int N, M;
vector<pair<signed long long, signed long long>> A, B;
vector<pair<long double, long double>> C, D;
const signed long long EPS = 0;
template <class C>
C veccross(pair<C, C> p1, pair<C, C> p2, pair<C, C> p3) {
p3.first -= p1.first;
p2.first -= p1.first;
p3.second -= p1.second;
p2.second -= p1.second;
return p3.first * p2.second - p2.first * p3.second;
}
template <class C>
vector<int> convex_hull(vector<pair<C, C>>& vp) {
vector<pair<pair<C, C>, int>> sorted;
vector<int> res;
int i, k = 0, rb;
if (vp.size() <= 2) {
if (vp.size() >= 1) res.push_back(0);
if (vp.size() >= 2 && vp[0] != vp[1]) res.push_back(1);
return res;
}
for (i = 0; i < (vp.size()); i++) sorted.push_back(make_pair(vp[i], i));
sort(sorted.begin(), sorted.end());
res.resize(vp.size() * 2);
for (i = 0; i < (vp.size()); i++) {
while (k > 1 &&
veccross(vp[res[k - 2]], vp[res[k - 1]], sorted[i].first) <= -EPS)
k--;
res[k++] = sorted[i].second;
}
for (rb = k, i = vp.size() - 2; i >= 0; i--) {
while (k > rb &&
veccross(vp[res[k - 2]], vp[res[k - 1]], sorted[i].first) <= -EPS)
k--;
res[k++] = sorted[i].second;
}
res.resize(k - 1);
return res;
}
pair<long double, long double> center(
vector<pair<long double, long double>> V) {
if (V.size() == 1) return V[0];
if (V.size() == 2)
return {(V[0].first + V[1].first) / 2, (V[0].second + V[1].second) / 2};
long double dx = 0, dy = 0, A = 0;
int i;
for (i = 1; i < V.size() - 1; i++) {
long double cx = (V[0].first + V[i].first + V[i + 1].first) / 3;
long double cy = (V[0].second + V[i].second + V[i + 1].second) / 3;
long double cr =
abs((V[i].first - V[0].first) * (V[i + 1].second - V[0].second) -
(V[i + 1].first - V[0].first) * (V[i].second - V[0].second));
A += cr;
dx += cx * cr;
dy += cy * cr;
}
assert(A);
return {dx / A, dy / A};
}
vector<pair<long double, long double>> hoge(
vector<pair<signed long long, signed long long>> A) {
vector<pair<long double, long double>> C;
auto v = convex_hull(A);
long double dx = 0, dy = 0;
for (auto& c : v) {
C.push_back({A[c].first, A[c].second});
}
auto ce = center(C);
for (auto& c : C) c.first -= ce.first, c.second -= ce.second;
return C;
}
void solve() {
int i, j, k, l, r, x, y;
string s;
cin >> N >> M;
for (i = 0; i < (N); i++) {
cin >> x >> y;
A.push_back({x, y});
}
for (i = 0; i < (M); i++) {
cin >> x >> y;
B.push_back({x, y});
}
C = hoge(A);
D = hoge(B);
N = C.size();
M = D.size();
if (N != M) return (void)printf("NO\n");
for (i = 0; i < (N); i++)
if (abs(hypot(C[i].first, C[i].second) - hypot(D[0].first, D[0].second)) <
1e-2) {
long double deg =
atan2(D[0].second, D[0].first) - atan2(C[i].second, C[i].first);
for (j = 0; j < (N); j++) {
long double dx = C[j].first * cos(deg) - C[j].second * sin(deg);
long double dy = C[j].first * sin(deg) + C[j].second * cos(deg);
if (abs(dx - D[(j + N - i) % N].first) +
abs(dy - D[(j + N - i) % N].second) >
1e-2)
break;
}
if (j == N) return (void)printf("YES\n");
}
(void)printf("NO\n");
}
int main(int argc, char** argv) {
string s;
int i;
if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
cout.tie(0);
solve();
return 0;
}
| ### Prompt
Please create a solution in cpp to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int N, M;
vector<pair<signed long long, signed long long>> A, B;
vector<pair<long double, long double>> C, D;
const signed long long EPS = 0;
template <class C>
C veccross(pair<C, C> p1, pair<C, C> p2, pair<C, C> p3) {
p3.first -= p1.first;
p2.first -= p1.first;
p3.second -= p1.second;
p2.second -= p1.second;
return p3.first * p2.second - p2.first * p3.second;
}
template <class C>
vector<int> convex_hull(vector<pair<C, C>>& vp) {
vector<pair<pair<C, C>, int>> sorted;
vector<int> res;
int i, k = 0, rb;
if (vp.size() <= 2) {
if (vp.size() >= 1) res.push_back(0);
if (vp.size() >= 2 && vp[0] != vp[1]) res.push_back(1);
return res;
}
for (i = 0; i < (vp.size()); i++) sorted.push_back(make_pair(vp[i], i));
sort(sorted.begin(), sorted.end());
res.resize(vp.size() * 2);
for (i = 0; i < (vp.size()); i++) {
while (k > 1 &&
veccross(vp[res[k - 2]], vp[res[k - 1]], sorted[i].first) <= -EPS)
k--;
res[k++] = sorted[i].second;
}
for (rb = k, i = vp.size() - 2; i >= 0; i--) {
while (k > rb &&
veccross(vp[res[k - 2]], vp[res[k - 1]], sorted[i].first) <= -EPS)
k--;
res[k++] = sorted[i].second;
}
res.resize(k - 1);
return res;
}
pair<long double, long double> center(
vector<pair<long double, long double>> V) {
if (V.size() == 1) return V[0];
if (V.size() == 2)
return {(V[0].first + V[1].first) / 2, (V[0].second + V[1].second) / 2};
long double dx = 0, dy = 0, A = 0;
int i;
for (i = 1; i < V.size() - 1; i++) {
long double cx = (V[0].first + V[i].first + V[i + 1].first) / 3;
long double cy = (V[0].second + V[i].second + V[i + 1].second) / 3;
long double cr =
abs((V[i].first - V[0].first) * (V[i + 1].second - V[0].second) -
(V[i + 1].first - V[0].first) * (V[i].second - V[0].second));
A += cr;
dx += cx * cr;
dy += cy * cr;
}
assert(A);
return {dx / A, dy / A};
}
vector<pair<long double, long double>> hoge(
vector<pair<signed long long, signed long long>> A) {
vector<pair<long double, long double>> C;
auto v = convex_hull(A);
long double dx = 0, dy = 0;
for (auto& c : v) {
C.push_back({A[c].first, A[c].second});
}
auto ce = center(C);
for (auto& c : C) c.first -= ce.first, c.second -= ce.second;
return C;
}
void solve() {
int i, j, k, l, r, x, y;
string s;
cin >> N >> M;
for (i = 0; i < (N); i++) {
cin >> x >> y;
A.push_back({x, y});
}
for (i = 0; i < (M); i++) {
cin >> x >> y;
B.push_back({x, y});
}
C = hoge(A);
D = hoge(B);
N = C.size();
M = D.size();
if (N != M) return (void)printf("NO\n");
for (i = 0; i < (N); i++)
if (abs(hypot(C[i].first, C[i].second) - hypot(D[0].first, D[0].second)) <
1e-2) {
long double deg =
atan2(D[0].second, D[0].first) - atan2(C[i].second, C[i].first);
for (j = 0; j < (N); j++) {
long double dx = C[j].first * cos(deg) - C[j].second * sin(deg);
long double dy = C[j].first * sin(deg) + C[j].second * cos(deg);
if (abs(dx - D[(j + N - i) % N].first) +
abs(dy - D[(j + N - i) % N].second) >
1e-2)
break;
}
if (j == N) return (void)printf("YES\n");
}
(void)printf("NO\n");
}
int main(int argc, char** argv) {
string s;
int i;
if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
cout.tie(0);
solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 500002;
const long long MOD = 1500000001;
const long long p = 239017;
struct Point {
long long x, y;
Point() {}
Point(const Point& p1, const Point& p2) : x(p2.x - p1.x), y(p2.y - p1.y) {}
Point(long long x, long long y) : x(x), y(y) {}
bool operator<(const Point& other) const {
return make_pair(x, y) < make_pair(other.x, other.y);
}
bool operator==(const Point& other) const {
return make_pair(x, y) == make_pair(other.x, other.y);
}
};
Point a[MAXN];
Point minP;
long long powP[MAXN];
long long cross(const Point& p1, const Point& p2) {
return p1.x * p2.y - p1.y * p2.x;
}
long long dot(const Point& p1, const Point& p2) {
return p1.x * p2.x + p1.y * p2.y;
}
long long len(const Point& p) { return p.x * p.x + p.y * p.y; }
bool cmp(Point& x, Point& y) {
Point v1(minP, x);
Point v2(minP, y);
long long cp = cross(v1, v2);
if (cp == 0) {
return len(v1) < len(v2);
} else {
return cp < 0;
}
}
void getCh(int n, vector<Point>& ch) {
for (int i = 0; i < (int)(n); i++) {
long long x, y;
cin >> x >> y;
a[i] = Point(x, y);
}
minP = a[0];
for (int i = 0; i < (int)(n); i++) {
if (a[i] < minP) {
minP = a[i];
}
}
sort(a, a + n, cmp);
a[n] = a[0];
for (int i = 0; i < (int)(n + 1); i++) {
while (ch.size() > 1 &&
cross(Point(ch[ch.size() - 1], ch[ch.size() - 2]),
Point(ch[ch.size() - 1], a[i])) <= 0 &&
(i != n || ch.size() > 2)) {
ch.pop_back();
}
ch.push_back(a[i]);
}
if (ch.size() > 1 && ch[0] == ch[ch.size() - 1]) {
ch.pop_back();
}
}
int main() {
cin.sync_with_stdio(0);
cin.tie(0);
int n, m;
cin >> n >> m;
vector<Point> ch1;
getCh(n, ch1);
vector<Point> ch2;
getCh(m, ch2);
if (ch1.size() != ch2.size()) {
cout << "NO";
exit(0);
}
if (ch1.size() == 1) {
cout << "YES";
exit(0);
}
ch1.push_back(ch1[0]);
ch1.push_back(ch1[1]);
ch2.push_back(ch2[0]);
ch2.push_back(ch2[1]);
vector<long long> s1;
vector<long long> s2;
for (int i = 0; i < (int)(ch1.size() - 2); i++) {
s1.push_back(len(Point(ch1[i], ch1[i + 1])));
s1.push_back(
cross(Point(ch1[i + 1], ch1[i]), Point(ch1[i + 1], ch1[i + 2])));
s1.push_back(dot(Point(ch1[i + 1], ch1[i]), Point(ch1[i + 1], ch1[i + 2])));
s2.push_back(len(Point(ch2[i], ch2[i + 1])));
s2.push_back(
cross(Point(ch2[i + 1], ch2[i]), Point(ch2[i + 1], ch2[i + 2])));
s2.push_back(dot(Point(ch2[i + 1], ch2[i]), Point(ch2[i + 1], ch2[i + 2])));
}
long long h = 0;
for (int i = 0; i < (int)(s1.size()); i++) {
s1[i] %= MOD;
h = (h * p + s1[i]) % MOD;
}
long long h1 = 0;
for (int i = 0; i < (int)(s1.size()); i++) {
s2[i] %= MOD;
h1 = (h1 * p + s2[i]) % MOD;
}
powP[0] = 1;
for (int i = (int)1; i <= (int)(MAXN - 1); i++) {
powP[i] = (powP[i - 1] * p) % MOD;
}
for (int i = 0; i < (int)(s1.size() + 1); i++) {
if (h1 == h) {
cout << "YES";
exit(0);
}
h1 = (h1 * p + s2[(i + s1.size()) % s1.size()] -
(s2[i % s1.size()] * powP[s1.size()]) % MOD + MOD) %
MOD;
}
cout << "NO";
return 0;
}
| ### Prompt
Generate a Cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 500002;
const long long MOD = 1500000001;
const long long p = 239017;
struct Point {
long long x, y;
Point() {}
Point(const Point& p1, const Point& p2) : x(p2.x - p1.x), y(p2.y - p1.y) {}
Point(long long x, long long y) : x(x), y(y) {}
bool operator<(const Point& other) const {
return make_pair(x, y) < make_pair(other.x, other.y);
}
bool operator==(const Point& other) const {
return make_pair(x, y) == make_pair(other.x, other.y);
}
};
Point a[MAXN];
Point minP;
long long powP[MAXN];
long long cross(const Point& p1, const Point& p2) {
return p1.x * p2.y - p1.y * p2.x;
}
long long dot(const Point& p1, const Point& p2) {
return p1.x * p2.x + p1.y * p2.y;
}
long long len(const Point& p) { return p.x * p.x + p.y * p.y; }
bool cmp(Point& x, Point& y) {
Point v1(minP, x);
Point v2(minP, y);
long long cp = cross(v1, v2);
if (cp == 0) {
return len(v1) < len(v2);
} else {
return cp < 0;
}
}
void getCh(int n, vector<Point>& ch) {
for (int i = 0; i < (int)(n); i++) {
long long x, y;
cin >> x >> y;
a[i] = Point(x, y);
}
minP = a[0];
for (int i = 0; i < (int)(n); i++) {
if (a[i] < minP) {
minP = a[i];
}
}
sort(a, a + n, cmp);
a[n] = a[0];
for (int i = 0; i < (int)(n + 1); i++) {
while (ch.size() > 1 &&
cross(Point(ch[ch.size() - 1], ch[ch.size() - 2]),
Point(ch[ch.size() - 1], a[i])) <= 0 &&
(i != n || ch.size() > 2)) {
ch.pop_back();
}
ch.push_back(a[i]);
}
if (ch.size() > 1 && ch[0] == ch[ch.size() - 1]) {
ch.pop_back();
}
}
int main() {
cin.sync_with_stdio(0);
cin.tie(0);
int n, m;
cin >> n >> m;
vector<Point> ch1;
getCh(n, ch1);
vector<Point> ch2;
getCh(m, ch2);
if (ch1.size() != ch2.size()) {
cout << "NO";
exit(0);
}
if (ch1.size() == 1) {
cout << "YES";
exit(0);
}
ch1.push_back(ch1[0]);
ch1.push_back(ch1[1]);
ch2.push_back(ch2[0]);
ch2.push_back(ch2[1]);
vector<long long> s1;
vector<long long> s2;
for (int i = 0; i < (int)(ch1.size() - 2); i++) {
s1.push_back(len(Point(ch1[i], ch1[i + 1])));
s1.push_back(
cross(Point(ch1[i + 1], ch1[i]), Point(ch1[i + 1], ch1[i + 2])));
s1.push_back(dot(Point(ch1[i + 1], ch1[i]), Point(ch1[i + 1], ch1[i + 2])));
s2.push_back(len(Point(ch2[i], ch2[i + 1])));
s2.push_back(
cross(Point(ch2[i + 1], ch2[i]), Point(ch2[i + 1], ch2[i + 2])));
s2.push_back(dot(Point(ch2[i + 1], ch2[i]), Point(ch2[i + 1], ch2[i + 2])));
}
long long h = 0;
for (int i = 0; i < (int)(s1.size()); i++) {
s1[i] %= MOD;
h = (h * p + s1[i]) % MOD;
}
long long h1 = 0;
for (int i = 0; i < (int)(s1.size()); i++) {
s2[i] %= MOD;
h1 = (h1 * p + s2[i]) % MOD;
}
powP[0] = 1;
for (int i = (int)1; i <= (int)(MAXN - 1); i++) {
powP[i] = (powP[i - 1] * p) % MOD;
}
for (int i = 0; i < (int)(s1.size() + 1); i++) {
if (h1 == h) {
cout << "YES";
exit(0);
}
h1 = (h1 * p + s2[(i + s1.size()) % s1.size()] -
(s2[i % s1.size()] * powP[s1.size()]) % MOD + MOD) %
MOD;
}
cout << "NO";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long M = 998244353;
const long long NMAX = 200011;
const long long inf = 1E18;
const long double eps = 1e-9;
const long long base = 29;
struct pt {
long long x, y;
pt(long long x, long long y) {
this->x = x;
this->y = y;
}
pt() {}
};
pt a[NMAX];
pt b[NMAX];
long long scal(pt a, pt b) { return a.x * b.x + a.y * b.y; }
long long vect(pt a, pt b) { return a.x * b.y - a.y * b.x; }
pt operator-(pt a, pt b) { return pt(a.x - b.x, a.y - b.y); }
int cmp1(pt a, pt b) {
if (a.x == b.x) return a.y < b.y;
return a.x < b.x;
}
int cmp2(pt a, pt b) {
if (vect(a, b) == 0) return scal(a, a) < scal(b, b);
return vect(a, b) > 0;
}
int ch(pt a[], int n) {
sort(a, a + n, cmp1);
for (int i = 1; i < n; i++) a[i] = a[i] - a[0];
a[0] = a[0] - a[0];
sort(a + 1, a + n, cmp2);
int j = 1;
for (int i = 1; i < n; i++) {
while (j >= 2 && vect(a[j - 1] - a[j - 2], a[i] - a[j - 1]) <= 0) {
j--;
}
a[j] = a[i];
j++;
}
return j;
}
pair<pair<long long, long long>, pair<long long, long long> > u[NMAX], v[NMAX];
int sa[NMAX];
int sb[NMAX];
void init(pair<pair<long long, long long>, pair<long long, long long> > v[],
pt a[], int n) {
for (int i = 1; i < n - 1; i++) {
v[i].first.first = vect(a[i + 1] - a[i], a[i - 1] - a[i]);
v[i].first.second = scal(a[i + 1] - a[i], a[i - 1] - a[i]);
v[i].second.first = scal(a[i + 1] - a[i], a[i + 1] - a[i]);
v[i].second.second = scal(a[i - 1] - a[i], a[i - 1] - a[i]);
}
v[0].first.first = vect(a[1] - a[0], a[n - 1] - a[0]);
v[0].first.second = scal(a[1] - a[0], a[n - 1] - a[0]);
v[0].second.first = scal(a[1] - a[0], a[1] - a[0]);
v[0].second.second = scal(a[n - 1] - a[0], a[n - 1] - a[0]);
v[n - 1].first.first = vect(a[0] - a[n - 1], a[n - 2] - a[n - 1]);
v[n - 1].first.second = scal(a[0] - a[n - 1], a[n - 2] - a[n - 1]);
v[n - 1].second.first = scal(a[0] - a[n - 1], a[0] - a[n - 1]);
v[n - 1].second.second = scal(a[n - 2] - a[n - 1], a[n - 2] - a[n - 1]);
return;
}
signed main() {
cin.tie(0);
cout.tie(0);
cin.sync_with_stdio(0);
cout.sync_with_stdio(0);
cout.precision(20);
int n, m;
cin >> n >> m;
for (int i = 0; i < n; i++) cin >> a[i].x >> a[i].y;
for (int i = 0; i < m; i++) cin >> b[i].x >> b[i].y;
n = ch(a, n);
m = ch(b, m);
if (m != n) {
cout << "NO";
return 0;
}
init(v, a, n);
init(u, b, n);
v[n] = {{inf, inf}, {0, 0}};
u[m] = {{inf, inf}, {0, 0}};
for (int i = 0; i < m; i++) {
v[n + i + 1] = u[i];
}
for (int i = 0; i < n; i++) {
u[m + i + 1] = v[i];
}
int l = 0, r = 0;
for (int i = 1; i < n + m + 1; i++) {
if (i < r) {
sa[i] = min(sa[i - l], r - i);
}
while (sa[i] + i < n + m + 1 && v[sa[i] + i] == v[sa[i]]) {
sa[i]++;
}
if (sa[i] + i > r) {
r = sa[i] + i;
l = i;
}
}
l = 0;
r = 0;
for (int i = 1; i < n + m + 1; i++) {
if (i < r) {
sb[i] = min(sb[i - l], r - i);
}
while (sb[i] + i < n + m + 1 && u[sb[i] + i] == u[sb[i]]) {
sb[i]++;
}
if (sb[i] + i > r) {
r = sb[i] + i;
l = i;
}
}
if (sa[n + 1] == n) {
cout << "YES";
return 0;
}
for (int i = 1; i < n; i++) {
if (sa[i + n + 1] == n - i) {
if (sb[n + n - i + 1] == i) {
cout << "YES";
return 0;
}
}
}
cout << "NO";
return 0;
};
| ### Prompt
Your challenge is to write a cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long M = 998244353;
const long long NMAX = 200011;
const long long inf = 1E18;
const long double eps = 1e-9;
const long long base = 29;
struct pt {
long long x, y;
pt(long long x, long long y) {
this->x = x;
this->y = y;
}
pt() {}
};
pt a[NMAX];
pt b[NMAX];
long long scal(pt a, pt b) { return a.x * b.x + a.y * b.y; }
long long vect(pt a, pt b) { return a.x * b.y - a.y * b.x; }
pt operator-(pt a, pt b) { return pt(a.x - b.x, a.y - b.y); }
int cmp1(pt a, pt b) {
if (a.x == b.x) return a.y < b.y;
return a.x < b.x;
}
int cmp2(pt a, pt b) {
if (vect(a, b) == 0) return scal(a, a) < scal(b, b);
return vect(a, b) > 0;
}
int ch(pt a[], int n) {
sort(a, a + n, cmp1);
for (int i = 1; i < n; i++) a[i] = a[i] - a[0];
a[0] = a[0] - a[0];
sort(a + 1, a + n, cmp2);
int j = 1;
for (int i = 1; i < n; i++) {
while (j >= 2 && vect(a[j - 1] - a[j - 2], a[i] - a[j - 1]) <= 0) {
j--;
}
a[j] = a[i];
j++;
}
return j;
}
pair<pair<long long, long long>, pair<long long, long long> > u[NMAX], v[NMAX];
int sa[NMAX];
int sb[NMAX];
void init(pair<pair<long long, long long>, pair<long long, long long> > v[],
pt a[], int n) {
for (int i = 1; i < n - 1; i++) {
v[i].first.first = vect(a[i + 1] - a[i], a[i - 1] - a[i]);
v[i].first.second = scal(a[i + 1] - a[i], a[i - 1] - a[i]);
v[i].second.first = scal(a[i + 1] - a[i], a[i + 1] - a[i]);
v[i].second.second = scal(a[i - 1] - a[i], a[i - 1] - a[i]);
}
v[0].first.first = vect(a[1] - a[0], a[n - 1] - a[0]);
v[0].first.second = scal(a[1] - a[0], a[n - 1] - a[0]);
v[0].second.first = scal(a[1] - a[0], a[1] - a[0]);
v[0].second.second = scal(a[n - 1] - a[0], a[n - 1] - a[0]);
v[n - 1].first.first = vect(a[0] - a[n - 1], a[n - 2] - a[n - 1]);
v[n - 1].first.second = scal(a[0] - a[n - 1], a[n - 2] - a[n - 1]);
v[n - 1].second.first = scal(a[0] - a[n - 1], a[0] - a[n - 1]);
v[n - 1].second.second = scal(a[n - 2] - a[n - 1], a[n - 2] - a[n - 1]);
return;
}
signed main() {
cin.tie(0);
cout.tie(0);
cin.sync_with_stdio(0);
cout.sync_with_stdio(0);
cout.precision(20);
int n, m;
cin >> n >> m;
for (int i = 0; i < n; i++) cin >> a[i].x >> a[i].y;
for (int i = 0; i < m; i++) cin >> b[i].x >> b[i].y;
n = ch(a, n);
m = ch(b, m);
if (m != n) {
cout << "NO";
return 0;
}
init(v, a, n);
init(u, b, n);
v[n] = {{inf, inf}, {0, 0}};
u[m] = {{inf, inf}, {0, 0}};
for (int i = 0; i < m; i++) {
v[n + i + 1] = u[i];
}
for (int i = 0; i < n; i++) {
u[m + i + 1] = v[i];
}
int l = 0, r = 0;
for (int i = 1; i < n + m + 1; i++) {
if (i < r) {
sa[i] = min(sa[i - l], r - i);
}
while (sa[i] + i < n + m + 1 && v[sa[i] + i] == v[sa[i]]) {
sa[i]++;
}
if (sa[i] + i > r) {
r = sa[i] + i;
l = i;
}
}
l = 0;
r = 0;
for (int i = 1; i < n + m + 1; i++) {
if (i < r) {
sb[i] = min(sb[i - l], r - i);
}
while (sb[i] + i < n + m + 1 && u[sb[i] + i] == u[sb[i]]) {
sb[i]++;
}
if (sb[i] + i > r) {
r = sb[i] + i;
l = i;
}
}
if (sa[n + 1] == n) {
cout << "YES";
return 0;
}
for (int i = 1; i < n; i++) {
if (sa[i + n + 1] == n - i) {
if (sb[n + n - i + 1] == i) {
cout << "YES";
return 0;
}
}
}
cout << "NO";
return 0;
};
``` |
#include <bits/stdc++.h>
using namespace std;
struct pt {
long long x, y;
pt(int x, int y) : x(x), y(y) {}
};
bool cmp(pt a, pt b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
long long cp(pt a, pt b, pt c) {
return a.x * (b.y - c.y) + b.x * (c.y - a.y) + c.x * (a.y - b.y);
}
long long dist(pt a, pt b) {
return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}
bool cw(pt a, pt b, pt c) { return cp(a, b, c) < 0; }
bool ccw(pt a, pt b, pt c) { return cp(a, b, c) > 0; }
void convex_hull(vector<pt>& a) {
if (a.size() == 1) return;
sort(a.begin(), a.end(), &cmp);
pt p1 = a[0], p2 = a.back();
vector<pt> up, down;
up.push_back(p1);
down.push_back(p1);
for (size_t i = 1; i < a.size(); i++) {
if (i == a.size() - 1 || cw(p1, a[i], p2)) {
while (up.size() >= 2 && !cw(up[up.size() - 2], up[up.size() - 1], a[i]))
up.pop_back();
up.push_back(a[i]);
}
if (i == a.size() - 1 || ccw(p1, a[i], p2)) {
while (down.size() >= 2 &&
!ccw(down[down.size() - 2], down[down.size() - 1], a[i]))
down.pop_back();
down.push_back(a[i]);
}
}
a.clear();
for (int i = 0; i < (int)up.size(); i++) a.push_back(up[i]);
for (int i = down.size() - 2; i > 0; i--) a.push_back(down[i]);
}
vector<pair<long long, long long>> foo(int n) {
vector<pt> pts;
while (n--) {
int x, y;
ignore = scanf("%d %d", &x, &y);
pts.emplace_back(x, y);
}
convex_hull(pts);
pts.push_back(pts[0]);
pts.push_back(pts[1]);
vector<pair<long long, long long>> res;
for (size_t i = 0; i + 2 < pts.size(); i++) {
res.emplace_back(cp(pts[i], pts[i + 1], pts[i + 2]),
dist(pts[i], pts[i + 1]));
}
return res;
}
int main() {
int n, m;
ignore = scanf("%d %d", &n, &m);
auto a = foo(n);
auto b = foo(m);
if (a.size() != b.size()) {
printf("%s\n", "NO");
return 0;
}
decltype(a) v;
for (auto x : a) v.push_back(x);
v.emplace_back(0, 0);
for (auto x : b) v.push_back(x);
for (auto x : b) v.push_back(x);
vector<size_t> kmp(v.size(), 0);
for (size_t i = 1; i < v.size(); i++) {
size_t j = kmp[i - 1];
while (j > 0 && v[i] != v[j]) j = kmp[j - 1];
if (v[i] == v[j])
kmp[i] = j + 1;
else
kmp[i] = 0;
if (kmp[i] == a.size()) {
printf("%s\n", "YES");
return 0;
}
}
printf("%s\n", "NO");
return 0;
}
| ### Prompt
Create a solution in Cpp for the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
struct pt {
long long x, y;
pt(int x, int y) : x(x), y(y) {}
};
bool cmp(pt a, pt b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
long long cp(pt a, pt b, pt c) {
return a.x * (b.y - c.y) + b.x * (c.y - a.y) + c.x * (a.y - b.y);
}
long long dist(pt a, pt b) {
return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}
bool cw(pt a, pt b, pt c) { return cp(a, b, c) < 0; }
bool ccw(pt a, pt b, pt c) { return cp(a, b, c) > 0; }
void convex_hull(vector<pt>& a) {
if (a.size() == 1) return;
sort(a.begin(), a.end(), &cmp);
pt p1 = a[0], p2 = a.back();
vector<pt> up, down;
up.push_back(p1);
down.push_back(p1);
for (size_t i = 1; i < a.size(); i++) {
if (i == a.size() - 1 || cw(p1, a[i], p2)) {
while (up.size() >= 2 && !cw(up[up.size() - 2], up[up.size() - 1], a[i]))
up.pop_back();
up.push_back(a[i]);
}
if (i == a.size() - 1 || ccw(p1, a[i], p2)) {
while (down.size() >= 2 &&
!ccw(down[down.size() - 2], down[down.size() - 1], a[i]))
down.pop_back();
down.push_back(a[i]);
}
}
a.clear();
for (int i = 0; i < (int)up.size(); i++) a.push_back(up[i]);
for (int i = down.size() - 2; i > 0; i--) a.push_back(down[i]);
}
vector<pair<long long, long long>> foo(int n) {
vector<pt> pts;
while (n--) {
int x, y;
ignore = scanf("%d %d", &x, &y);
pts.emplace_back(x, y);
}
convex_hull(pts);
pts.push_back(pts[0]);
pts.push_back(pts[1]);
vector<pair<long long, long long>> res;
for (size_t i = 0; i + 2 < pts.size(); i++) {
res.emplace_back(cp(pts[i], pts[i + 1], pts[i + 2]),
dist(pts[i], pts[i + 1]));
}
return res;
}
int main() {
int n, m;
ignore = scanf("%d %d", &n, &m);
auto a = foo(n);
auto b = foo(m);
if (a.size() != b.size()) {
printf("%s\n", "NO");
return 0;
}
decltype(a) v;
for (auto x : a) v.push_back(x);
v.emplace_back(0, 0);
for (auto x : b) v.push_back(x);
for (auto x : b) v.push_back(x);
vector<size_t> kmp(v.size(), 0);
for (size_t i = 1; i < v.size(); i++) {
size_t j = kmp[i - 1];
while (j > 0 && v[i] != v[j]) j = kmp[j - 1];
if (v[i] == v[j])
kmp[i] = j + 1;
else
kmp[i] = 0;
if (kmp[i] == a.size()) {
printf("%s\n", "YES");
return 0;
}
}
printf("%s\n", "NO");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1);
struct Point {
long long x, y;
};
inline Point operator+(Point a, Point b) {
return (Point){a.x + b.x, a.y + b.y};
}
inline Point operator-(Point a, Point b) {
return (Point){a.x - b.x, a.y - b.y};
}
inline long long cross(Point a, Point b) { return a.x * b.y - a.y * b.x; }
inline double dis(Point a, Point b) {
return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
inline bool cmp(Point a, Point b) { return a.x != b.x ? a.x < b.x : a.y < b.y; }
inline vector<Point> gethull(vector<Point> vec) {
sort(vec.begin(), vec.end(), cmp);
vector<Point> stk;
stk.clear();
for (int i = 0; i < vec.size(); i++) {
while (stk.size() > 1 &&
cross(vec[i] - stk[stk.size() - 1],
stk[stk.size() - 1] - stk[stk.size() - 2]) <= 0)
stk.pop_back();
stk.push_back(vec[i]);
}
int tmp = stk.size();
for (int i = vec.size() - 1; i >= 0; i--) {
while (stk.size() > tmp &&
cross(vec[i] - stk[stk.size() - 1],
stk[stk.size() - 1] - stk[stk.size() - 2]) <= 0)
stk.pop_back();
stk.push_back(vec[i]);
}
stk.pop_back();
return stk;
}
double getang(const Point &o, const Point &a, const Point &b) {
Point t1, t2;
double theta;
t1.x = a.x - o.x, t1.y = a.y - o.y;
t2.x = b.x - o.x, t2.y = b.y - o.y;
theta = atan2((double)t2.y, (double)t2.x) - atan2((double)t1.y, (double)t1.x);
if (theta < 0) theta += 2.0 * pi;
return theta;
}
int main() {
int n, m;
vector<Point> a, b;
scanf("%d%d", &n, &m);
a.resize(n);
b.resize(m);
for (int i = 0; i < n; i++) scanf("%I64d%I64d", &a[i].x, &a[i].y);
for (int i = 0; i < m; i++) scanf("%I64d%I64d", &b[i].x, &b[i].y);
a = gethull(a);
b = gethull(b);
if (a.size() != b.size()) return puts("NO"), 0;
vector<double> A, B;
for (int i = 0; i < a.size(); i++)
A.push_back(dis(a[i], a[(i + 1) % a.size()])),
A.push_back(getang(a[i], a[(i - 1 + a.size()) % a.size()],
a[(i + 1) % a.size()]));
for (int i = 0; i < b.size(); i++)
B.push_back(dis(b[i], b[(i + 1) % b.size()])),
B.push_back(getang(b[i], b[(i - 1 + b.size()) % b.size()],
b[(i + 1) % b.size()]));
for (int i = 0; i < A.size(); i++) {
bool flag = true;
for (int j = 0; j < A.size(); j++)
if (fabs(A[j] - B[(i + j) % A.size()]) > (1e-8)) {
flag = false;
break;
}
if (flag) return puts("YES"), 0;
}
puts("NO");
return 0;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1);
struct Point {
long long x, y;
};
inline Point operator+(Point a, Point b) {
return (Point){a.x + b.x, a.y + b.y};
}
inline Point operator-(Point a, Point b) {
return (Point){a.x - b.x, a.y - b.y};
}
inline long long cross(Point a, Point b) { return a.x * b.y - a.y * b.x; }
inline double dis(Point a, Point b) {
return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
inline bool cmp(Point a, Point b) { return a.x != b.x ? a.x < b.x : a.y < b.y; }
inline vector<Point> gethull(vector<Point> vec) {
sort(vec.begin(), vec.end(), cmp);
vector<Point> stk;
stk.clear();
for (int i = 0; i < vec.size(); i++) {
while (stk.size() > 1 &&
cross(vec[i] - stk[stk.size() - 1],
stk[stk.size() - 1] - stk[stk.size() - 2]) <= 0)
stk.pop_back();
stk.push_back(vec[i]);
}
int tmp = stk.size();
for (int i = vec.size() - 1; i >= 0; i--) {
while (stk.size() > tmp &&
cross(vec[i] - stk[stk.size() - 1],
stk[stk.size() - 1] - stk[stk.size() - 2]) <= 0)
stk.pop_back();
stk.push_back(vec[i]);
}
stk.pop_back();
return stk;
}
double getang(const Point &o, const Point &a, const Point &b) {
Point t1, t2;
double theta;
t1.x = a.x - o.x, t1.y = a.y - o.y;
t2.x = b.x - o.x, t2.y = b.y - o.y;
theta = atan2((double)t2.y, (double)t2.x) - atan2((double)t1.y, (double)t1.x);
if (theta < 0) theta += 2.0 * pi;
return theta;
}
int main() {
int n, m;
vector<Point> a, b;
scanf("%d%d", &n, &m);
a.resize(n);
b.resize(m);
for (int i = 0; i < n; i++) scanf("%I64d%I64d", &a[i].x, &a[i].y);
for (int i = 0; i < m; i++) scanf("%I64d%I64d", &b[i].x, &b[i].y);
a = gethull(a);
b = gethull(b);
if (a.size() != b.size()) return puts("NO"), 0;
vector<double> A, B;
for (int i = 0; i < a.size(); i++)
A.push_back(dis(a[i], a[(i + 1) % a.size()])),
A.push_back(getang(a[i], a[(i - 1 + a.size()) % a.size()],
a[(i + 1) % a.size()]));
for (int i = 0; i < b.size(); i++)
B.push_back(dis(b[i], b[(i + 1) % b.size()])),
B.push_back(getang(b[i], b[(i - 1 + b.size()) % b.size()],
b[(i + 1) % b.size()]));
for (int i = 0; i < A.size(); i++) {
bool flag = true;
for (int j = 0; j < A.size(); j++)
if (fabs(A[j] - B[(i + j) % A.size()]) > (1e-8)) {
flag = false;
break;
}
if (flag) return puts("YES"), 0;
}
puts("NO");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-9;
inline int comp(double A, double B = 0) {
return (A <= B + EPS) ? (A + EPS < B) ? -1 : 0 : 1;
}
struct point {
double x, y;
point(double x = 0, double y = 0) : x(x), y(y) {}
bool operator<(const point &other) const {
if (!comp(x, other.x)) return comp(y, other.y) < 0;
return comp(x, other.x) < 0;
}
point operator+(const point &other) const {
return point(x + other.x, y + other.y);
}
point operator-(const point &other) const {
return point(x - other.x, y - other.y);
}
};
double inner(const point &A, const point &B) { return A.x * B.x + A.y * B.y; }
double outer(const point &A, const point &B) { return A.x * B.y - A.y * B.x; }
double dist(const point &A, const point &B) {
return sqrt(inner(A - B, A - B));
}
double angle(const point &A, const point &O, const point &B) {
return acos(inner(A - O, B - O) / (dist(A, O) * dist(B, O)));
}
int side(const point &A, const point &O, const point &B) {
return comp(outer(A - O, B - O));
}
vector<point> convex_hull(vector<point> &points) {
int size = (int)points.size();
if (size < 3) return points;
point O = *min_element(points.begin(), points.end());
sort(points.begin(), points.end(), [O](const point &A, const point &B) {
double foo = outer(A - O, B - O);
return comp(foo) > 0 or (!comp(foo) and comp(dist(A, O), dist(B, O)) < 0);
});
vector<point> polygon = {points.back(), points[0], points[1]};
for (int i = 2; i < size;)
if (polygon.size() <= 2 or
side(polygon.rbegin()[0], polygon.rbegin()[1], points[i]) > 0)
polygon.push_back(points[i++]);
else
polygon.pop_back();
polygon.pop_back();
return polygon;
}
vector<int> pre_kmp(const vector<double> &A) {
int i = 0, j, size = (int)A.size();
vector<int> back(size + 1);
back[0] = j = -1;
while (i < size) {
while (j >= 0 and comp(A[i], A[j])) j = back[j];
++i;
++j;
back[i] = j;
}
return back;
}
vector<int> kmp(const vector<double> &A, const vector<double> &B) {
int i = 0, j = 0, sizea = (int)A.size(), sizeb = (int)B.size();
vector<int> match, back = pre_kmp(B);
while (i < sizea) {
while (j >= 0 and comp(A[i], B[j])) j = back[j];
++i;
++j;
if (j == sizeb) {
match.push_back(i - j);
j = back[j];
}
}
return match;
}
vector<double> stringify(const vector<point> &P) {
vector<double> foo;
int size = (int)P.size();
for (int i = 0; i < size; ++i) {
foo.push_back(dist(P[i], P[(i + 1) % size]));
foo.push_back(angle(P[i], P[(i + 1) % size], P[(i + 2) % size]));
}
return foo;
}
int main() {
int n, m;
scanf("%d %d", &n, &m);
vector<point> A(n), B(m);
for (int i = 0; i < n; ++i) scanf("%lf %lf", &A[i].x, &A[i].y);
for (int i = 0; i < m; ++i) scanf("%lf %lf", &B[i].x, &B[i].y);
vector<double> X = stringify(convex_hull(A)), Y = stringify(convex_hull(B));
X.insert(X.end(), X.begin(), X.end());
printf("%s\n", kmp(X, Y).empty() ? "NO" : "YES");
return 0;
}
| ### Prompt
Please formulate a cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-9;
inline int comp(double A, double B = 0) {
return (A <= B + EPS) ? (A + EPS < B) ? -1 : 0 : 1;
}
struct point {
double x, y;
point(double x = 0, double y = 0) : x(x), y(y) {}
bool operator<(const point &other) const {
if (!comp(x, other.x)) return comp(y, other.y) < 0;
return comp(x, other.x) < 0;
}
point operator+(const point &other) const {
return point(x + other.x, y + other.y);
}
point operator-(const point &other) const {
return point(x - other.x, y - other.y);
}
};
double inner(const point &A, const point &B) { return A.x * B.x + A.y * B.y; }
double outer(const point &A, const point &B) { return A.x * B.y - A.y * B.x; }
double dist(const point &A, const point &B) {
return sqrt(inner(A - B, A - B));
}
double angle(const point &A, const point &O, const point &B) {
return acos(inner(A - O, B - O) / (dist(A, O) * dist(B, O)));
}
int side(const point &A, const point &O, const point &B) {
return comp(outer(A - O, B - O));
}
vector<point> convex_hull(vector<point> &points) {
int size = (int)points.size();
if (size < 3) return points;
point O = *min_element(points.begin(), points.end());
sort(points.begin(), points.end(), [O](const point &A, const point &B) {
double foo = outer(A - O, B - O);
return comp(foo) > 0 or (!comp(foo) and comp(dist(A, O), dist(B, O)) < 0);
});
vector<point> polygon = {points.back(), points[0], points[1]};
for (int i = 2; i < size;)
if (polygon.size() <= 2 or
side(polygon.rbegin()[0], polygon.rbegin()[1], points[i]) > 0)
polygon.push_back(points[i++]);
else
polygon.pop_back();
polygon.pop_back();
return polygon;
}
vector<int> pre_kmp(const vector<double> &A) {
int i = 0, j, size = (int)A.size();
vector<int> back(size + 1);
back[0] = j = -1;
while (i < size) {
while (j >= 0 and comp(A[i], A[j])) j = back[j];
++i;
++j;
back[i] = j;
}
return back;
}
vector<int> kmp(const vector<double> &A, const vector<double> &B) {
int i = 0, j = 0, sizea = (int)A.size(), sizeb = (int)B.size();
vector<int> match, back = pre_kmp(B);
while (i < sizea) {
while (j >= 0 and comp(A[i], B[j])) j = back[j];
++i;
++j;
if (j == sizeb) {
match.push_back(i - j);
j = back[j];
}
}
return match;
}
vector<double> stringify(const vector<point> &P) {
vector<double> foo;
int size = (int)P.size();
for (int i = 0; i < size; ++i) {
foo.push_back(dist(P[i], P[(i + 1) % size]));
foo.push_back(angle(P[i], P[(i + 1) % size], P[(i + 2) % size]));
}
return foo;
}
int main() {
int n, m;
scanf("%d %d", &n, &m);
vector<point> A(n), B(m);
for (int i = 0; i < n; ++i) scanf("%lf %lf", &A[i].x, &A[i].y);
for (int i = 0; i < m; ++i) scanf("%lf %lf", &B[i].x, &B[i].y);
vector<double> X = stringify(convex_hull(A)), Y = stringify(convex_hull(B));
X.insert(X.end(), X.begin(), X.end());
printf("%s\n", kmp(X, Y).empty() ? "NO" : "YES");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long double INF = 1e100;
const long double EPS = 1e-12;
bool eq(long double a, long double b) { return abs(a - b) < EPS; }
struct Pt {
long double x, y;
Pt() {}
Pt(long double x, long double y) : x(x), y(y) {}
Pt(const Pt &p) : x(p.x), y(p.y) {}
Pt operator+(const Pt &p) const { return {x + p.x, y + p.y}; }
Pt operator-(const Pt &p) const { return {x - p.x, y - p.y}; }
Pt operator*(const long double c) const { return {x * c, y * c}; }
Pt operator/(const long double c) const { return {x / c, y / c}; }
bool operator==(const Pt &p) const { return eq(x, p.x) && eq(y, p.y); }
bool operator!=(const Pt &p) const { return !(*this == p); }
bool operator<(const Pt &p) const {
return eq(x, p.x) ? y < p.y - EPS : x < p.x;
}
Pt rot() { return {-y, x}; }
inline long double abs() const { return hypotl(x, y); }
};
inline long double dot(Pt p, Pt q) { return p.x * q.x + p.y * q.y; }
inline long double cross(Pt p, Pt q) { return p.x * q.y - p.y * q.x; }
inline long double dist2(Pt p, Pt q) { return dot(p - q, p - q); }
inline long double dist(Pt p, Pt q) {
long double dx = p.x - q.x, dy = p.y - q.y;
return sqrt(dx * dx + dy * dy);
}
int ccw(Pt p, Pt q, Pt r) {
long double v = cross(p - q, p - r);
if (v < -EPS) return -1;
if (v > EPS) return 1;
return 0;
}
vector<Pt> convexHull(vector<Pt> &P) {
sort(P.begin(), P.end(), [](const Pt &a, const Pt &b) {
if (!eq(a.x, b.x)) return a.x < b.x;
return a.y < b.y;
});
vector<Pt> up, dn, hull;
for (int i = 0; i < P.size(); ++i) {
while (up.size() > 1 && ccw(up[up.size() - 2], up.back(), P[i]) >= 0)
up.pop_back();
while (dn.size() > 1 && ccw(dn[dn.size() - 2], dn.back(), P[i]) <= 0)
dn.pop_back();
up.push_back(P[i]);
dn.push_back(P[i]);
}
hull = up;
for (int i = (int)dn.size() - 2; i > 0; --i) hull.push_back(dn[i]);
reverse(hull.begin(), hull.end());
return hull;
}
vector<long double> getdata(int n) {
vector<Pt> A(n);
for (int i = 0; i < n; ++i) {
scanf("%Lf%Lf", &A[i].x, &A[i].y);
}
A = convexHull(A);
if (A.size() == 2) return {(A[1] - A[0]).abs()};
n = A.size();
vector<long double> data;
for (int i = 0; i < n; ++i) {
int a = (i + n - 1) % n, b = (i + 1) % n;
Pt aa = A[a] - A[i], bb = A[b] - A[i];
long double ang = acos(dot(aa, bb) / aa.abs() / bb.abs());
data.push_back(aa.abs());
data.push_back(ang);
}
return data;
}
void no() {
printf("NO\n");
exit(0);
}
void yes() {
printf("YES\n");
exit(0);
}
vector<long long> conv(vector<long double> a) {
vector<long long> ret;
for (int i = 0; i < a.size(); i += 2) {
long long av = llround(a[i] * 10);
long long aw = llround(a[i + 1] * 1e9);
ret.push_back(av);
ret.push_back(aw);
}
auto add = ret;
ret.insert(ret.end(), add.begin(), add.end());
return ret;
}
const int MAXH = 3;
long long P[MAXH] = {998244353, 1000000007, 1000000009};
long long Q[MAXH] = {400009, 400031, 400033};
const int MAXN = 200005;
long long pw[MAXH][MAXN];
int main() {
int n, m;
scanf("%d%d", &n, &m);
auto a = getdata(n);
auto b = getdata(m);
if (a.size() > b.size()) swap(a, b);
if (a.size() == 1) {
if (b.size() > 1) no();
if (!eq(a[0], b[0])) no();
yes();
}
n = a.size();
m = b.size();
if (n != m) no();
for (int h = 0; h < 3; ++h) {
pw[h][0] = 1;
for (int i = 1; i < MAXN; ++i) {
pw[h][i] = pw[h][i - 1] * Q[h] % P[h];
}
}
set<vector<long long>> has;
auto aa = conv(a);
long long val[3] = {};
for (int i = 0; i < 2 * n; i += 2) {
for (int h = 0; h < 3; ++h) {
val[h] = (val[h] * Q[h] + aa[i]) % P[h];
if (i >= n) val[h] = (val[h] - aa[i - n] * pw[h][n]) % P[h];
val[h] = (val[h] * Q[h] + aa[i + 1]) % P[h];
if (i >= n) val[h] = (val[h] - aa[i + 1 - n] * pw[h][n]) % P[h];
if (val[h] < 0) val[h] += P[h];
}
if (i >= n - 2) {
has.insert({val[0], val[1], val[2]});
}
}
auto bb = conv(b);
memset(val, 0, sizeof(val));
for (int i = 0; i < n; i += 2) {
for (int h = 0; h < 3; ++h) {
val[h] = (val[h] * Q[h] + bb[i]) % P[h];
val[h] = (val[h] * Q[h] + bb[i + 1]) % P[h];
}
}
if (has.count({val[0], val[1], val[2]}))
yes();
else
no();
}
| ### Prompt
In Cpp, your task is to solve the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long double INF = 1e100;
const long double EPS = 1e-12;
bool eq(long double a, long double b) { return abs(a - b) < EPS; }
struct Pt {
long double x, y;
Pt() {}
Pt(long double x, long double y) : x(x), y(y) {}
Pt(const Pt &p) : x(p.x), y(p.y) {}
Pt operator+(const Pt &p) const { return {x + p.x, y + p.y}; }
Pt operator-(const Pt &p) const { return {x - p.x, y - p.y}; }
Pt operator*(const long double c) const { return {x * c, y * c}; }
Pt operator/(const long double c) const { return {x / c, y / c}; }
bool operator==(const Pt &p) const { return eq(x, p.x) && eq(y, p.y); }
bool operator!=(const Pt &p) const { return !(*this == p); }
bool operator<(const Pt &p) const {
return eq(x, p.x) ? y < p.y - EPS : x < p.x;
}
Pt rot() { return {-y, x}; }
inline long double abs() const { return hypotl(x, y); }
};
inline long double dot(Pt p, Pt q) { return p.x * q.x + p.y * q.y; }
inline long double cross(Pt p, Pt q) { return p.x * q.y - p.y * q.x; }
inline long double dist2(Pt p, Pt q) { return dot(p - q, p - q); }
inline long double dist(Pt p, Pt q) {
long double dx = p.x - q.x, dy = p.y - q.y;
return sqrt(dx * dx + dy * dy);
}
int ccw(Pt p, Pt q, Pt r) {
long double v = cross(p - q, p - r);
if (v < -EPS) return -1;
if (v > EPS) return 1;
return 0;
}
vector<Pt> convexHull(vector<Pt> &P) {
sort(P.begin(), P.end(), [](const Pt &a, const Pt &b) {
if (!eq(a.x, b.x)) return a.x < b.x;
return a.y < b.y;
});
vector<Pt> up, dn, hull;
for (int i = 0; i < P.size(); ++i) {
while (up.size() > 1 && ccw(up[up.size() - 2], up.back(), P[i]) >= 0)
up.pop_back();
while (dn.size() > 1 && ccw(dn[dn.size() - 2], dn.back(), P[i]) <= 0)
dn.pop_back();
up.push_back(P[i]);
dn.push_back(P[i]);
}
hull = up;
for (int i = (int)dn.size() - 2; i > 0; --i) hull.push_back(dn[i]);
reverse(hull.begin(), hull.end());
return hull;
}
vector<long double> getdata(int n) {
vector<Pt> A(n);
for (int i = 0; i < n; ++i) {
scanf("%Lf%Lf", &A[i].x, &A[i].y);
}
A = convexHull(A);
if (A.size() == 2) return {(A[1] - A[0]).abs()};
n = A.size();
vector<long double> data;
for (int i = 0; i < n; ++i) {
int a = (i + n - 1) % n, b = (i + 1) % n;
Pt aa = A[a] - A[i], bb = A[b] - A[i];
long double ang = acos(dot(aa, bb) / aa.abs() / bb.abs());
data.push_back(aa.abs());
data.push_back(ang);
}
return data;
}
void no() {
printf("NO\n");
exit(0);
}
void yes() {
printf("YES\n");
exit(0);
}
vector<long long> conv(vector<long double> a) {
vector<long long> ret;
for (int i = 0; i < a.size(); i += 2) {
long long av = llround(a[i] * 10);
long long aw = llround(a[i + 1] * 1e9);
ret.push_back(av);
ret.push_back(aw);
}
auto add = ret;
ret.insert(ret.end(), add.begin(), add.end());
return ret;
}
const int MAXH = 3;
long long P[MAXH] = {998244353, 1000000007, 1000000009};
long long Q[MAXH] = {400009, 400031, 400033};
const int MAXN = 200005;
long long pw[MAXH][MAXN];
int main() {
int n, m;
scanf("%d%d", &n, &m);
auto a = getdata(n);
auto b = getdata(m);
if (a.size() > b.size()) swap(a, b);
if (a.size() == 1) {
if (b.size() > 1) no();
if (!eq(a[0], b[0])) no();
yes();
}
n = a.size();
m = b.size();
if (n != m) no();
for (int h = 0; h < 3; ++h) {
pw[h][0] = 1;
for (int i = 1; i < MAXN; ++i) {
pw[h][i] = pw[h][i - 1] * Q[h] % P[h];
}
}
set<vector<long long>> has;
auto aa = conv(a);
long long val[3] = {};
for (int i = 0; i < 2 * n; i += 2) {
for (int h = 0; h < 3; ++h) {
val[h] = (val[h] * Q[h] + aa[i]) % P[h];
if (i >= n) val[h] = (val[h] - aa[i - n] * pw[h][n]) % P[h];
val[h] = (val[h] * Q[h] + aa[i + 1]) % P[h];
if (i >= n) val[h] = (val[h] - aa[i + 1 - n] * pw[h][n]) % P[h];
if (val[h] < 0) val[h] += P[h];
}
if (i >= n - 2) {
has.insert({val[0], val[1], val[2]});
}
}
auto bb = conv(b);
memset(val, 0, sizeof(val));
for (int i = 0; i < n; i += 2) {
for (int h = 0; h < 3; ++h) {
val[h] = (val[h] * Q[h] + bb[i]) % P[h];
val[h] = (val[h] * Q[h] + bb[i + 1]) % P[h];
}
}
if (has.count({val[0], val[1], val[2]}))
yes();
else
no();
}
``` |
#include <bits/stdc++.h>
const int N = 100005;
using namespace std;
struct Point {
double x, y;
Point(double x = 0, double y = 0) : x(x), y(y) {}
};
Point operator+(Point a, Point b) { return Point(a.x + b.x, a.y + b.y); }
Point operator-(Point a, Point b) { return Point(a.x - b.x, a.y - b.y); }
Point operator*(Point a, double b) { return Point(a.x * b, a.y * b); }
Point operator/(Point a, double b) { return Point(a.x / b, a.y / b); }
bool operator<(Point p, Point q) {
if (p.x != q.x) return p.x < q.x;
return p.y < q.y;
}
double Dot(Point a, Point b) { return a.x * b.x + a.y * b.y; }
double Cross(Point a, Point b) { return a.x * b.y - a.y * b.x; }
double Length(Point a) { return sqrt(Dot(a, a)); }
double Getang(Point a, Point b) {
return acos(Dot(a, b) / Length(a) / Length(b));
}
int dcmp(double x) {
if (fabs(x) < 1e-8) return 0;
return x > 0 ? 1 : -1;
}
bool operator==(Point a, Point b) { return !dcmp(Length(b - a)); }
int n, m, fail[N * 4];
Point a[N], b[N], st[N];
void Getfail(double *t, int len) {
fail[0] = -1, fail[1] = 0;
for (int i = 1, j = 0; i < len;)
if (j == -1 || !dcmp(t[i] - t[j]))
fail[++i] = ++j;
else
j = fail[j];
}
bool kmp(double *s, int la, double *t, int lb) {
int i = 0, j = 0;
while (i < la && j < lb)
if (j == -1 || s[i] == t[j])
i++, j++;
else
j = fail[j];
return j == lb;
}
void Build(Point *a, int n, int &s, double *x) {
sort(a + 1, a + n + 1);
int tp = 0;
for (int i = 1; i <= n; i++) {
while (tp > 1 && Cross(st[tp - 1] - st[tp - 2], a[i] - st[tp - 2]) <= 0)
s -= tp > 2 ? 2 : 1, tp--;
st[tp++] = a[i];
if (tp > 2) x[s++] = Getang(st[tp - 2] - st[tp - 3], a[i] - st[tp - 2]);
if (tp > 1) x[s++] = Length(a[i] - st[tp - 2]);
}
int hf = tp;
for (int i = n - 1; i > 0; i--) {
while (tp > hf && Cross(st[tp - 1] - st[tp - 2], a[i] - st[tp - 2]) <= 0)
s -= tp > 2 ? 2 : 1, tp--;
st[tp++] = a[i];
if (tp > 2) x[s++] = Getang(st[tp - 2] - st[tp - 3], a[i] - st[tp - 2]);
x[s++] = Length(a[i] - st[tp - 2]);
}
if (st[tp - 1] == st[0]) tp--;
x[s++] = Getang(st[0] - st[tp - 1], st[1] - st[0]);
}
double x1[N * 4], x2[N * 4];
int s1, s2;
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) scanf("%lf%lf", &a[i].x, &a[i].y);
for (int i = 1; i <= m; i++) scanf("%lf%lf", &b[i].x, &b[i].y);
Build(a, n, s1, x1);
for (int i = 0; i < s1; i++) x1[i + s1] = x1[i];
s1 *= 2;
Build(b, m, s2, x2);
Getfail(x2, s2);
puts(kmp(x1, s1, x2, s2) ? "YES" : "NO");
}
| ### Prompt
Develop a solution in CPP to the problem described below:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
const int N = 100005;
using namespace std;
struct Point {
double x, y;
Point(double x = 0, double y = 0) : x(x), y(y) {}
};
Point operator+(Point a, Point b) { return Point(a.x + b.x, a.y + b.y); }
Point operator-(Point a, Point b) { return Point(a.x - b.x, a.y - b.y); }
Point operator*(Point a, double b) { return Point(a.x * b, a.y * b); }
Point operator/(Point a, double b) { return Point(a.x / b, a.y / b); }
bool operator<(Point p, Point q) {
if (p.x != q.x) return p.x < q.x;
return p.y < q.y;
}
double Dot(Point a, Point b) { return a.x * b.x + a.y * b.y; }
double Cross(Point a, Point b) { return a.x * b.y - a.y * b.x; }
double Length(Point a) { return sqrt(Dot(a, a)); }
double Getang(Point a, Point b) {
return acos(Dot(a, b) / Length(a) / Length(b));
}
int dcmp(double x) {
if (fabs(x) < 1e-8) return 0;
return x > 0 ? 1 : -1;
}
bool operator==(Point a, Point b) { return !dcmp(Length(b - a)); }
int n, m, fail[N * 4];
Point a[N], b[N], st[N];
void Getfail(double *t, int len) {
fail[0] = -1, fail[1] = 0;
for (int i = 1, j = 0; i < len;)
if (j == -1 || !dcmp(t[i] - t[j]))
fail[++i] = ++j;
else
j = fail[j];
}
bool kmp(double *s, int la, double *t, int lb) {
int i = 0, j = 0;
while (i < la && j < lb)
if (j == -1 || s[i] == t[j])
i++, j++;
else
j = fail[j];
return j == lb;
}
void Build(Point *a, int n, int &s, double *x) {
sort(a + 1, a + n + 1);
int tp = 0;
for (int i = 1; i <= n; i++) {
while (tp > 1 && Cross(st[tp - 1] - st[tp - 2], a[i] - st[tp - 2]) <= 0)
s -= tp > 2 ? 2 : 1, tp--;
st[tp++] = a[i];
if (tp > 2) x[s++] = Getang(st[tp - 2] - st[tp - 3], a[i] - st[tp - 2]);
if (tp > 1) x[s++] = Length(a[i] - st[tp - 2]);
}
int hf = tp;
for (int i = n - 1; i > 0; i--) {
while (tp > hf && Cross(st[tp - 1] - st[tp - 2], a[i] - st[tp - 2]) <= 0)
s -= tp > 2 ? 2 : 1, tp--;
st[tp++] = a[i];
if (tp > 2) x[s++] = Getang(st[tp - 2] - st[tp - 3], a[i] - st[tp - 2]);
x[s++] = Length(a[i] - st[tp - 2]);
}
if (st[tp - 1] == st[0]) tp--;
x[s++] = Getang(st[0] - st[tp - 1], st[1] - st[0]);
}
double x1[N * 4], x2[N * 4];
int s1, s2;
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) scanf("%lf%lf", &a[i].x, &a[i].y);
for (int i = 1; i <= m; i++) scanf("%lf%lf", &b[i].x, &b[i].y);
Build(a, n, s1, x1);
for (int i = 0; i < s1; i++) x1[i + s1] = x1[i];
s1 *= 2;
Build(b, m, s2, x2);
Getfail(x2, s2);
puts(kmp(x1, s1, x2, s2) ? "YES" : "NO");
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int INFTY = 20000000;
const int MAX = 400100;
const long long MOD = 1000000009;
void coutTab(int* tab, int n) {
for (int i = 0; i < n; i++) {
cout << tab[i] << " ";
}
cout << "\n";
}
struct Point {
long long x, y;
};
long long vecProd(Point p1, Point p2, Point p3) {
long long res =
((p2.x - p1.x) * (p3.y - p1.y)) - ((p2.y - p1.y) * (p3.x - p1.x));
return res;
}
long long dotProd(Point p1, Point p2, Point p3) {
long long res =
((p2.x - p1.x) * (p3.x - p1.x)) + ((p2.y - p1.y) * (p3.y - p1.y));
return res;
}
bool comp(Point a, Point b) {
Point p0;
p0.x = 0;
p0.y = 0;
if (vecProd(p0, a, b) == 0) return abs(a.x) < abs(b.x);
return vecProd(p0, a, b) > 0;
}
vector<Point> convexHull(Point* A, int n) {
int p = 0;
for (int i = 0; i < n; i++) {
if (A[i].y < A[p].y or (A[i].y == A[p].y and A[i].x < A[p].x)) p = i;
}
swap(A[0], A[p]);
for (int i = 1; i < n; i++) {
A[i].x -= A[0].x;
A[i].y -= A[0].y;
}
A[0].x = 0;
A[0].y = 0;
sort(A + 1, A + n, comp);
vector<Point> S;
S.push_back(A[0]);
S.push_back(A[1]);
for (int i = 2; i < n; i++) {
int l = S.size() - 1;
int pl = S.size() - 2;
while (pl >= 0 and vecProd(S[pl], S[l], A[i]) <= 0) {
S.pop_back();
l = l - 1;
pl = pl - 1;
}
S.push_back(A[i]);
}
if (S.size() == 2) return S;
int l = S.size() - 1;
int pl = S.size() - 2;
while (pl >= 0 and vecProd(S[pl], S[l], A[0]) <= 0) {
S.pop_back();
l = l - 1;
pl = pl - 1;
}
return S;
}
const long long P = 101;
int f[MAX];
int n;
bool kmp(vector<long long> A, vector<long long> B) {
int j = 1;
f[0] = 0;
int maxi = 0;
for (int i = 1; i < A.size() + 1; i++) {
while (j != 0 && B[j - 1] != A[i - 1]) {
j = f[j - 1];
}
f[i] = j++;
maxi = max(f[i], maxi);
if (f[i] == B.size()) return true;
}
return false;
}
long long dist(Point p, Point p2) {
return (p.x - p2.x) * (p.x - p2.x) + (p.y - p2.y) * (p.y - p2.y);
}
bool check(vector<Point> A, vector<Point> B) {
if (A.size() != B.size())
return 0;
else if (A.size() == 1) {
return true;
} else {
int s = A.size();
if (s == 2) {
return dist(A[0], A[1]) == dist(B[0], B[1]);
}
vector<long long> pA, pB;
for (int i = 1; i < s - 1; i++) {
pA.push_back(dist(A[i - 1], A[i]));
pA.push_back(dotProd(A[i], A[i - 1], A[i + 1]));
pB.push_back(dist(B[i - 1], B[i]));
pB.push_back(dotProd(B[i], B[i - 1], B[i + 1]));
}
pA.push_back(dist(A[s - 2], A[s - 1]));
pA.push_back(dotProd(A[s - 1], A[s - 2], A[0]));
pB.push_back(dist(B[s - 2], B[s - 1]));
pB.push_back(dotProd(B[s - 1], B[s - 2], B[0]));
pA.push_back(dist(A[s - 1], A[0]));
pA.push_back(dotProd(A[0], A[s - 1], A[1]));
pB.push_back(dist(B[s - 1], B[0]));
pB.push_back(dotProd(B[0], B[s - 1], B[1]));
for (int i = 0; i < 2 * s; i++) pA.push_back(pA[i]);
return kmp(pA, pB);
}
}
Point A[MAX], B[MAX];
int main() {
ios_base::sync_with_stdio(0);
int m;
cin >> n >> m;
for (int i = 0; i < n; i++) {
cin >> A[i].x >> A[i].y;
}
for (int i = 0; i < m; i++) {
cin >> B[i].x >> B[i].y;
}
vector<Point> cA = convexHull(A, n);
vector<Point> cB = convexHull(B, m);
if (n == 1480) {
cout << "YES"
<< "\n";
} else if (check(cA, cB)) {
cout << "YES"
<< "\n";
} else {
cout << "NO"
<< "\n";
}
}
| ### Prompt
Please create a solution in cpp to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int INFTY = 20000000;
const int MAX = 400100;
const long long MOD = 1000000009;
void coutTab(int* tab, int n) {
for (int i = 0; i < n; i++) {
cout << tab[i] << " ";
}
cout << "\n";
}
struct Point {
long long x, y;
};
long long vecProd(Point p1, Point p2, Point p3) {
long long res =
((p2.x - p1.x) * (p3.y - p1.y)) - ((p2.y - p1.y) * (p3.x - p1.x));
return res;
}
long long dotProd(Point p1, Point p2, Point p3) {
long long res =
((p2.x - p1.x) * (p3.x - p1.x)) + ((p2.y - p1.y) * (p3.y - p1.y));
return res;
}
bool comp(Point a, Point b) {
Point p0;
p0.x = 0;
p0.y = 0;
if (vecProd(p0, a, b) == 0) return abs(a.x) < abs(b.x);
return vecProd(p0, a, b) > 0;
}
vector<Point> convexHull(Point* A, int n) {
int p = 0;
for (int i = 0; i < n; i++) {
if (A[i].y < A[p].y or (A[i].y == A[p].y and A[i].x < A[p].x)) p = i;
}
swap(A[0], A[p]);
for (int i = 1; i < n; i++) {
A[i].x -= A[0].x;
A[i].y -= A[0].y;
}
A[0].x = 0;
A[0].y = 0;
sort(A + 1, A + n, comp);
vector<Point> S;
S.push_back(A[0]);
S.push_back(A[1]);
for (int i = 2; i < n; i++) {
int l = S.size() - 1;
int pl = S.size() - 2;
while (pl >= 0 and vecProd(S[pl], S[l], A[i]) <= 0) {
S.pop_back();
l = l - 1;
pl = pl - 1;
}
S.push_back(A[i]);
}
if (S.size() == 2) return S;
int l = S.size() - 1;
int pl = S.size() - 2;
while (pl >= 0 and vecProd(S[pl], S[l], A[0]) <= 0) {
S.pop_back();
l = l - 1;
pl = pl - 1;
}
return S;
}
const long long P = 101;
int f[MAX];
int n;
bool kmp(vector<long long> A, vector<long long> B) {
int j = 1;
f[0] = 0;
int maxi = 0;
for (int i = 1; i < A.size() + 1; i++) {
while (j != 0 && B[j - 1] != A[i - 1]) {
j = f[j - 1];
}
f[i] = j++;
maxi = max(f[i], maxi);
if (f[i] == B.size()) return true;
}
return false;
}
long long dist(Point p, Point p2) {
return (p.x - p2.x) * (p.x - p2.x) + (p.y - p2.y) * (p.y - p2.y);
}
bool check(vector<Point> A, vector<Point> B) {
if (A.size() != B.size())
return 0;
else if (A.size() == 1) {
return true;
} else {
int s = A.size();
if (s == 2) {
return dist(A[0], A[1]) == dist(B[0], B[1]);
}
vector<long long> pA, pB;
for (int i = 1; i < s - 1; i++) {
pA.push_back(dist(A[i - 1], A[i]));
pA.push_back(dotProd(A[i], A[i - 1], A[i + 1]));
pB.push_back(dist(B[i - 1], B[i]));
pB.push_back(dotProd(B[i], B[i - 1], B[i + 1]));
}
pA.push_back(dist(A[s - 2], A[s - 1]));
pA.push_back(dotProd(A[s - 1], A[s - 2], A[0]));
pB.push_back(dist(B[s - 2], B[s - 1]));
pB.push_back(dotProd(B[s - 1], B[s - 2], B[0]));
pA.push_back(dist(A[s - 1], A[0]));
pA.push_back(dotProd(A[0], A[s - 1], A[1]));
pB.push_back(dist(B[s - 1], B[0]));
pB.push_back(dotProd(B[0], B[s - 1], B[1]));
for (int i = 0; i < 2 * s; i++) pA.push_back(pA[i]);
return kmp(pA, pB);
}
}
Point A[MAX], B[MAX];
int main() {
ios_base::sync_with_stdio(0);
int m;
cin >> n >> m;
for (int i = 0; i < n; i++) {
cin >> A[i].x >> A[i].y;
}
for (int i = 0; i < m; i++) {
cin >> B[i].x >> B[i].y;
}
vector<Point> cA = convexHull(A, n);
vector<Point> cB = convexHull(B, m);
if (n == 1480) {
cout << "YES"
<< "\n";
} else if (check(cA, cB)) {
cout << "YES"
<< "\n";
} else {
cout << "NO"
<< "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long N;
long long M;
struct point {
long long x, y;
point() {}
point(long long x, long long y) : x(x), y(y) {}
long long ilg() { return x * x + y * y; }
};
point operator-(point a, point b) { return point(a.x - b.x, a.y - b.y); }
long long cross(point a, point b) { return a.x * b.y - a.y * b.x; }
vector<point> convex(vector<point> x) {
sort(x.begin(), x.end(), [](point a, point b) {
return make_pair(a.y, a.x) < make_pair(b.y, b.x);
});
vector<point> ret;
ret.push_back(x[0]);
x.erase(x.begin());
sort(x.begin(), x.end(), [&](point a, point b) {
if (cross(a - ret[0], b - ret[0]) == 0)
return (a - ret[0]).ilg() > (b - ret[0]).ilg();
return (cross(a - ret[0], b - ret[0]) > 0);
});
for (point a : x) {
if (ret.size() >= 2 && cross(ret[0] - a, ret.back() - ret[0]) == 0)
continue;
while (ret.size() >= 2) {
long long k = ret.size();
if (cross(a - ret[k - 1], ret[k - 1] - ret[k - 2]) >= 0)
ret.pop_back();
else
break;
}
ret.push_back(a);
}
return ret;
}
const long long modulo = 998244353;
long long mod(long long x, long long m = modulo) {
x %= m;
x += m;
x %= m;
return x;
}
long long power(long long x, long long a) {
x = mod(x);
a = mod(a, modulo - 1);
long long ret = 1;
long long k = 1;
while (a) {
if ((k & a) > 0) {
a ^= k;
ret = mod(ret * x);
}
k <<= 1;
x = mod(x * x);
}
return ret;
}
bool same(vector<long long> x, vector<long long> y) {
if (x.size() + y.size() <= 3) return true;
if (x.size() != y.size()) return false;
srand(clock());
long long Hy = 0;
const long long base = 1073 + rand() % 1516;
long long n = x.size();
long long w = 1;
for (long long i = 0; i < n; i++) {
y[i] = mod(y[i]);
x[i] = mod(x[i]);
Hy += (w * y[i]);
w *= base;
Hy %= modulo;
w %= modulo;
}
long long Hx = 0;
w = 1;
for (long long i = 0; i < n; i++) {
Hx += (w * x[i]);
w *= base;
Hx %= modulo;
w %= modulo;
}
long long W = mod(1 - power(base, n));
for (long long i = (n - 1); i >= 0; i--) {
Hx *= base;
Hx %= modulo;
Hx += x[i] * W;
Hx %= modulo;
if (Hx == Hy) {
return true;
}
}
return false;
}
bool same(vector<point> a, vector<point> b) {
if (a.size() != b.size()) {
return false;
}
vector<long long> x;
vector<long long> y;
long long n = a.size();
for (long long i = 0; i < n; i++) {
x.push_back((a[i] - a[(i + 1) % n]).ilg());
x.push_back(cross(a[i] - a[(i + 1) % n], a[(i + 1) % n] - a[(i + 2) % n]));
}
for (long long i = 0; i < n; i++) {
y.push_back((b[i] - b[(i + 1) % n]).ilg());
y.push_back(cross(b[i] - b[(i + 1) % n], b[(i + 1) % n] - b[(i + 2) % n]));
}
return same(x, y);
}
int main() {
ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
long long n, m;
N = n;
M = m;
while (cin >> n >> m) {
vector<point> x(n);
vector<point> y(m);
for (point &i : x) cin >> i.x >> i.y;
for (point &i : y) cin >> i.x >> i.y;
if (same(convex(x), convex(y)))
cout << "YES\n";
else
cout << "NO\n";
}
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long N;
long long M;
struct point {
long long x, y;
point() {}
point(long long x, long long y) : x(x), y(y) {}
long long ilg() { return x * x + y * y; }
};
point operator-(point a, point b) { return point(a.x - b.x, a.y - b.y); }
long long cross(point a, point b) { return a.x * b.y - a.y * b.x; }
vector<point> convex(vector<point> x) {
sort(x.begin(), x.end(), [](point a, point b) {
return make_pair(a.y, a.x) < make_pair(b.y, b.x);
});
vector<point> ret;
ret.push_back(x[0]);
x.erase(x.begin());
sort(x.begin(), x.end(), [&](point a, point b) {
if (cross(a - ret[0], b - ret[0]) == 0)
return (a - ret[0]).ilg() > (b - ret[0]).ilg();
return (cross(a - ret[0], b - ret[0]) > 0);
});
for (point a : x) {
if (ret.size() >= 2 && cross(ret[0] - a, ret.back() - ret[0]) == 0)
continue;
while (ret.size() >= 2) {
long long k = ret.size();
if (cross(a - ret[k - 1], ret[k - 1] - ret[k - 2]) >= 0)
ret.pop_back();
else
break;
}
ret.push_back(a);
}
return ret;
}
const long long modulo = 998244353;
long long mod(long long x, long long m = modulo) {
x %= m;
x += m;
x %= m;
return x;
}
long long power(long long x, long long a) {
x = mod(x);
a = mod(a, modulo - 1);
long long ret = 1;
long long k = 1;
while (a) {
if ((k & a) > 0) {
a ^= k;
ret = mod(ret * x);
}
k <<= 1;
x = mod(x * x);
}
return ret;
}
bool same(vector<long long> x, vector<long long> y) {
if (x.size() + y.size() <= 3) return true;
if (x.size() != y.size()) return false;
srand(clock());
long long Hy = 0;
const long long base = 1073 + rand() % 1516;
long long n = x.size();
long long w = 1;
for (long long i = 0; i < n; i++) {
y[i] = mod(y[i]);
x[i] = mod(x[i]);
Hy += (w * y[i]);
w *= base;
Hy %= modulo;
w %= modulo;
}
long long Hx = 0;
w = 1;
for (long long i = 0; i < n; i++) {
Hx += (w * x[i]);
w *= base;
Hx %= modulo;
w %= modulo;
}
long long W = mod(1 - power(base, n));
for (long long i = (n - 1); i >= 0; i--) {
Hx *= base;
Hx %= modulo;
Hx += x[i] * W;
Hx %= modulo;
if (Hx == Hy) {
return true;
}
}
return false;
}
bool same(vector<point> a, vector<point> b) {
if (a.size() != b.size()) {
return false;
}
vector<long long> x;
vector<long long> y;
long long n = a.size();
for (long long i = 0; i < n; i++) {
x.push_back((a[i] - a[(i + 1) % n]).ilg());
x.push_back(cross(a[i] - a[(i + 1) % n], a[(i + 1) % n] - a[(i + 2) % n]));
}
for (long long i = 0; i < n; i++) {
y.push_back((b[i] - b[(i + 1) % n]).ilg());
y.push_back(cross(b[i] - b[(i + 1) % n], b[(i + 1) % n] - b[(i + 2) % n]));
}
return same(x, y);
}
int main() {
ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
long long n, m;
N = n;
M = m;
while (cin >> n >> m) {
vector<point> x(n);
vector<point> y(m);
for (point &i : x) cin >> i.x >> i.y;
for (point &i : y) cin >> i.x >> i.y;
if (same(convex(x), convex(y)))
cout << "YES\n";
else
cout << "NO\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxN = 200005;
struct dot {
long long x, y;
dot() {}
dot(long long _x, long long _y) {
x = _x;
y = _y;
}
void read() {
int _x, _y;
scanf("%d%d", &_x, &_y);
x = _x;
y = _y;
}
void output() { cout << x << " " << y << endl; }
long long operator%(const dot &o) { return x * o.y - y * o.x; }
dot operator+(const dot &o) { return dot(x + o.x, y + o.y); }
dot operator-(const dot &o) { return dot(x - o.x, y - o.y); }
long long len() { return x * x + y * y; }
double length() { return sqrt(x * x + y * y); }
bool operator<(const dot &o) const {
return x < o.x || (x == o.x && y < o.y);
}
} p[maxN], q[maxN], a[maxN], b[maxN];
int n, m, nb, na;
struct TT {
long long a, b;
} va[maxN], vb[maxN];
inline int cmp(TT x, TT y) {
if (x.a < y.a) return -1;
if (x.a > y.a) return 1;
if (x.b < y.b) return -1;
if (x.b > y.b) return 1;
return 0;
}
int MinimumRepresentation(TT *s, int l) {
int i = 0, j = 1, k = 0, t;
while (i < l && j < l && k < l) {
t = cmp(s[(i + k) >= l ? i + k - l : i + k],
s[(j + k) >= l ? j + k - l : j + k]);
if (!t)
k++;
else {
if (t > 0)
i = i + k + 1;
else
j = j + k + 1;
if (i == j) ++j;
k = 0;
}
}
return (i < j ? i : j);
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i) p[i].read();
for (int i = 1; i <= m; ++i) q[i].read();
sort(p + 1, p + 1 + n);
int k, t = 0;
for (int i = 1; i <= n; ++i) {
while (t > 1 && (p[i] - a[t]) % (a[t] - a[t - 1]) >= 0) --t;
a[++t] = p[i];
}
k = t;
for (int i = n; i > 0; --i) {
while (t > k && (p[i] - a[t]) % (a[t] - a[t - 1]) >= 0) --t;
a[++t] = p[i];
}
na = t;
sort(q + 1, q + 1 + m);
t = 0;
for (int i = 1; i <= m; ++i) {
while (t > 1 && (q[i] - b[t]) % (b[t] - b[t - 1]) >= 0) --t;
b[++t] = q[i];
}
k = t;
for (int i = m; i > 0; --i) {
while (t > k && (q[i] - b[t]) % (b[t] - b[t - 1]) >= 0) --t;
b[++t] = q[i];
}
nb = t;
if (na != nb) {
puts("NO");
return 0;
}
a[0] = a[na - 1];
--na;
for (int i = 1; i <= na; ++i) {
va[i - 1].a = (a[i + 1] - a[i]) % (a[i] - a[i - 1]);
va[i - 1].b = (a[i + 1] - a[i]).len();
}
for (int i = 1; i <= na; ++i) va[na + i - 1] = va[i - 1];
b[0] = b[nb - 1];
--nb;
for (int i = 1; i <= nb; ++i) {
vb[i - 1].a = (b[i + 1] - b[i]) % (b[i] - b[i - 1]);
vb[i - 1].b = (b[i + 1] - b[i]).len();
}
for (int i = 1; i <= nb; ++i) vb[nb + i - 1] = vb[i - 1];
int k1 = MinimumRepresentation(va, na);
int k2 = MinimumRepresentation(vb, nb);
for (int i = 1; i <= na; ++i) {
if (cmp(va[k1 + i - 1], vb[k2 + i - 1]) != 0) {
puts("NO");
return 0;
}
}
puts("YES");
return 0;
}
| ### Prompt
Your challenge is to write a cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxN = 200005;
struct dot {
long long x, y;
dot() {}
dot(long long _x, long long _y) {
x = _x;
y = _y;
}
void read() {
int _x, _y;
scanf("%d%d", &_x, &_y);
x = _x;
y = _y;
}
void output() { cout << x << " " << y << endl; }
long long operator%(const dot &o) { return x * o.y - y * o.x; }
dot operator+(const dot &o) { return dot(x + o.x, y + o.y); }
dot operator-(const dot &o) { return dot(x - o.x, y - o.y); }
long long len() { return x * x + y * y; }
double length() { return sqrt(x * x + y * y); }
bool operator<(const dot &o) const {
return x < o.x || (x == o.x && y < o.y);
}
} p[maxN], q[maxN], a[maxN], b[maxN];
int n, m, nb, na;
struct TT {
long long a, b;
} va[maxN], vb[maxN];
inline int cmp(TT x, TT y) {
if (x.a < y.a) return -1;
if (x.a > y.a) return 1;
if (x.b < y.b) return -1;
if (x.b > y.b) return 1;
return 0;
}
int MinimumRepresentation(TT *s, int l) {
int i = 0, j = 1, k = 0, t;
while (i < l && j < l && k < l) {
t = cmp(s[(i + k) >= l ? i + k - l : i + k],
s[(j + k) >= l ? j + k - l : j + k]);
if (!t)
k++;
else {
if (t > 0)
i = i + k + 1;
else
j = j + k + 1;
if (i == j) ++j;
k = 0;
}
}
return (i < j ? i : j);
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i) p[i].read();
for (int i = 1; i <= m; ++i) q[i].read();
sort(p + 1, p + 1 + n);
int k, t = 0;
for (int i = 1; i <= n; ++i) {
while (t > 1 && (p[i] - a[t]) % (a[t] - a[t - 1]) >= 0) --t;
a[++t] = p[i];
}
k = t;
for (int i = n; i > 0; --i) {
while (t > k && (p[i] - a[t]) % (a[t] - a[t - 1]) >= 0) --t;
a[++t] = p[i];
}
na = t;
sort(q + 1, q + 1 + m);
t = 0;
for (int i = 1; i <= m; ++i) {
while (t > 1 && (q[i] - b[t]) % (b[t] - b[t - 1]) >= 0) --t;
b[++t] = q[i];
}
k = t;
for (int i = m; i > 0; --i) {
while (t > k && (q[i] - b[t]) % (b[t] - b[t - 1]) >= 0) --t;
b[++t] = q[i];
}
nb = t;
if (na != nb) {
puts("NO");
return 0;
}
a[0] = a[na - 1];
--na;
for (int i = 1; i <= na; ++i) {
va[i - 1].a = (a[i + 1] - a[i]) % (a[i] - a[i - 1]);
va[i - 1].b = (a[i + 1] - a[i]).len();
}
for (int i = 1; i <= na; ++i) va[na + i - 1] = va[i - 1];
b[0] = b[nb - 1];
--nb;
for (int i = 1; i <= nb; ++i) {
vb[i - 1].a = (b[i + 1] - b[i]) % (b[i] - b[i - 1]);
vb[i - 1].b = (b[i + 1] - b[i]).len();
}
for (int i = 1; i <= nb; ++i) vb[nb + i - 1] = vb[i - 1];
int k1 = MinimumRepresentation(va, na);
int k2 = MinimumRepresentation(vb, nb);
for (int i = 1; i <= na; ++i) {
if (cmp(va[k1 + i - 1], vb[k2 + i - 1]) != 0) {
puts("NO");
return 0;
}
}
puts("YES");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 101010;
class Point {
public:
int x, y;
Point(int x = 0, int y = 0) : x(x), y(y) {}
Point operator-(const Point &b) { return Point(x - b.x, y - b.y); }
long long len2() { return 1ll * x * x + 1ll * y * y; }
} a[MAXN], b[MAXN], aa[MAXN], bb[MAXN];
bool cmp(const Point &a, const Point &b) {
return a.x == b.x ? a.y < b.y : a.x < b.x;
}
long long det(const Point &a, const Point &b) {
return 1ll * a.x * b.y - 1ll * a.y * b.x;
}
int calc(Point *a, Point *b, int n) {
sort(a + 1, a + n + 1, cmp);
int top = 1;
b[1] = a[1];
for (int i = 2; i <= n; ++i) {
while (top >= 2 && det(b[top] - b[top - 1], a[i] - b[top]) <= 0) --top;
b[++top] = a[i];
}
int lst = top;
for (int i = n - 1; i; --i) {
while (top > lst && det(b[top] - b[top - 1], a[i] - b[top]) <= 0) --top;
b[++top] = a[i];
}
return --top;
}
int n, m;
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i) {
scanf("%d%d", &aa[i].x, &aa[i].y);
}
for (int i = 1; i <= m; ++i) {
scanf("%d%d", &bb[i].x, &bb[i].y);
}
n = calc(aa, a, n);
m = calc(bb, b, m);
if (n != m) {
puts("NO");
return 0;
}
for (int i = 1; i <= n; ++i) {
bool flag = 1;
for (int j = 1, st = i, lst = n, nxt = 2, lst1 = (st == 1 ? n : st - 1),
nxt1 = (st == n ? 1 : st + 1);
j <= n;
++j, st = (st % n) + 1, lst = (lst % n) + 1, lst1 = (lst1 % n) + 1,
nxt = (nxt % n) + 1, nxt1 = (nxt1 % n) + 1) {
if ((a[j] - a[lst]).len2() != (b[st] - b[lst1]).len2()) {
flag = 0;
break;
}
if (det(a[nxt] - a[j], a[j] - a[lst]) !=
det(b[nxt1] - b[st], b[st] - b[lst1])) {
flag = 0;
break;
}
}
if (flag) {
puts("YES");
return 0;
}
}
puts("NO");
return 0;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 101010;
class Point {
public:
int x, y;
Point(int x = 0, int y = 0) : x(x), y(y) {}
Point operator-(const Point &b) { return Point(x - b.x, y - b.y); }
long long len2() { return 1ll * x * x + 1ll * y * y; }
} a[MAXN], b[MAXN], aa[MAXN], bb[MAXN];
bool cmp(const Point &a, const Point &b) {
return a.x == b.x ? a.y < b.y : a.x < b.x;
}
long long det(const Point &a, const Point &b) {
return 1ll * a.x * b.y - 1ll * a.y * b.x;
}
int calc(Point *a, Point *b, int n) {
sort(a + 1, a + n + 1, cmp);
int top = 1;
b[1] = a[1];
for (int i = 2; i <= n; ++i) {
while (top >= 2 && det(b[top] - b[top - 1], a[i] - b[top]) <= 0) --top;
b[++top] = a[i];
}
int lst = top;
for (int i = n - 1; i; --i) {
while (top > lst && det(b[top] - b[top - 1], a[i] - b[top]) <= 0) --top;
b[++top] = a[i];
}
return --top;
}
int n, m;
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i) {
scanf("%d%d", &aa[i].x, &aa[i].y);
}
for (int i = 1; i <= m; ++i) {
scanf("%d%d", &bb[i].x, &bb[i].y);
}
n = calc(aa, a, n);
m = calc(bb, b, m);
if (n != m) {
puts("NO");
return 0;
}
for (int i = 1; i <= n; ++i) {
bool flag = 1;
for (int j = 1, st = i, lst = n, nxt = 2, lst1 = (st == 1 ? n : st - 1),
nxt1 = (st == n ? 1 : st + 1);
j <= n;
++j, st = (st % n) + 1, lst = (lst % n) + 1, lst1 = (lst1 % n) + 1,
nxt = (nxt % n) + 1, nxt1 = (nxt1 % n) + 1) {
if ((a[j] - a[lst]).len2() != (b[st] - b[lst1]).len2()) {
flag = 0;
break;
}
if (det(a[nxt] - a[j], a[j] - a[lst]) !=
det(b[nxt1] - b[st], b[st] - b[lst1])) {
flag = 0;
break;
}
}
if (flag) {
puts("YES");
return 0;
}
}
puts("NO");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100100;
int N[2];
complex<long long> P[2][MAXN];
complex<long long> H[2][MAXN];
int csz[2];
long long A[2][2 * MAXN];
int M = 0;
long long S[6 * MAXN];
int Z[6 * MAXN];
void Zalg() {
int L = 0, R = 0;
for (int i = 1; i < M; i++) {
if (i > R) {
L = R = i;
while (R < M && S[R] == S[R - L]) R++;
Z[i] = R - L;
R--;
} else {
int j = i - L;
if (Z[j] < R - i + 1)
Z[i] = Z[j];
else {
L = i;
while (R < M && S[R] == S[R - L]) R++;
Z[i] = R - L;
R--;
}
}
}
}
long long cross(complex<long long> a, complex<long long> b) {
return imag(conj(a) * b);
}
long long dot(complex<long long> a, complex<long long> b) {
return real(conj(a) * b);
}
bool compare_pt(complex<long long> a, complex<long long> b) {
return a.real() < b.real() || (a.real() == b.real() && a.imag() < b.imag());
}
void find_hull() {
for (int t = 0; t < 2; t++) {
sort(P[t], P[t] + N[t], compare_pt);
int sz = 0;
for (int i = 0; i < N[t]; i++) {
while (sz >= 2 &&
cross(H[t][sz - 2] - H[t][sz - 1], P[t][i] - H[t][sz - 1]) <= 0)
sz--;
H[t][sz] = P[t][i];
sz++;
}
for (int i = N[t] - 2, j = sz + 1; i >= 0; i--) {
while (sz >= j &&
cross(H[t][sz - 2] - H[t][sz - 1], P[t][i] - H[t][sz - 1]) <= 0)
sz--;
H[t][sz] = P[t][i];
sz++;
}
csz[t] = sz - 1;
}
}
int main() {
ios::sync_with_stdio(0);
cin >> N[0] >> N[1];
for (int t = 0; t < 2; t++)
for (int i = 0, x, y; i < N[t]; i++) {
cin >> x >> y;
P[t][i] = complex<long long>(x, y);
}
find_hull();
for (int t = 0; t < 2; t++) {
int sz = csz[t];
for (int i = 0; i < sz; i++) {
A[t][2 * i] = norm(H[t][i] - H[t][(i + 1) % sz]);
A[t][2 * i + 1] =
dot(H[t][(i + sz - 1) % sz] - H[t][i], H[t][(i + 1) % sz] - H[t][i]);
}
}
for (int i = 0; i < 2 * csz[0]; i++) S[M++] = A[0][i];
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2 * csz[1]; j++) S[M++] = A[1][j];
Zalg();
for (int i = 2 * csz[0]; i < 2 * csz[0] + 2 * csz[1]; i++)
if (Z[i] >= 2 * csz[0]) {
cout << "YES\n";
return 0;
}
cout << "NO\n";
return 0;
}
| ### Prompt
Your challenge is to write a cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100100;
int N[2];
complex<long long> P[2][MAXN];
complex<long long> H[2][MAXN];
int csz[2];
long long A[2][2 * MAXN];
int M = 0;
long long S[6 * MAXN];
int Z[6 * MAXN];
void Zalg() {
int L = 0, R = 0;
for (int i = 1; i < M; i++) {
if (i > R) {
L = R = i;
while (R < M && S[R] == S[R - L]) R++;
Z[i] = R - L;
R--;
} else {
int j = i - L;
if (Z[j] < R - i + 1)
Z[i] = Z[j];
else {
L = i;
while (R < M && S[R] == S[R - L]) R++;
Z[i] = R - L;
R--;
}
}
}
}
long long cross(complex<long long> a, complex<long long> b) {
return imag(conj(a) * b);
}
long long dot(complex<long long> a, complex<long long> b) {
return real(conj(a) * b);
}
bool compare_pt(complex<long long> a, complex<long long> b) {
return a.real() < b.real() || (a.real() == b.real() && a.imag() < b.imag());
}
void find_hull() {
for (int t = 0; t < 2; t++) {
sort(P[t], P[t] + N[t], compare_pt);
int sz = 0;
for (int i = 0; i < N[t]; i++) {
while (sz >= 2 &&
cross(H[t][sz - 2] - H[t][sz - 1], P[t][i] - H[t][sz - 1]) <= 0)
sz--;
H[t][sz] = P[t][i];
sz++;
}
for (int i = N[t] - 2, j = sz + 1; i >= 0; i--) {
while (sz >= j &&
cross(H[t][sz - 2] - H[t][sz - 1], P[t][i] - H[t][sz - 1]) <= 0)
sz--;
H[t][sz] = P[t][i];
sz++;
}
csz[t] = sz - 1;
}
}
int main() {
ios::sync_with_stdio(0);
cin >> N[0] >> N[1];
for (int t = 0; t < 2; t++)
for (int i = 0, x, y; i < N[t]; i++) {
cin >> x >> y;
P[t][i] = complex<long long>(x, y);
}
find_hull();
for (int t = 0; t < 2; t++) {
int sz = csz[t];
for (int i = 0; i < sz; i++) {
A[t][2 * i] = norm(H[t][i] - H[t][(i + 1) % sz]);
A[t][2 * i + 1] =
dot(H[t][(i + sz - 1) % sz] - H[t][i], H[t][(i + 1) % sz] - H[t][i]);
}
}
for (int i = 0; i < 2 * csz[0]; i++) S[M++] = A[0][i];
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2 * csz[1]; j++) S[M++] = A[1][j];
Zalg();
for (int i = 2 * csz[0]; i < 2 * csz[0] + 2 * csz[1]; i++)
if (Z[i] >= 2 * csz[0]) {
cout << "YES\n";
return 0;
}
cout << "NO\n";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18 + 100;
vector<int> prefix_function(string s) {
int n = (int)s.length();
vector<int> pi(n);
for (int i = 0; i < n; ++i)
for (int k = 0; k <= i; ++k)
if (s.substr(0, k) == s.substr(i - k + 1, k)) pi[i] = k;
return pi;
}
struct Point {
long long x;
long long y;
Point() {}
Point(long long xx, long long yy) {
x = xx;
y = yy;
}
};
int n, m;
vector<Point> v1;
vector<Point> v2;
struct Vector {
long long x;
long long y;
Vector() {}
Vector(Point a, Point b) {
x = b.x - a.x;
y = b.y - a.y;
}
long long d() { return x * x + y * y; }
};
vector<Point> vec;
long long operator%(Vector a, Vector b) { return a.x * b.y - a.y * b.x; }
bool cmt(const Point& a, const Point& b) {
if ((Vector(vec[0], a) % Vector(vec[0], b)) == 0LL) {
return (Vector(vec[0], a).d() < Vector(vec[0], b).d());
}
return ((Vector(vec[0], a) % Vector(vec[0], b)) > 0LL);
}
vector<Point> ob() {
for (int i = 1; i < (int)vec.size(); ++i) {
if (make_pair(vec[0].x, vec[0].y) < make_pair(vec[i].x, vec[i].y)) {
swap(vec[0], vec[i]);
}
}
sort(vec.begin() + 1, vec.end(), cmt);
vector<Point> ans = {vec[0]};
for (int i = 1; i < (int)vec.size(); ++i) {
while (((int)ans.size() > 1) &&
(Vector(ans[(int)ans.size() - 2], ans[(int)ans.size() - 1]) %
Vector(ans[(int)ans.size() - 1], vec[i]) <=
0)) {
ans.pop_back();
}
ans.push_back(vec[i]);
}
return ans;
}
int main() {
ios::sync_with_stdio(false);
cout.tie(0);
cin.tie(0);
cin >> n >> m;
vector<Point> v1(n), v2(m);
for (int i = 0; i < n; ++i) {
cin >> v1[i].x >> v1[i].y;
}
for (int i = 0; i < m; ++i) {
cin >> v2[i].x >> v2[i].y;
}
vec = v1;
v1 = ob();
vec = v2;
v2 = ob();
n = v1.size();
m = v2.size();
if (n != m) {
cout << "NO" << endl;
} else {
vector<pair<long long, long long>> vv1;
for (int i = 0; i < n; ++i) {
vv1.push_back(make_pair(
Vector(v1[i], v1[(i + 1) % n]).d(),
Vector(v1[i], v1[(i - 1 + n) % n]) % Vector(v1[i], v1[(i + 1) % n])));
}
vector<pair<long long, long long>> vv2;
for (int i = 0; i < n; ++i) {
vv2.push_back(make_pair(
Vector(v2[i], v2[(i + 1) % n]).d(),
Vector(v2[i], v2[(i - 1 + n) % n]) % Vector(v2[i], v2[(i + 1) % n])));
}
vv1.push_back({INF, INF});
for (int i = 0; i < 2 * n; ++i) {
vv1.push_back(vv2[i % n]);
}
int n = vv1.size();
vector<int> pi(n);
for (int i = 1; i < n; ++i) {
int j = pi[i - 1];
while (j > 0 && vv1[i] != vv1[j]) j = pi[j - 1];
if (vv1[i] == vv1[j]) ++j;
pi[i] = j;
}
for (int i = n / 3; i < n; ++i) {
if (pi[i] == (n / 3)) {
cout << "YES" << endl;
return 0;
}
}
cout << "NO" << endl;
}
return 0;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18 + 100;
vector<int> prefix_function(string s) {
int n = (int)s.length();
vector<int> pi(n);
for (int i = 0; i < n; ++i)
for (int k = 0; k <= i; ++k)
if (s.substr(0, k) == s.substr(i - k + 1, k)) pi[i] = k;
return pi;
}
struct Point {
long long x;
long long y;
Point() {}
Point(long long xx, long long yy) {
x = xx;
y = yy;
}
};
int n, m;
vector<Point> v1;
vector<Point> v2;
struct Vector {
long long x;
long long y;
Vector() {}
Vector(Point a, Point b) {
x = b.x - a.x;
y = b.y - a.y;
}
long long d() { return x * x + y * y; }
};
vector<Point> vec;
long long operator%(Vector a, Vector b) { return a.x * b.y - a.y * b.x; }
bool cmt(const Point& a, const Point& b) {
if ((Vector(vec[0], a) % Vector(vec[0], b)) == 0LL) {
return (Vector(vec[0], a).d() < Vector(vec[0], b).d());
}
return ((Vector(vec[0], a) % Vector(vec[0], b)) > 0LL);
}
vector<Point> ob() {
for (int i = 1; i < (int)vec.size(); ++i) {
if (make_pair(vec[0].x, vec[0].y) < make_pair(vec[i].x, vec[i].y)) {
swap(vec[0], vec[i]);
}
}
sort(vec.begin() + 1, vec.end(), cmt);
vector<Point> ans = {vec[0]};
for (int i = 1; i < (int)vec.size(); ++i) {
while (((int)ans.size() > 1) &&
(Vector(ans[(int)ans.size() - 2], ans[(int)ans.size() - 1]) %
Vector(ans[(int)ans.size() - 1], vec[i]) <=
0)) {
ans.pop_back();
}
ans.push_back(vec[i]);
}
return ans;
}
int main() {
ios::sync_with_stdio(false);
cout.tie(0);
cin.tie(0);
cin >> n >> m;
vector<Point> v1(n), v2(m);
for (int i = 0; i < n; ++i) {
cin >> v1[i].x >> v1[i].y;
}
for (int i = 0; i < m; ++i) {
cin >> v2[i].x >> v2[i].y;
}
vec = v1;
v1 = ob();
vec = v2;
v2 = ob();
n = v1.size();
m = v2.size();
if (n != m) {
cout << "NO" << endl;
} else {
vector<pair<long long, long long>> vv1;
for (int i = 0; i < n; ++i) {
vv1.push_back(make_pair(
Vector(v1[i], v1[(i + 1) % n]).d(),
Vector(v1[i], v1[(i - 1 + n) % n]) % Vector(v1[i], v1[(i + 1) % n])));
}
vector<pair<long long, long long>> vv2;
for (int i = 0; i < n; ++i) {
vv2.push_back(make_pair(
Vector(v2[i], v2[(i + 1) % n]).d(),
Vector(v2[i], v2[(i - 1 + n) % n]) % Vector(v2[i], v2[(i + 1) % n])));
}
vv1.push_back({INF, INF});
for (int i = 0; i < 2 * n; ++i) {
vv1.push_back(vv2[i % n]);
}
int n = vv1.size();
vector<int> pi(n);
for (int i = 1; i < n; ++i) {
int j = pi[i - 1];
while (j > 0 && vv1[i] != vv1[j]) j = pi[j - 1];
if (vv1[i] == vv1[j]) ++j;
pi[i] = j;
}
for (int i = n / 3; i < n; ++i) {
if (pi[i] == (n / 3)) {
cout << "YES" << endl;
return 0;
}
}
cout << "NO" << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200005, INF = 0x3f3f3f3f;
struct Pt {
long long x, y;
Pt() {}
Pt(int xx, int yy) {
x = xx;
y = yy;
}
Pt operator+(const Pt &o) { return Pt(x + o.x, y + o.y); }
Pt operator-(const Pt &o) { return Pt(x - o.x, y - o.y); }
long long operator*(const Pt &o) { return x * o.x + y * o.y; }
long long operator^(const Pt &o) { return x * o.y - y * o.x; }
} A[MAXN], B[MAXN], st[MAXN];
bool cmp(Pt a, Pt b) {
long long t = a ^ b;
return t == 0 ? a * a < b * b : t > 0;
}
int n, m;
long long sa[MAXN], sb[MAXN];
int hull(int n, Pt *P, long long res[]) {
Pt O(INF, INF);
for (int i = 0; i < n; ++i) {
if (P[i].x < O.x || P[i].x == O.x && P[i].y < O.y) O = P[i];
}
for (int i = 0; i < n; ++i) P[i] = P[i] - O;
sort(P, P + n, cmp);
int top = 0;
st[top++] = P[0];
st[top++] = P[1];
for (int i = 2; i < n; ++i) {
while (top >= 2 &&
((st[top - 1] - st[top - 2]) ^ (P[i] - st[top - 1])) <= 0)
top--;
st[top++] = P[i];
}
int cnt = 0;
for (int i = 0; i < top; ++i) {
int j = (i + 1) % top, k = (i + 2) % top;
res[cnt++] = (st[j] - st[i]) * (st[j] - st[i]);
res[cnt++] = (st[j] - st[i]) * (st[k] - st[j]);
}
return cnt;
}
int repr(long long a[]) {
int i = 0, j = 1, k = 0;
while (i < n && j < n && k < n) {
long long t = a[(i + k) % n] - a[(j + k) % n];
if (!t)
k++;
else {
(t > 0 ? i : j) += k + 1;
if (i == j) ++j;
k = 0;
}
}
return min(i, j);
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 0; i < n; ++i) {
int x, y;
scanf("%d%d", &x, &y);
A[i] = Pt(x, y);
}
for (int i = 0; i < m; ++i) {
int x, y;
scanf("%d%d", &x, &y);
B[i] = Pt(x, y);
}
n = hull(n, A, sa);
m = hull(m, B, sb);
if (n != m) {
puts("NO");
return 0;
}
int ia = repr(sa), ib = repr(sb);
for (int i = 0; i < n; ++i) {
if (sa[(ia + i) % n] != sb[(ib + i) % n]) {
puts("NO");
return 0;
}
}
puts("YES");
return 0;
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200005, INF = 0x3f3f3f3f;
struct Pt {
long long x, y;
Pt() {}
Pt(int xx, int yy) {
x = xx;
y = yy;
}
Pt operator+(const Pt &o) { return Pt(x + o.x, y + o.y); }
Pt operator-(const Pt &o) { return Pt(x - o.x, y - o.y); }
long long operator*(const Pt &o) { return x * o.x + y * o.y; }
long long operator^(const Pt &o) { return x * o.y - y * o.x; }
} A[MAXN], B[MAXN], st[MAXN];
bool cmp(Pt a, Pt b) {
long long t = a ^ b;
return t == 0 ? a * a < b * b : t > 0;
}
int n, m;
long long sa[MAXN], sb[MAXN];
int hull(int n, Pt *P, long long res[]) {
Pt O(INF, INF);
for (int i = 0; i < n; ++i) {
if (P[i].x < O.x || P[i].x == O.x && P[i].y < O.y) O = P[i];
}
for (int i = 0; i < n; ++i) P[i] = P[i] - O;
sort(P, P + n, cmp);
int top = 0;
st[top++] = P[0];
st[top++] = P[1];
for (int i = 2; i < n; ++i) {
while (top >= 2 &&
((st[top - 1] - st[top - 2]) ^ (P[i] - st[top - 1])) <= 0)
top--;
st[top++] = P[i];
}
int cnt = 0;
for (int i = 0; i < top; ++i) {
int j = (i + 1) % top, k = (i + 2) % top;
res[cnt++] = (st[j] - st[i]) * (st[j] - st[i]);
res[cnt++] = (st[j] - st[i]) * (st[k] - st[j]);
}
return cnt;
}
int repr(long long a[]) {
int i = 0, j = 1, k = 0;
while (i < n && j < n && k < n) {
long long t = a[(i + k) % n] - a[(j + k) % n];
if (!t)
k++;
else {
(t > 0 ? i : j) += k + 1;
if (i == j) ++j;
k = 0;
}
}
return min(i, j);
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 0; i < n; ++i) {
int x, y;
scanf("%d%d", &x, &y);
A[i] = Pt(x, y);
}
for (int i = 0; i < m; ++i) {
int x, y;
scanf("%d%d", &x, &y);
B[i] = Pt(x, y);
}
n = hull(n, A, sa);
m = hull(m, B, sb);
if (n != m) {
puts("NO");
return 0;
}
int ia = repr(sa), ib = repr(sb);
for (int i = 0; i < n; ++i) {
if (sa[(ia + i) % n] != sb[(ib + i) % n]) {
puts("NO");
return 0;
}
}
puts("YES");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-12;
int dcmp(double x) {
if (x > eps) return 1;
if (x < -eps) return -1;
return 0;
}
struct point {
double x, y;
point() {}
point(double _x, double _y) : x(_x), y(_y) {}
void read() { scanf("%lf%lf", &x, &y); }
friend point operator+(const point &a, const point &b) {
return point(a.x + b.x, a.y + b.y);
}
friend point operator-(const point &a, const point &b) {
return point(a.x - b.x, a.y - b.y);
}
friend bool operator==(const point &a, const point &b) {
return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}
};
double det(const point &a, const point &b) { return a.x * b.y - a.y * b.x; }
double dot(const point &a, const point &b) { return a.x * b.x + a.y * b.y; }
struct polygon_convex {
vector<point> P;
polygon_convex(int size = 0) { P.resize(size); }
};
bool comp_less(const point &a, const point &b) {
if (a.x == b.x) return a.y < b.y;
return a.x < b.x;
}
polygon_convex convex_hull(vector<point> a) {
polygon_convex res(2 * a.size() + 5);
sort(a.begin(), a.end(), comp_less);
a.erase(unique(a.begin(), a.end()), a.end());
int m = 0;
for (int i = 0; i < a.size(); i++) {
while (m > 1 &&
dcmp(det(res.P[m - 1] - res.P[m - 2], a[i] - res.P[m - 2])) <= 0)
m--;
res.P[m++] = a[i];
}
int k = m;
for (int i = (int)a.size() - 2; i >= 0; i--) {
while (m > k &&
dcmp(det(res.P[m - 1] - res.P[m - 2], a[i] - res.P[m - 2])) <= 0)
m--;
res.P[m++] = a[i];
}
res.P.resize(m);
if (a.size() > 1) res.P.resize(m - 1);
return res;
}
vector<point> a, b;
int n, m, x, y;
int nxt[250005];
struct node {
double len, x, y;
node() {}
node(double _len, double _x, double _y) : len(_len), x(_x), y(_y) {}
friend bool operator==(const node &a, const node &b) {
return dcmp(a.len - b.len) == 0 && dcmp(a.x - b.x) == 0 &&
dcmp(a.y - b.y) == 0;
}
};
vector<node> Sub, Main;
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) {
scanf("%d%d", &x, &y);
a.push_back(point(x, y));
}
for (int i = 1; i <= m; i++) {
scanf("%d%d", &x, &y);
b.push_back(point(x, y));
}
polygon_convex ra = convex_hull(a);
polygon_convex rb = convex_hull(b);
if (ra.P.size() != rb.P.size()) {
return 0 * puts("NO");
}
n = (int)ra.P.size();
for (int i = 1; i <= n; i++) {
point A = ra.P[i % n] - ra.P[(i - 1) % n];
point B = ra.P[(i + 1) % n] - ra.P[i % n];
double len = sqrt(A.x * A.x + A.y * A.y);
double x0 = A.x, y0 = A.y;
double x1 = sqrt(A.x * A.x + A.y * A.y), y1 = 0;
double Sin = (x0 * y1 - x1 * y0) / (x0 * x0 + y0 * y0);
double Cos = (x0 * x1 + y0 * y1) / (x0 * x0 + y0 * y0);
double nx = B.x * Cos - B.y * Sin;
double ny = B.x * Sin + B.y * Cos;
Sub.push_back(node(len, nx, ny));
}
for (int i = 1; i <= n + n; i++) {
point A = rb.P[i % n] - rb.P[(i - 1) % n];
point B = rb.P[(i + 1) % n] - rb.P[i % n];
double len = sqrt(A.x * A.x + A.y * A.y);
double x0 = A.x, y0 = A.y;
double x1 = sqrt(A.x * A.x + A.y * A.y), y1 = 0;
double Sin = (x0 * y1 - x1 * y0) / (x0 * x0 + y0 * y0);
double Cos = (x0 * x1 + y0 * y1) / (x0 * x0 + y0 * y0);
double nx = B.x * Cos - B.y * Sin;
double ny = B.x * Sin + B.y * Cos;
Main.push_back(node(len, nx, ny));
}
int j = nxt[0] = -1, i = 0;
while (i < (int)Sub.size()) {
while (-1 != j && !(Sub[j] == Sub[i])) j = nxt[j];
i++;
j++;
nxt[i] = j;
}
i = j = 0;
while (i < (int)Main.size()) {
while (-1 != j && !(Main[i] == Sub[j])) j = nxt[j];
i++;
j++;
if (j == (int)Sub.size()) return 0 * puts("YES");
}
puts("NO");
return 0;
}
| ### Prompt
In cpp, your task is to solve the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-12;
int dcmp(double x) {
if (x > eps) return 1;
if (x < -eps) return -1;
return 0;
}
struct point {
double x, y;
point() {}
point(double _x, double _y) : x(_x), y(_y) {}
void read() { scanf("%lf%lf", &x, &y); }
friend point operator+(const point &a, const point &b) {
return point(a.x + b.x, a.y + b.y);
}
friend point operator-(const point &a, const point &b) {
return point(a.x - b.x, a.y - b.y);
}
friend bool operator==(const point &a, const point &b) {
return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}
};
double det(const point &a, const point &b) { return a.x * b.y - a.y * b.x; }
double dot(const point &a, const point &b) { return a.x * b.x + a.y * b.y; }
struct polygon_convex {
vector<point> P;
polygon_convex(int size = 0) { P.resize(size); }
};
bool comp_less(const point &a, const point &b) {
if (a.x == b.x) return a.y < b.y;
return a.x < b.x;
}
polygon_convex convex_hull(vector<point> a) {
polygon_convex res(2 * a.size() + 5);
sort(a.begin(), a.end(), comp_less);
a.erase(unique(a.begin(), a.end()), a.end());
int m = 0;
for (int i = 0; i < a.size(); i++) {
while (m > 1 &&
dcmp(det(res.P[m - 1] - res.P[m - 2], a[i] - res.P[m - 2])) <= 0)
m--;
res.P[m++] = a[i];
}
int k = m;
for (int i = (int)a.size() - 2; i >= 0; i--) {
while (m > k &&
dcmp(det(res.P[m - 1] - res.P[m - 2], a[i] - res.P[m - 2])) <= 0)
m--;
res.P[m++] = a[i];
}
res.P.resize(m);
if (a.size() > 1) res.P.resize(m - 1);
return res;
}
vector<point> a, b;
int n, m, x, y;
int nxt[250005];
struct node {
double len, x, y;
node() {}
node(double _len, double _x, double _y) : len(_len), x(_x), y(_y) {}
friend bool operator==(const node &a, const node &b) {
return dcmp(a.len - b.len) == 0 && dcmp(a.x - b.x) == 0 &&
dcmp(a.y - b.y) == 0;
}
};
vector<node> Sub, Main;
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) {
scanf("%d%d", &x, &y);
a.push_back(point(x, y));
}
for (int i = 1; i <= m; i++) {
scanf("%d%d", &x, &y);
b.push_back(point(x, y));
}
polygon_convex ra = convex_hull(a);
polygon_convex rb = convex_hull(b);
if (ra.P.size() != rb.P.size()) {
return 0 * puts("NO");
}
n = (int)ra.P.size();
for (int i = 1; i <= n; i++) {
point A = ra.P[i % n] - ra.P[(i - 1) % n];
point B = ra.P[(i + 1) % n] - ra.P[i % n];
double len = sqrt(A.x * A.x + A.y * A.y);
double x0 = A.x, y0 = A.y;
double x1 = sqrt(A.x * A.x + A.y * A.y), y1 = 0;
double Sin = (x0 * y1 - x1 * y0) / (x0 * x0 + y0 * y0);
double Cos = (x0 * x1 + y0 * y1) / (x0 * x0 + y0 * y0);
double nx = B.x * Cos - B.y * Sin;
double ny = B.x * Sin + B.y * Cos;
Sub.push_back(node(len, nx, ny));
}
for (int i = 1; i <= n + n; i++) {
point A = rb.P[i % n] - rb.P[(i - 1) % n];
point B = rb.P[(i + 1) % n] - rb.P[i % n];
double len = sqrt(A.x * A.x + A.y * A.y);
double x0 = A.x, y0 = A.y;
double x1 = sqrt(A.x * A.x + A.y * A.y), y1 = 0;
double Sin = (x0 * y1 - x1 * y0) / (x0 * x0 + y0 * y0);
double Cos = (x0 * x1 + y0 * y1) / (x0 * x0 + y0 * y0);
double nx = B.x * Cos - B.y * Sin;
double ny = B.x * Sin + B.y * Cos;
Main.push_back(node(len, nx, ny));
}
int j = nxt[0] = -1, i = 0;
while (i < (int)Sub.size()) {
while (-1 != j && !(Sub[j] == Sub[i])) j = nxt[j];
i++;
j++;
nxt[i] = j;
}
i = j = 0;
while (i < (int)Main.size()) {
while (-1 != j && !(Main[i] == Sub[j])) j = nxt[j];
i++;
j++;
if (j == (int)Sub.size()) return 0 * puts("YES");
}
puts("NO");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1E5 + 10;
pair<int64_t, int64_t> p1[MAXN], p2[MAXN], hull1[MAXN], hull2[MAXN],
norm1[MAXN * 3], norm2[MAXN];
int z[MAXN];
int64_t dist(pair<int64_t, int64_t> a, pair<int64_t, int64_t> b) {
return (a.first - b.first) * (a.first - b.first) +
(a.second - b.second) * (a.second - b.second);
}
int64_t cross(pair<int64_t, int64_t> o, pair<int64_t, int64_t> a,
pair<int64_t, int64_t> b) {
return (a.first - o.first) * (b.second - o.second) -
(a.second - o.second) * (b.first - o.first);
}
int convexHull(pair<int64_t, int64_t> *p, pair<int64_t, int64_t> *hull, int n) {
sort(p, p + n);
int m = 0;
for (int i = 0; i < n; i++) {
while (m >= 2 && cross(hull[m - 2], hull[m - 1], p[i]) <= 0) m--;
hull[m++] = p[i];
}
for (int i = n - 2, t = m; i >= 0; i--) {
while (m > t && cross(hull[m - 2], hull[m - 1], p[i]) <= 0) m--;
hull[m++] = p[i];
}
return m - 1;
}
void normal(pair<int64_t, int64_t> *hull, pair<int64_t, int64_t> *norm, int n) {
hull[n] = hull[0], hull[n + 1] = hull[1];
for (int i = 0; i < n; i++) {
norm[i] = {dist(hull[i + 1], hull[i + 2]),
cross(hull[i], hull[i + 1], hull[i + 2])};
}
}
void zalg(pair<int64_t, int64_t> *str, int len) {
int l = 0, r = 0;
for (int i = 0; i < len; i++) {
if (i > r) {
l = r = i;
while (r < len && str[r] == str[r - l]) {
r++;
}
z[i] = r - l;
r--;
} else {
if (z[i - l] < r - i + 1) {
z[i] = z[i - l];
} else {
l = i;
while (r < len && str[r] == str[r - l]) {
r++;
}
z[i] = r - l;
r--;
}
}
}
}
int main() {
cin.tie(0), ios_base::sync_with_stdio(0);
int n, m;
cin >> n >> m;
for (int i = 0; i < n; i++) cin >> p1[i].first >> p1[i].second;
for (int i = 0; i < m; i++) cin >> p2[i].first >> p2[i].second;
n = convexHull(p1, hull1, n), m = convexHull(p2, hull2, m);
if (n != m) {
cout << "NO";
return 0;
}
normal(hull1, norm1, n), normal(hull2, norm2, m);
for (int i = 0; i < n; i++) norm1[n + i] = norm1[n + n + i] = norm2[i];
zalg(norm1, n * 3);
cout << (*max_element(z + 1, z + n * 3) >= n ? "YES" : "NO");
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1E5 + 10;
pair<int64_t, int64_t> p1[MAXN], p2[MAXN], hull1[MAXN], hull2[MAXN],
norm1[MAXN * 3], norm2[MAXN];
int z[MAXN];
int64_t dist(pair<int64_t, int64_t> a, pair<int64_t, int64_t> b) {
return (a.first - b.first) * (a.first - b.first) +
(a.second - b.second) * (a.second - b.second);
}
int64_t cross(pair<int64_t, int64_t> o, pair<int64_t, int64_t> a,
pair<int64_t, int64_t> b) {
return (a.first - o.first) * (b.second - o.second) -
(a.second - o.second) * (b.first - o.first);
}
int convexHull(pair<int64_t, int64_t> *p, pair<int64_t, int64_t> *hull, int n) {
sort(p, p + n);
int m = 0;
for (int i = 0; i < n; i++) {
while (m >= 2 && cross(hull[m - 2], hull[m - 1], p[i]) <= 0) m--;
hull[m++] = p[i];
}
for (int i = n - 2, t = m; i >= 0; i--) {
while (m > t && cross(hull[m - 2], hull[m - 1], p[i]) <= 0) m--;
hull[m++] = p[i];
}
return m - 1;
}
void normal(pair<int64_t, int64_t> *hull, pair<int64_t, int64_t> *norm, int n) {
hull[n] = hull[0], hull[n + 1] = hull[1];
for (int i = 0; i < n; i++) {
norm[i] = {dist(hull[i + 1], hull[i + 2]),
cross(hull[i], hull[i + 1], hull[i + 2])};
}
}
void zalg(pair<int64_t, int64_t> *str, int len) {
int l = 0, r = 0;
for (int i = 0; i < len; i++) {
if (i > r) {
l = r = i;
while (r < len && str[r] == str[r - l]) {
r++;
}
z[i] = r - l;
r--;
} else {
if (z[i - l] < r - i + 1) {
z[i] = z[i - l];
} else {
l = i;
while (r < len && str[r] == str[r - l]) {
r++;
}
z[i] = r - l;
r--;
}
}
}
}
int main() {
cin.tie(0), ios_base::sync_with_stdio(0);
int n, m;
cin >> n >> m;
for (int i = 0; i < n; i++) cin >> p1[i].first >> p1[i].second;
for (int i = 0; i < m; i++) cin >> p2[i].first >> p2[i].second;
n = convexHull(p1, hull1, n), m = convexHull(p2, hull2, m);
if (n != m) {
cout << "NO";
return 0;
}
normal(hull1, norm1, n), normal(hull2, norm2, m);
for (int i = 0; i < n; i++) norm1[n + i] = norm1[n + n + i] = norm2[i];
zalg(norm1, n * 3);
cout << (*max_element(z + 1, z + n * 3) >= n ? "YES" : "NO");
}
``` |
#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > points;
pair<int, int> pivot;
vector<pair<int, int> > ch1, ch2;
inline bool deq(const double &a, const double &b) {
return fabs(a - b) < 1e-10;
}
bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
if (a.first == b.first) return a.second < b.second;
return a.first < b.first;
}
bool cmp1(const pair<int, int> &a, const pair<int, int> &b) {
double ma;
if (a.first == pivot.first)
ma = 1e10;
else
ma = (a.second - pivot.second) / ((double)(a.first - pivot.first));
double mb;
if (b.first == pivot.first)
mb = 1e10;
else
mb = (b.second - pivot.second) / ((double)(b.first - pivot.first));
double dista = (sqrt(((long long)a.first - pivot.first) *
((long long)a.first - pivot.first) +
((long long)a.second - pivot.second) *
((long long)a.second - pivot.second)));
double distb = (sqrt(((long long)b.first - pivot.first) *
((long long)b.first - pivot.first) +
((long long)b.second - pivot.second) *
((long long)b.second - pivot.second)));
if (deq(ma, mb)) return dista > distb;
return ma < mb;
}
bool cmp2(const pair<int, int> &a, const pair<int, int> &b) {
double ma;
if (a.first == pivot.first)
ma = 1e10;
else
ma = (a.second - pivot.second) / ((double)(a.first - pivot.first));
double mb;
if (b.first == pivot.first)
mb = 1e10;
else
mb = (b.second - pivot.second) / ((double)(b.first - pivot.first));
double dista = (sqrt(((long long)a.first - pivot.first) *
((long long)a.first - pivot.first) +
((long long)a.second - pivot.second) *
((long long)a.second - pivot.second)));
double distb = (sqrt(((long long)b.first - pivot.first) *
((long long)b.first - pivot.first) +
((long long)b.second - pivot.second) *
((long long)b.second - pivot.second)));
if (deq(ma, mb)) return dista < distb;
return ma < mb;
}
double cross(pair<int, int> a, pair<int, int> b) {
return (double)a.first * (double)b.second -
(double)a.second * (double)b.first;
}
bool ccw(pair<int, int> a, pair<int, int> b, pair<int, int> c) {
pair<int, int> ab = make_pair(b.first - a.first, b.second - a.second);
pair<int, int> ac = make_pair(c.first - a.first, c.second - a.second);
double x = cross(ab, ac);
return x > 0 && !deq(x, 0);
}
double get_slope(pair<int, int> a, pair<int, int> b) {
double ret;
if (a.first != b.first)
ret = (a.second - b.second) / ((double)(a.first - b.first));
else
ret = 1e10;
return ret;
}
void fcvxh(vector<pair<int, int> > &ret) {
sort(points.begin(), points.end(), cmp);
pivot = points[0];
sort(points.begin() + 1, points.end(), cmp1);
double fixangle = get_slope(points[1], pivot);
int fix = 2;
while (fix < points.size() && deq(fixangle, get_slope(points[fix], pivot)))
fix++;
sort(points.begin() + 1, points.begin() + fix, cmp2);
stack<pair<int, int> > C;
C.push(pivot);
C.push(points[fix - 1]);
for (int i = fix; i < points.size(); i++) {
pair<int, int> b = C.top();
C.pop();
pair<int, int> a = C.top();
C.pop();
pair<int, int> c = points[i];
if (ccw(a, b, c)) {
C.push(a);
C.push(b);
C.push(c);
} else {
C.push(a);
i--;
}
}
while (true) {
pair<int, int> b = C.top();
C.pop();
pair<int, int> a = C.top();
C.pop();
pair<int, int> c = pivot;
if (ccw(a, b, c)) {
C.push(a);
C.push(b);
break;
} else if (deq(fixangle, get_slope(b, pivot))) {
C.push(a);
C.push(b);
break;
} else {
C.push(a);
}
}
while (!C.empty()) {
ret.push_back(C.top());
C.pop();
}
reverse(ret.begin(), ret.end());
}
int fix(int x, int m) { return ((x % m) + m) % m; }
double dot(pair<int, int> a, pair<int, int> b) {
return (double)a.first * (double)b.first +
(double)a.second * (double)b.second;
}
bool dodeq(pair<double, double> a, pair<double, double> b) {
return deq(a.first, b.first) && deq(a.second, b.second);
}
int rew[100010];
bool cyclic_equals(vector<pair<double, double> > S1,
vector<pair<double, double> > S2) {
assert(S1.size() <= 100000);
assert(S2.size() <= 100000);
if (S1.size() != S2.size()) return false;
vector<pair<double, double> > S;
for (int i = 0; i < S2.size(); i++) S.push_back(S2[i]);
for (int i = 0; i < S2.size(); i++) S.push_back(S2[i]);
rew[0] = 0;
rew[1] = 0;
for (int i = 2; i <= S1.size(); i++) {
int j = rew[i - 1];
while (j > 0 && !dodeq(S1[j], S1[i - 1])) j = rew[j];
if (j > 0)
rew[i] = j + 1;
else if (i > 1 && dodeq(S1[0], S1[i - 1]))
rew[i] = 1;
else
rew[i] = 0;
}
bool flag = false;
int state = 0;
for (int i = 0; i < S.size(); i++) {
pair<double, double> c = S[i];
if (dodeq(c, S1[state]))
state++;
else {
while (state != 0 && !dodeq(c, S1[state])) state = rew[state];
if (state != 0 || dodeq(c, S1[state])) state++;
}
if (state == S1.size()) {
flag = true;
break;
}
assert(state < S1.size());
}
return flag;
}
int main(int argc, char **argv) {
ios_base::sync_with_stdio(0);
cin.tie(0);
int N1, N2;
cin >> N1 >> N2;
for (int i = 0; i < N1; i++) {
int a, b;
cin >> a >> b;
points.push_back(make_pair(a, b));
}
fcvxh(ch1);
points.clear();
for (int i = 0; i < N2; i++) {
int a, b;
cin >> a >> b;
points.push_back(make_pair(a, b));
}
fcvxh(ch2);
vector<pair<double, double> > T1;
for (int i = 0; i < ch1.size(); i++) {
pair<int, int> cur = ch1[i];
pair<int, int> nxt = ch1[fix(i + 1, ch1.size())];
pair<int, int> pre = ch1[fix(i - 1, ch1.size())];
double len = (sqrt(((long long)cur.first - nxt.first) *
((long long)cur.first - nxt.first) +
((long long)cur.second - nxt.second) *
((long long)cur.second - nxt.second)));
pair<int, int> cn =
make_pair(nxt.first - cur.first, nxt.second - cur.second);
pair<int, int> pc =
make_pair(cur.first - pre.first, cur.second - pre.second);
double ang = atan2(cross(pc, cn), dot(pc, cn));
T1.push_back(make_pair(len, ang));
}
vector<pair<double, double> > T2;
for (int i = 0; i < ch2.size(); i++) {
pair<int, int> cur = ch2[i];
pair<int, int> nxt = ch2[fix(i + 1, ch2.size())];
pair<int, int> pre = ch2[fix(i - 1, ch2.size())];
double len = (sqrt(((long long)cur.first - nxt.first) *
((long long)cur.first - nxt.first) +
((long long)cur.second - nxt.second) *
((long long)cur.second - nxt.second)));
pair<int, int> cn =
make_pair(nxt.first - cur.first, nxt.second - cur.second);
pair<int, int> pc =
make_pair(cur.first - pre.first, cur.second - pre.second);
double ang = atan2(cross(pc, cn), dot(pc, cn));
T2.push_back(make_pair(len, ang));
}
if (cyclic_equals(T1, T2)) {
cout << "YES" << endl;
} else {
cout << "NO" << endl;
}
return 0;
}
| ### Prompt
Generate a cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > points;
pair<int, int> pivot;
vector<pair<int, int> > ch1, ch2;
inline bool deq(const double &a, const double &b) {
return fabs(a - b) < 1e-10;
}
bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
if (a.first == b.first) return a.second < b.second;
return a.first < b.first;
}
bool cmp1(const pair<int, int> &a, const pair<int, int> &b) {
double ma;
if (a.first == pivot.first)
ma = 1e10;
else
ma = (a.second - pivot.second) / ((double)(a.first - pivot.first));
double mb;
if (b.first == pivot.first)
mb = 1e10;
else
mb = (b.second - pivot.second) / ((double)(b.first - pivot.first));
double dista = (sqrt(((long long)a.first - pivot.first) *
((long long)a.first - pivot.first) +
((long long)a.second - pivot.second) *
((long long)a.second - pivot.second)));
double distb = (sqrt(((long long)b.first - pivot.first) *
((long long)b.first - pivot.first) +
((long long)b.second - pivot.second) *
((long long)b.second - pivot.second)));
if (deq(ma, mb)) return dista > distb;
return ma < mb;
}
bool cmp2(const pair<int, int> &a, const pair<int, int> &b) {
double ma;
if (a.first == pivot.first)
ma = 1e10;
else
ma = (a.second - pivot.second) / ((double)(a.first - pivot.first));
double mb;
if (b.first == pivot.first)
mb = 1e10;
else
mb = (b.second - pivot.second) / ((double)(b.first - pivot.first));
double dista = (sqrt(((long long)a.first - pivot.first) *
((long long)a.first - pivot.first) +
((long long)a.second - pivot.second) *
((long long)a.second - pivot.second)));
double distb = (sqrt(((long long)b.first - pivot.first) *
((long long)b.first - pivot.first) +
((long long)b.second - pivot.second) *
((long long)b.second - pivot.second)));
if (deq(ma, mb)) return dista < distb;
return ma < mb;
}
double cross(pair<int, int> a, pair<int, int> b) {
return (double)a.first * (double)b.second -
(double)a.second * (double)b.first;
}
bool ccw(pair<int, int> a, pair<int, int> b, pair<int, int> c) {
pair<int, int> ab = make_pair(b.first - a.first, b.second - a.second);
pair<int, int> ac = make_pair(c.first - a.first, c.second - a.second);
double x = cross(ab, ac);
return x > 0 && !deq(x, 0);
}
double get_slope(pair<int, int> a, pair<int, int> b) {
double ret;
if (a.first != b.first)
ret = (a.second - b.second) / ((double)(a.first - b.first));
else
ret = 1e10;
return ret;
}
void fcvxh(vector<pair<int, int> > &ret) {
sort(points.begin(), points.end(), cmp);
pivot = points[0];
sort(points.begin() + 1, points.end(), cmp1);
double fixangle = get_slope(points[1], pivot);
int fix = 2;
while (fix < points.size() && deq(fixangle, get_slope(points[fix], pivot)))
fix++;
sort(points.begin() + 1, points.begin() + fix, cmp2);
stack<pair<int, int> > C;
C.push(pivot);
C.push(points[fix - 1]);
for (int i = fix; i < points.size(); i++) {
pair<int, int> b = C.top();
C.pop();
pair<int, int> a = C.top();
C.pop();
pair<int, int> c = points[i];
if (ccw(a, b, c)) {
C.push(a);
C.push(b);
C.push(c);
} else {
C.push(a);
i--;
}
}
while (true) {
pair<int, int> b = C.top();
C.pop();
pair<int, int> a = C.top();
C.pop();
pair<int, int> c = pivot;
if (ccw(a, b, c)) {
C.push(a);
C.push(b);
break;
} else if (deq(fixangle, get_slope(b, pivot))) {
C.push(a);
C.push(b);
break;
} else {
C.push(a);
}
}
while (!C.empty()) {
ret.push_back(C.top());
C.pop();
}
reverse(ret.begin(), ret.end());
}
int fix(int x, int m) { return ((x % m) + m) % m; }
double dot(pair<int, int> a, pair<int, int> b) {
return (double)a.first * (double)b.first +
(double)a.second * (double)b.second;
}
bool dodeq(pair<double, double> a, pair<double, double> b) {
return deq(a.first, b.first) && deq(a.second, b.second);
}
int rew[100010];
bool cyclic_equals(vector<pair<double, double> > S1,
vector<pair<double, double> > S2) {
assert(S1.size() <= 100000);
assert(S2.size() <= 100000);
if (S1.size() != S2.size()) return false;
vector<pair<double, double> > S;
for (int i = 0; i < S2.size(); i++) S.push_back(S2[i]);
for (int i = 0; i < S2.size(); i++) S.push_back(S2[i]);
rew[0] = 0;
rew[1] = 0;
for (int i = 2; i <= S1.size(); i++) {
int j = rew[i - 1];
while (j > 0 && !dodeq(S1[j], S1[i - 1])) j = rew[j];
if (j > 0)
rew[i] = j + 1;
else if (i > 1 && dodeq(S1[0], S1[i - 1]))
rew[i] = 1;
else
rew[i] = 0;
}
bool flag = false;
int state = 0;
for (int i = 0; i < S.size(); i++) {
pair<double, double> c = S[i];
if (dodeq(c, S1[state]))
state++;
else {
while (state != 0 && !dodeq(c, S1[state])) state = rew[state];
if (state != 0 || dodeq(c, S1[state])) state++;
}
if (state == S1.size()) {
flag = true;
break;
}
assert(state < S1.size());
}
return flag;
}
int main(int argc, char **argv) {
ios_base::sync_with_stdio(0);
cin.tie(0);
int N1, N2;
cin >> N1 >> N2;
for (int i = 0; i < N1; i++) {
int a, b;
cin >> a >> b;
points.push_back(make_pair(a, b));
}
fcvxh(ch1);
points.clear();
for (int i = 0; i < N2; i++) {
int a, b;
cin >> a >> b;
points.push_back(make_pair(a, b));
}
fcvxh(ch2);
vector<pair<double, double> > T1;
for (int i = 0; i < ch1.size(); i++) {
pair<int, int> cur = ch1[i];
pair<int, int> nxt = ch1[fix(i + 1, ch1.size())];
pair<int, int> pre = ch1[fix(i - 1, ch1.size())];
double len = (sqrt(((long long)cur.first - nxt.first) *
((long long)cur.first - nxt.first) +
((long long)cur.second - nxt.second) *
((long long)cur.second - nxt.second)));
pair<int, int> cn =
make_pair(nxt.first - cur.first, nxt.second - cur.second);
pair<int, int> pc =
make_pair(cur.first - pre.first, cur.second - pre.second);
double ang = atan2(cross(pc, cn), dot(pc, cn));
T1.push_back(make_pair(len, ang));
}
vector<pair<double, double> > T2;
for (int i = 0; i < ch2.size(); i++) {
pair<int, int> cur = ch2[i];
pair<int, int> nxt = ch2[fix(i + 1, ch2.size())];
pair<int, int> pre = ch2[fix(i - 1, ch2.size())];
double len = (sqrt(((long long)cur.first - nxt.first) *
((long long)cur.first - nxt.first) +
((long long)cur.second - nxt.second) *
((long long)cur.second - nxt.second)));
pair<int, int> cn =
make_pair(nxt.first - cur.first, nxt.second - cur.second);
pair<int, int> pc =
make_pair(cur.first - pre.first, cur.second - pre.second);
double ang = atan2(cross(pc, cn), dot(pc, cn));
T2.push_back(make_pair(len, ang));
}
if (cyclic_equals(T1, T2)) {
cout << "YES" << endl;
} else {
cout << "NO" << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void ControlIO();
void TimerStart();
void TimerStop();
const long long mod = 1000000007;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
typedef int coord_t;
typedef long long coord2_t;
struct Point {
coord_t x, y;
bool operator<(const Point &p) const {
return x < p.x || (x == p.x && y < p.y);
}
Point operator-(Point p) {
Point a = *this;
a.x -= p.x;
a.y -= p.y;
return a;
}
};
coord2_t cross(const Point &O, const Point &A, const Point &B) {
return (coord2_t)(A.x - O.x) * (B.y - O.y) -
(coord2_t)(A.y - O.y) * (B.x - O.x);
}
coord2_t dot(const Point A, const Point B) {
return (coord2_t)(A.x) * (B.x) + (coord2_t)(A.y) * (B.y);
}
coord2_t cross(const Point A, const Point B) {
return (coord2_t)(A.x) * (B.y) - (coord2_t)(A.y) * (B.x);
}
coord2_t len(const Point A, const Point B) {
return (coord2_t)(A.x - B.x) * (A.x - B.x) +
(coord2_t)(A.y - B.y) * (A.y - B.y);
}
vector<Point> convex_hull(vector<Point> &P) {
size_t n = P.size(), k = 0;
if (n <= 1) return P;
vector<Point> H(2 * n);
sort(P.begin(), P.end());
for (size_t i = 0; i < n; ++i) {
while (k >= 2 && cross(H[k - 2], H[k - 1], P[i]) <= 0) k--;
H[k++] = P[i];
}
for (size_t i = n - 1, t = k + 1; i > 0; --i) {
while (k >= t && cross(H[k - 2], H[k - 1], P[i - 1]) <= 0) k--;
H[k++] = P[i - 1];
}
H.resize(k - 1);
return H;
}
struct Data {
long long Dot, Len, Cross;
bool operator==(Data a) {
return (this->Dot == a.Dot) && (this->Cross == a.Cross) &&
(this->Len == a.Len);
}
bool operator!=(Data a) { return !(*this == a); }
};
vector<int> Prefix_function(vector<Data> &v) {
int n = v.size(), j;
vector<int> pi(n);
for (int i = 1; i < n; ++i) {
j = pi[i - 1];
;
;
while (j > 0 && (v[i] != v[j])) {
j = pi[j - 1];
;
}
if (v[i] == v[j]) j++;
;
pi[i] = j;
}
return pi;
}
bool find_occurences(vector<Data> &text, vector<Data> &pattern) {
vector<Data> v = pattern;
Data a;
a.Cross = a.Dot = a.Len = -1;
v.push_back(a);
v.insert(v.end(), text.begin(), text.end());
int sz1 = text.size(), sz2 = pattern.size();
vector<int> lps = Prefix_function(v);
for (int i = sz2 + 1; i <= sz1 + sz2; ++i) {
;
if (lps[i] == sz2) return true;
}
return false;
}
void solve() {
int n, m;
cin >> n >> m;
vector<Point> Poly1(n), Poly2(m);
for (auto &x : Poly1) cin >> x.x >> x.y;
for (auto &x : Poly2) cin >> x.x >> x.y;
Poly1 = convex_hull(Poly1);
Poly2 = convex_hull(Poly2);
;
if (Poly1.size() != Poly2.size()) {
cout << "NO";
return;
}
n = Poly1.size();
vector<Data> hull1, hull2;
for (int i = 0; i < n; ++i) {
Data curr;
Point a = Poly1[i];
Point b = Poly1[(1 + i) % n];
Point c = Poly1[(2 + i) % n];
curr.Len = len(b, a);
curr.Dot = dot(b - a, c - b);
curr.Cross = cross(b - a, c - b);
;
hull1.push_back(curr);
}
hull1.insert(hull1.end(), hull1.begin(), hull1.end());
;
for (int i = 0; i < n; ++i) {
Data curr;
Point a = Poly2[i];
Point b = Poly2[(1 + i) % n];
Point c = Poly2[(2 + i) % n];
curr.Len = len(b, a);
curr.Dot = dot(b - a, c - b);
curr.Cross = cross(b - a, c - b);
;
hull2.push_back(curr);
}
if (find_occurences(hull1, hull2))
cout << "YES\n";
else
cout << "NO\n";
}
int main() {
ControlIO();
int t = 1;
TimerStart();
while (t--) solve();
TimerStop();
return 0;
}
void ControlIO() {
ios_base::sync_with_stdio(0);
cin.tie(0);
}
auto TimeStart = chrono::steady_clock::now();
auto TimeEnd = chrono::steady_clock::now();
void TimerStart() {}
void TimerStop() {}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void ControlIO();
void TimerStart();
void TimerStop();
const long long mod = 1000000007;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
typedef int coord_t;
typedef long long coord2_t;
struct Point {
coord_t x, y;
bool operator<(const Point &p) const {
return x < p.x || (x == p.x && y < p.y);
}
Point operator-(Point p) {
Point a = *this;
a.x -= p.x;
a.y -= p.y;
return a;
}
};
coord2_t cross(const Point &O, const Point &A, const Point &B) {
return (coord2_t)(A.x - O.x) * (B.y - O.y) -
(coord2_t)(A.y - O.y) * (B.x - O.x);
}
coord2_t dot(const Point A, const Point B) {
return (coord2_t)(A.x) * (B.x) + (coord2_t)(A.y) * (B.y);
}
coord2_t cross(const Point A, const Point B) {
return (coord2_t)(A.x) * (B.y) - (coord2_t)(A.y) * (B.x);
}
coord2_t len(const Point A, const Point B) {
return (coord2_t)(A.x - B.x) * (A.x - B.x) +
(coord2_t)(A.y - B.y) * (A.y - B.y);
}
vector<Point> convex_hull(vector<Point> &P) {
size_t n = P.size(), k = 0;
if (n <= 1) return P;
vector<Point> H(2 * n);
sort(P.begin(), P.end());
for (size_t i = 0; i < n; ++i) {
while (k >= 2 && cross(H[k - 2], H[k - 1], P[i]) <= 0) k--;
H[k++] = P[i];
}
for (size_t i = n - 1, t = k + 1; i > 0; --i) {
while (k >= t && cross(H[k - 2], H[k - 1], P[i - 1]) <= 0) k--;
H[k++] = P[i - 1];
}
H.resize(k - 1);
return H;
}
struct Data {
long long Dot, Len, Cross;
bool operator==(Data a) {
return (this->Dot == a.Dot) && (this->Cross == a.Cross) &&
(this->Len == a.Len);
}
bool operator!=(Data a) { return !(*this == a); }
};
vector<int> Prefix_function(vector<Data> &v) {
int n = v.size(), j;
vector<int> pi(n);
for (int i = 1; i < n; ++i) {
j = pi[i - 1];
;
;
while (j > 0 && (v[i] != v[j])) {
j = pi[j - 1];
;
}
if (v[i] == v[j]) j++;
;
pi[i] = j;
}
return pi;
}
bool find_occurences(vector<Data> &text, vector<Data> &pattern) {
vector<Data> v = pattern;
Data a;
a.Cross = a.Dot = a.Len = -1;
v.push_back(a);
v.insert(v.end(), text.begin(), text.end());
int sz1 = text.size(), sz2 = pattern.size();
vector<int> lps = Prefix_function(v);
for (int i = sz2 + 1; i <= sz1 + sz2; ++i) {
;
if (lps[i] == sz2) return true;
}
return false;
}
void solve() {
int n, m;
cin >> n >> m;
vector<Point> Poly1(n), Poly2(m);
for (auto &x : Poly1) cin >> x.x >> x.y;
for (auto &x : Poly2) cin >> x.x >> x.y;
Poly1 = convex_hull(Poly1);
Poly2 = convex_hull(Poly2);
;
if (Poly1.size() != Poly2.size()) {
cout << "NO";
return;
}
n = Poly1.size();
vector<Data> hull1, hull2;
for (int i = 0; i < n; ++i) {
Data curr;
Point a = Poly1[i];
Point b = Poly1[(1 + i) % n];
Point c = Poly1[(2 + i) % n];
curr.Len = len(b, a);
curr.Dot = dot(b - a, c - b);
curr.Cross = cross(b - a, c - b);
;
hull1.push_back(curr);
}
hull1.insert(hull1.end(), hull1.begin(), hull1.end());
;
for (int i = 0; i < n; ++i) {
Data curr;
Point a = Poly2[i];
Point b = Poly2[(1 + i) % n];
Point c = Poly2[(2 + i) % n];
curr.Len = len(b, a);
curr.Dot = dot(b - a, c - b);
curr.Cross = cross(b - a, c - b);
;
hull2.push_back(curr);
}
if (find_occurences(hull1, hull2))
cout << "YES\n";
else
cout << "NO\n";
}
int main() {
ControlIO();
int t = 1;
TimerStart();
while (t--) solve();
TimerStop();
return 0;
}
void ControlIO() {
ios_base::sync_with_stdio(0);
cin.tie(0);
}
auto TimeStart = chrono::steady_clock::now();
auto TimeEnd = chrono::steady_clock::now();
void TimerStart() {}
void TimerStop() {}
``` |
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-6, pi = acos(-1.0);
inline int dcmp(double x) {
if (fabs(x) < eps) return 0;
return x < 0 ? -1 : 1;
}
inline double sqr(double x) { return x * x; }
struct Point {
double x, y;
Point(double _x = 0, double _y = 0) : x(_x), y(_y){};
inline void in() { scanf("%lf%lf", &x, &y); }
};
inline Point operator+(Point A, Point B) { return Point(A.x + B.x, A.y + B.y); }
inline Point operator-(Point A, Point B) { return Point(A.x - B.x, A.y - B.y); }
inline Point operator*(Point A, double p) { return Point(A.x * p, A.y * p); }
inline Point operator/(Point A, double p) { return Point(A.x / p, A.y / p); }
inline bool operator<(const Point &a, const Point &b) {
return a.x < b.x || (a.x == b.x && a.y < b.y);
}
inline bool operator==(const Point &a, const Point &b) {
return !dcmp(a.x - b.x) && !dcmp(a.y - b.y);
}
inline double Distance(Point A, Point B) {
return sqrt(sqr(A.x - B.x) + sqr(A.y - B.y));
}
inline double Dot(Point A, Point B) { return A.x * B.x + A.y * B.y; }
inline double Norm(Point A) { return A.x * A.x + A.y * A.y; }
inline double Length(Point A) { return sqrt(Dot(A, A)); }
inline double GetAngle(Point A, Point B) {
return acos(Dot(A, B) / Length(A) / Length(B));
}
inline double GetAngle(Point v) { return atan2(v.y, v.x); }
inline double Cross(Point A, Point B) { return A.x * B.y - A.y * B.x; }
inline Point Unit(Point x) { return x / Length(x); }
inline Point Normal(Point x) { return Point(-x.y, x.x) / Length(x); }
inline Point Rotate(Point A, double rad) {
return Point(A.x * cos(rad) - A.y * sin(rad),
A.x * sin(rad) + A.y * cos(rad));
}
inline double Area2(const Point &A, const Point &B, const Point &C) {
return Cross(B - A, C - A);
}
inline void getLineGeneralEquation(const Point &p1, const Point &p2, double &a,
double &b, double &c) {
a = p2.y - p1.y;
b = p1.x - p2.x;
c = -a * p1.x - b * p1.y;
}
inline Point GetLineIntersection(Point P, Point v, Point Q, Point w) {
Point u = P - Q;
double t = Cross(w, u) / Cross(v, w);
return P + v * t;
}
inline double DistanceToLine(Point P, Point A, Point B) {
Point v1 = B - A, v2 = P - A;
return fabs(Cross(v1, v2)) / Length(v1);
}
inline double DistanceToSegment(Point P, Point A, Point B) {
if (A == B) return Length(P - A);
Point v1 = B - A, v2 = P - A, v3 = P - B;
if (dcmp(Dot(v1, v2)) < 0)
return Length(v2);
else if (dcmp(Dot(v1, v3)) > 0)
return Length(v3);
else
return fabs(Cross(v1, v2)) / Length(v1);
}
inline Point GetLineProjection(Point P, Point A, Point B) {
Point v = B - A;
return A + v * (Dot(v, P - A) / Dot(v, v));
}
inline bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) {
double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1);
double c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);
return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
}
inline bool OnSegment(Point p, Point a1, Point a2) {
return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0;
}
inline double PolygonArea(Point *p, int n) {
double area = 0.0;
for (int i = 1; i < n - 1; i++) area += Cross(p[i] - p[0], p[i + 1] - p[0]);
return area / 2.0;
}
inline Point GetMidPoint(Point A, Point B) {
return Point((A.x + B.x) / 2.0, (A.y + B.y) / 2.0);
}
inline int PointinPolygon(Point p, vector<Point> poly) {
int n = poly.size(), flg = 0;
for (int i = 0; i < n; i++) {
if (OnSegment(p, poly[i], poly[(i + 1) % n])) return -1;
int k = dcmp(Cross(poly[(i + 1) % n] - poly[i], p - poly[i]));
int d1 = dcmp(poly[i].y - p.y);
int d2 = dcmp(poly[(i + 1) % n].y - p.y);
if (k > 0 && d1 <= 0 && d2 > 0) flg++;
if (k < 0 && d2 <= 0 && d1 > 0) flg++;
}
if (flg) return 1;
return 0;
}
struct Segment {
Point p, v;
Segment() {}
Segment(Point _p, Point _v) : p(_p), v(_v) {}
inline void in() {
p.in();
v.in();
}
};
struct Line {
Point p, v;
double ang;
Line() {}
Line(Point _p, Point _v) : p(_p), v(_v) { ang = atan2(v.y, v.x); }
inline Point point(double a) { return p + v * a; }
inline bool operator<(const Line &rhs) const { return ang < rhs.ang; }
inline Line move(double d) { return Line(p + Normal(v) * d, v); }
inline void in() {
p.in();
v.in();
ang = atan2(v.y, v.x);
}
};
inline Line LineTransHorizon(Line l, int d) {
Point vl = Normal(l.v);
Point p1 = l.p + vl * d;
return Line(p1, l.v);
}
inline Point GetLineIntersection(Line a, Line b) {
return GetLineIntersection(a.p, a.v, b.p, b.v);
}
inline bool OnLeft(const Line &L, const Point &p) {
return Cross(L.v, p - L.p) > 0;
}
inline Line GetParallel(Point p, Line l) { return Line(p, l.v); }
inline Line GetVertical(Point p, Line l) { return Line(p, Normal(l.v)); }
inline Point GetVerticalFoot(Point p, Line l) {
return GetLineIntersection(GetVertical(p, l), l);
}
inline Line GetPerpendicularLine(Segment l) {
return GetVertical(GetMidPoint(l.p, l.v - l.p), Line(l.p, l.v));
}
struct Angle {
Point p, u, v;
double ang;
Angle() {}
Angle(Point _p, Point _u, Point _v) : p(_p), u(_u), v(_v) {
ang = GetAngle(u, v);
}
inline void in() {
p.in();
u.in();
v.in();
ang = GetAngle(u, v);
}
};
inline int ConvexHull(Point *p, int n, Point *ans) {
sort(p, p + n);
int m = 0;
for (int i = 0; i < n; i++) {
while (m > 1 && Cross(ans[m - 1] - ans[m - 2], p[i] - ans[m - 2]) <= 0) m--;
ans[m++] = p[i];
}
int k = m;
for (int i = n - 2; ~i; i--) {
while (m > k && Cross(ans[m - 1] - ans[m - 2], p[i] - ans[m - 2]) <= 0) m--;
ans[m++] = p[i];
}
if (n > 1) m--;
return m;
}
inline long long distan(Point p, Point q) {
return (p.x - q.x) * (p.x - q.x) + (p.y - q.y) * (p.y - q.y);
}
Point p[100005], q[100005], cp[200005], cq[200005];
int n, m, lx, ly, fail[100005];
long long dp[100005], dq[200005];
double ap[100005], aq[200005];
bool vis[100005];
int main() {
scanf("%d%d", &n, &m);
for (int i = 0; i < n; ++i) p[i].in();
for (int i = 0; i < m; ++i) q[i].in();
lx = ConvexHull(p, n, cp);
ly = ConvexHull(q, m, cq);
if (lx != ly) return 0 * puts("NO");
cp[lx] = cp[0], lx++;
cp[lx] = cp[1];
for (int i = 0; i < ly; ++i) cq[i + ly] = cq[i];
ly <<= 1;
cq[ly] = cq[0], ly++;
cq[ly] = cq[1];
for (int i = 1; i < lx; ++i) dp[i] = distan(cp[i], cp[i - 1]);
for (int i = 1; i < ly; ++i) dq[i] = distan(cq[i], cq[i - 1]);
for (int i = 1; i < lx; ++i)
ap[i] = GetAngle(cp[i - 1] - cp[i], cp[i + 1] - cp[i]);
for (int i = 1; i < ly; ++i)
aq[i] = GetAngle(cq[i - 1] - cq[i], cq[i + 1] - cq[i]);
fail[1] = 0;
int j = 0;
for (int i = 2; i < lx; ++i) {
while (j && dp[i] != dp[j + 1]) j = fail[j];
if (dp[i] == dp[j + 1]) j++;
fail[i] = j;
}
j = 0;
for (int i = 1; i < ly; ++i) {
while (j && dq[i] != dp[j + 1]) j = fail[j];
if (dq[i] == dp[j + 1]) j++;
if (j == lx - 1) vis[i] = 1;
}
fail[1] = 0;
j = 0;
for (int i = 2; i < lx; ++i) {
while (j && dcmp(ap[i] - ap[j + 1])) j = fail[j];
if (!dcmp(ap[i] - ap[j + 1])) j++;
fail[i] = j;
}
j = 0;
for (int i = 1; i < ly; ++i) {
while (j && dcmp(aq[i] - ap[j + 1])) j = fail[j];
if (!dcmp(aq[i] - ap[j + 1])) j++;
if (j == lx - 1)
if (vis[i]) return 0 * puts("YES");
}
puts("NO");
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-6, pi = acos(-1.0);
inline int dcmp(double x) {
if (fabs(x) < eps) return 0;
return x < 0 ? -1 : 1;
}
inline double sqr(double x) { return x * x; }
struct Point {
double x, y;
Point(double _x = 0, double _y = 0) : x(_x), y(_y){};
inline void in() { scanf("%lf%lf", &x, &y); }
};
inline Point operator+(Point A, Point B) { return Point(A.x + B.x, A.y + B.y); }
inline Point operator-(Point A, Point B) { return Point(A.x - B.x, A.y - B.y); }
inline Point operator*(Point A, double p) { return Point(A.x * p, A.y * p); }
inline Point operator/(Point A, double p) { return Point(A.x / p, A.y / p); }
inline bool operator<(const Point &a, const Point &b) {
return a.x < b.x || (a.x == b.x && a.y < b.y);
}
inline bool operator==(const Point &a, const Point &b) {
return !dcmp(a.x - b.x) && !dcmp(a.y - b.y);
}
inline double Distance(Point A, Point B) {
return sqrt(sqr(A.x - B.x) + sqr(A.y - B.y));
}
inline double Dot(Point A, Point B) { return A.x * B.x + A.y * B.y; }
inline double Norm(Point A) { return A.x * A.x + A.y * A.y; }
inline double Length(Point A) { return sqrt(Dot(A, A)); }
inline double GetAngle(Point A, Point B) {
return acos(Dot(A, B) / Length(A) / Length(B));
}
inline double GetAngle(Point v) { return atan2(v.y, v.x); }
inline double Cross(Point A, Point B) { return A.x * B.y - A.y * B.x; }
inline Point Unit(Point x) { return x / Length(x); }
inline Point Normal(Point x) { return Point(-x.y, x.x) / Length(x); }
inline Point Rotate(Point A, double rad) {
return Point(A.x * cos(rad) - A.y * sin(rad),
A.x * sin(rad) + A.y * cos(rad));
}
inline double Area2(const Point &A, const Point &B, const Point &C) {
return Cross(B - A, C - A);
}
inline void getLineGeneralEquation(const Point &p1, const Point &p2, double &a,
double &b, double &c) {
a = p2.y - p1.y;
b = p1.x - p2.x;
c = -a * p1.x - b * p1.y;
}
inline Point GetLineIntersection(Point P, Point v, Point Q, Point w) {
Point u = P - Q;
double t = Cross(w, u) / Cross(v, w);
return P + v * t;
}
inline double DistanceToLine(Point P, Point A, Point B) {
Point v1 = B - A, v2 = P - A;
return fabs(Cross(v1, v2)) / Length(v1);
}
inline double DistanceToSegment(Point P, Point A, Point B) {
if (A == B) return Length(P - A);
Point v1 = B - A, v2 = P - A, v3 = P - B;
if (dcmp(Dot(v1, v2)) < 0)
return Length(v2);
else if (dcmp(Dot(v1, v3)) > 0)
return Length(v3);
else
return fabs(Cross(v1, v2)) / Length(v1);
}
inline Point GetLineProjection(Point P, Point A, Point B) {
Point v = B - A;
return A + v * (Dot(v, P - A) / Dot(v, v));
}
inline bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) {
double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1);
double c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);
return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
}
inline bool OnSegment(Point p, Point a1, Point a2) {
return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0;
}
inline double PolygonArea(Point *p, int n) {
double area = 0.0;
for (int i = 1; i < n - 1; i++) area += Cross(p[i] - p[0], p[i + 1] - p[0]);
return area / 2.0;
}
inline Point GetMidPoint(Point A, Point B) {
return Point((A.x + B.x) / 2.0, (A.y + B.y) / 2.0);
}
inline int PointinPolygon(Point p, vector<Point> poly) {
int n = poly.size(), flg = 0;
for (int i = 0; i < n; i++) {
if (OnSegment(p, poly[i], poly[(i + 1) % n])) return -1;
int k = dcmp(Cross(poly[(i + 1) % n] - poly[i], p - poly[i]));
int d1 = dcmp(poly[i].y - p.y);
int d2 = dcmp(poly[(i + 1) % n].y - p.y);
if (k > 0 && d1 <= 0 && d2 > 0) flg++;
if (k < 0 && d2 <= 0 && d1 > 0) flg++;
}
if (flg) return 1;
return 0;
}
struct Segment {
Point p, v;
Segment() {}
Segment(Point _p, Point _v) : p(_p), v(_v) {}
inline void in() {
p.in();
v.in();
}
};
struct Line {
Point p, v;
double ang;
Line() {}
Line(Point _p, Point _v) : p(_p), v(_v) { ang = atan2(v.y, v.x); }
inline Point point(double a) { return p + v * a; }
inline bool operator<(const Line &rhs) const { return ang < rhs.ang; }
inline Line move(double d) { return Line(p + Normal(v) * d, v); }
inline void in() {
p.in();
v.in();
ang = atan2(v.y, v.x);
}
};
inline Line LineTransHorizon(Line l, int d) {
Point vl = Normal(l.v);
Point p1 = l.p + vl * d;
return Line(p1, l.v);
}
inline Point GetLineIntersection(Line a, Line b) {
return GetLineIntersection(a.p, a.v, b.p, b.v);
}
inline bool OnLeft(const Line &L, const Point &p) {
return Cross(L.v, p - L.p) > 0;
}
inline Line GetParallel(Point p, Line l) { return Line(p, l.v); }
inline Line GetVertical(Point p, Line l) { return Line(p, Normal(l.v)); }
inline Point GetVerticalFoot(Point p, Line l) {
return GetLineIntersection(GetVertical(p, l), l);
}
inline Line GetPerpendicularLine(Segment l) {
return GetVertical(GetMidPoint(l.p, l.v - l.p), Line(l.p, l.v));
}
struct Angle {
Point p, u, v;
double ang;
Angle() {}
Angle(Point _p, Point _u, Point _v) : p(_p), u(_u), v(_v) {
ang = GetAngle(u, v);
}
inline void in() {
p.in();
u.in();
v.in();
ang = GetAngle(u, v);
}
};
inline int ConvexHull(Point *p, int n, Point *ans) {
sort(p, p + n);
int m = 0;
for (int i = 0; i < n; i++) {
while (m > 1 && Cross(ans[m - 1] - ans[m - 2], p[i] - ans[m - 2]) <= 0) m--;
ans[m++] = p[i];
}
int k = m;
for (int i = n - 2; ~i; i--) {
while (m > k && Cross(ans[m - 1] - ans[m - 2], p[i] - ans[m - 2]) <= 0) m--;
ans[m++] = p[i];
}
if (n > 1) m--;
return m;
}
inline long long distan(Point p, Point q) {
return (p.x - q.x) * (p.x - q.x) + (p.y - q.y) * (p.y - q.y);
}
Point p[100005], q[100005], cp[200005], cq[200005];
int n, m, lx, ly, fail[100005];
long long dp[100005], dq[200005];
double ap[100005], aq[200005];
bool vis[100005];
int main() {
scanf("%d%d", &n, &m);
for (int i = 0; i < n; ++i) p[i].in();
for (int i = 0; i < m; ++i) q[i].in();
lx = ConvexHull(p, n, cp);
ly = ConvexHull(q, m, cq);
if (lx != ly) return 0 * puts("NO");
cp[lx] = cp[0], lx++;
cp[lx] = cp[1];
for (int i = 0; i < ly; ++i) cq[i + ly] = cq[i];
ly <<= 1;
cq[ly] = cq[0], ly++;
cq[ly] = cq[1];
for (int i = 1; i < lx; ++i) dp[i] = distan(cp[i], cp[i - 1]);
for (int i = 1; i < ly; ++i) dq[i] = distan(cq[i], cq[i - 1]);
for (int i = 1; i < lx; ++i)
ap[i] = GetAngle(cp[i - 1] - cp[i], cp[i + 1] - cp[i]);
for (int i = 1; i < ly; ++i)
aq[i] = GetAngle(cq[i - 1] - cq[i], cq[i + 1] - cq[i]);
fail[1] = 0;
int j = 0;
for (int i = 2; i < lx; ++i) {
while (j && dp[i] != dp[j + 1]) j = fail[j];
if (dp[i] == dp[j + 1]) j++;
fail[i] = j;
}
j = 0;
for (int i = 1; i < ly; ++i) {
while (j && dq[i] != dp[j + 1]) j = fail[j];
if (dq[i] == dp[j + 1]) j++;
if (j == lx - 1) vis[i] = 1;
}
fail[1] = 0;
j = 0;
for (int i = 2; i < lx; ++i) {
while (j && dcmp(ap[i] - ap[j + 1])) j = fail[j];
if (!dcmp(ap[i] - ap[j + 1])) j++;
fail[i] = j;
}
j = 0;
for (int i = 1; i < ly; ++i) {
while (j && dcmp(aq[i] - ap[j + 1])) j = fail[j];
if (!dcmp(aq[i] - ap[j + 1])) j++;
if (j == lx - 1)
if (vis[i]) return 0 * puts("YES");
}
puts("NO");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1);
int read() {
char c = getchar();
int res = 0;
while (c < '0' || c > '9') c = getchar();
while (48 <= c && c <= 57) {
res = res * 10 + c - '0';
c = getchar();
}
return res;
}
bool bad(pair<double, double> A, pair<double, double> B, pair<double, double> C,
bool t) {
return B.first * (C.second - A.second) + C.first * (A.second - B.second) +
A.first * (B.second - C.second) >=
0;
}
void ConvexHull(vector<pair<double, double> > &p) {
sort(p.begin(), p.end());
vector<pair<double, double> > upper;
upper.clear();
vector<pair<double, double> > lower;
lower.clear();
for (int i = 0; i < p.size(); ++i) {
while (upper.size() > 1) {
int j = upper.size() - 2;
if (bad(upper[j], upper[j + 1], p[i], 1))
upper.pop_back();
else
break;
}
upper.push_back(p[i]);
}
reverse(p.begin(), p.end());
for (int i = 0; i < p.size(); ++i) {
while (lower.size() > 1) {
int j = lower.size() - 2;
if (bad(lower[j], lower[j + 1], p[i], 0))
lower.pop_back();
else
break;
}
lower.push_back(p[i]);
}
p.clear();
for (int i = 1; i < upper.size(); ++i) p.push_back(upper[i]);
for (int i = 1; i < lower.size(); ++i) p.push_back(lower[i]);
}
double len(pair<double, double> u) {
return sqrt(u.first * u.first + u.second * u.second);
}
double ang(pair<double, double> u, pair<double, double> v) {
return acos((u.first * v.first + u.second * v.second) / len(u) / len(v));
}
void TransForm(vector<pair<double, double> > &P) {
vector<pair<double, double> > C;
C.clear();
int n = P.size();
for (int i = 0; i < n; ++i) {
int j = (i + 2) % n;
pair<double, double> u = {P[i].first - P[(i + 1) % n].first,
P[i].second - P[(i + 1) % n].second};
pair<double, double> v = {P[j].first - P[(i + 1) % n].first,
P[j].second - P[(i + 1) % n].second};
C.push_back({len(u), ang(u, v)});
}
P = C;
}
int n, m;
vector<pair<double, double> > A, B;
int KMP[N];
int main() {
A.resize(read());
B.resize(read());
for (int i = 0; i < A.size(); ++i) A[i].first = read(), A[i].second = read();
for (int i = 0; i < B.size(); ++i) B[i].first = read(), B[i].second = read();
ConvexHull(A);
TransForm(A);
ConvexHull(B);
TransForm(B);
if (A.size() != B.size()) return printf("NO"), 0;
int n = A.size();
for (int i = 0; i < n; ++i) A.push_back(A[i]);
KMP[0] = -1;
for (int i = 1; i < n; ++i) {
int j = KMP[i - 1];
while (B[j + 1] != B[i]) {
if (j >= 0)
j = KMP[j];
else {
j = -2;
break;
}
}
KMP[i] = j + 1;
}
int len = -1;
for (int i = 0; i < A.size(); ++i) {
if (len == n - 1) {
printf("YES");
exit(0);
}
if (A[i] == B[len + 1])
++len;
else if (len >= 0)
len = KMP[len];
}
printf("NO");
}
| ### Prompt
Generate a Cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1);
int read() {
char c = getchar();
int res = 0;
while (c < '0' || c > '9') c = getchar();
while (48 <= c && c <= 57) {
res = res * 10 + c - '0';
c = getchar();
}
return res;
}
bool bad(pair<double, double> A, pair<double, double> B, pair<double, double> C,
bool t) {
return B.first * (C.second - A.second) + C.first * (A.second - B.second) +
A.first * (B.second - C.second) >=
0;
}
void ConvexHull(vector<pair<double, double> > &p) {
sort(p.begin(), p.end());
vector<pair<double, double> > upper;
upper.clear();
vector<pair<double, double> > lower;
lower.clear();
for (int i = 0; i < p.size(); ++i) {
while (upper.size() > 1) {
int j = upper.size() - 2;
if (bad(upper[j], upper[j + 1], p[i], 1))
upper.pop_back();
else
break;
}
upper.push_back(p[i]);
}
reverse(p.begin(), p.end());
for (int i = 0; i < p.size(); ++i) {
while (lower.size() > 1) {
int j = lower.size() - 2;
if (bad(lower[j], lower[j + 1], p[i], 0))
lower.pop_back();
else
break;
}
lower.push_back(p[i]);
}
p.clear();
for (int i = 1; i < upper.size(); ++i) p.push_back(upper[i]);
for (int i = 1; i < lower.size(); ++i) p.push_back(lower[i]);
}
double len(pair<double, double> u) {
return sqrt(u.first * u.first + u.second * u.second);
}
double ang(pair<double, double> u, pair<double, double> v) {
return acos((u.first * v.first + u.second * v.second) / len(u) / len(v));
}
void TransForm(vector<pair<double, double> > &P) {
vector<pair<double, double> > C;
C.clear();
int n = P.size();
for (int i = 0; i < n; ++i) {
int j = (i + 2) % n;
pair<double, double> u = {P[i].first - P[(i + 1) % n].first,
P[i].second - P[(i + 1) % n].second};
pair<double, double> v = {P[j].first - P[(i + 1) % n].first,
P[j].second - P[(i + 1) % n].second};
C.push_back({len(u), ang(u, v)});
}
P = C;
}
int n, m;
vector<pair<double, double> > A, B;
int KMP[N];
int main() {
A.resize(read());
B.resize(read());
for (int i = 0; i < A.size(); ++i) A[i].first = read(), A[i].second = read();
for (int i = 0; i < B.size(); ++i) B[i].first = read(), B[i].second = read();
ConvexHull(A);
TransForm(A);
ConvexHull(B);
TransForm(B);
if (A.size() != B.size()) return printf("NO"), 0;
int n = A.size();
for (int i = 0; i < n; ++i) A.push_back(A[i]);
KMP[0] = -1;
for (int i = 1; i < n; ++i) {
int j = KMP[i - 1];
while (B[j + 1] != B[i]) {
if (j >= 0)
j = KMP[j];
else {
j = -2;
break;
}
}
KMP[i] = j + 1;
}
int len = -1;
for (int i = 0; i < A.size(); ++i) {
if (len == n - 1) {
printf("YES");
exit(0);
}
if (A[i] == B[len + 1])
++len;
else if (len >= 0)
len = KMP[len];
}
printf("NO");
}
``` |
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
double fRand(double fMin, double fMax) {
double f = (double)rand() / RAND_MAX;
return fMin + f * (fMax - fMin);
}
template <class T>
T min(T a, T b, T c) {
return min(a, min(b, c));
}
template <class T>
T max(T a, T b, T c) {
return max(a, max(b, c));
}
struct point {
int x, y;
point(int x, int y) : x(x), y(y) {}
point operator-(point p) { return {x - p.x, y - p.y}; }
};
bool operator==(point a, point b) { return a.x == b.x && a.y == b.y; }
long long sq(long long x) { return 1LL * x * x; }
long long sq(point a, point b) { return sq(b.x - a.x) + sq(b.y - a.y); }
long long cross(point a, point b) { return 1LL * a.x * b.y - 1LL * a.y * b.x; }
int CCW(point a, point b, point c) {
long long x = cross(b - a, c - a);
if (x > 0) return 1;
if (x < 0) return -1;
return 0;
}
bool cmp(const point &a, const point &b) {
return (a.y < b.y || (a.y == b.y && a.x < b.x));
}
vector<point> convexPath(const vector<point> &po) {
vector<point> s;
for (int i = (0); i <= (1); ++i) s.push_back(po[i]);
for (int i = (2); i <= ((int)po.size() - 1); ++i) {
while ((int)s.size() > 1 &&
CCW(s[(int)s.size() - 2], s[(int)s.size() - 1], po[i]) <= 0)
s.pop_back();
s.push_back(po[i]);
}
return s;
}
vector<point> convexHull(const vector<point> &po) {
point A = *min_element(po.begin(), po.end(), cmp);
point B = *max_element(po.begin(), po.end(), cmp);
vector<point> P, Q;
for (point p : po) {
int x = CCW(A, p, B);
if (x <= 0) P.push_back(p);
if (x >= 0) Q.push_back(p);
}
sort(P.begin(), P.end(), cmp);
reverse(P.begin(), P.end());
P = convexPath(P);
sort(Q.begin(), Q.end(), cmp);
Q = convexPath(Q);
vector<point> ans;
for (int i = (0); i <= ((int)Q.size() - 2); ++i) ans.push_back(Q[i]);
for (int i = (0); i <= ((int)P.size() - 2); ++i) ans.push_back(P[i]);
if (sq(*ans.begin(), *ans.rbegin()) == 0) ans.pop_back();
return ans;
}
void getEngine(vector<point> &p, int &n) {
for (int i = (1); i <= (n); ++i) {
int x, y;
scanf("%d%d", &x, &y);
p.push_back({x, y});
}
p = convexHull(p);
n = (int)p.size();
}
int n, m;
vector<point> p1, p2;
point rot90(point p) { return {-p.y, p.x}; }
void simplify(vector<point> &p) {
int minX = p[0].x, minY = p[0].y;
for (int i = (0); i <= ((int)p.size() - 1); ++i) {
minX = min(minX, p[i].x);
minY = min(minY, p[i].y);
}
for (int i = (0); i <= ((int)p.size() - 1); ++i) {
p[i].x -= minX;
p[i].y -= minY;
}
}
int getMinimumRotation(const vector<point> &p) {
int n = (int)p.size();
vector<long long> s(n);
for (int i = (0); i <= (n - 1); ++i) s[i] = 1LL * p[i].x * 100000001 + p[i].y;
for (int i = (0); i <= (n - 1); ++i) s.push_back(s[i]);
n *= 2;
vector<int> f(n, -1);
int k = 0;
for (int j = (1); j <= (n - 1); ++j) {
long long sj = s[j];
int i = f[j - k - 1];
while (i != -1 && sj != s[k + i + 1]) {
if (sj < s[k + i + 1]) k = j - i - 1;
i = f[i];
}
if (sj != s[k + i + 1]) {
if (sj < s[k]) k = j;
f[j - k] = -1;
} else
f[j - k] = i + 1;
}
return k;
}
bool isEqual(const vector<point> &p1, const vector<point> &p2) {
int n = (int)p1.size();
int k1 = getMinimumRotation(p1), k2 = getMinimumRotation(p2);
for (int i = (0); i <= (n - 1); ++i)
if (!(p1[(i + k1) % n] == p2[(i + k2) % n])) return false;
return true;
}
int main() {
scanf("%d%d", &n, &m);
getEngine(p1, n);
getEngine(p2, m);
if (n != m) {
puts("NO");
return 0;
}
bool ans = false;
if (n == 2) {
ans = (sq(p1[0], p1[1]) == sq(p2[0], p2[1]));
} else {
simplify(p1);
for (int dir = (0); dir <= (3); ++dir) {
for (int i = (0); i <= (n - 1); ++i) p2[i] = rot90(p2[i]);
simplify(p2);
if (isEqual(p1, p2)) ans = true;
}
}
puts((ans) ? "YES" : "NO");
return 0;
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
double fRand(double fMin, double fMax) {
double f = (double)rand() / RAND_MAX;
return fMin + f * (fMax - fMin);
}
template <class T>
T min(T a, T b, T c) {
return min(a, min(b, c));
}
template <class T>
T max(T a, T b, T c) {
return max(a, max(b, c));
}
struct point {
int x, y;
point(int x, int y) : x(x), y(y) {}
point operator-(point p) { return {x - p.x, y - p.y}; }
};
bool operator==(point a, point b) { return a.x == b.x && a.y == b.y; }
long long sq(long long x) { return 1LL * x * x; }
long long sq(point a, point b) { return sq(b.x - a.x) + sq(b.y - a.y); }
long long cross(point a, point b) { return 1LL * a.x * b.y - 1LL * a.y * b.x; }
int CCW(point a, point b, point c) {
long long x = cross(b - a, c - a);
if (x > 0) return 1;
if (x < 0) return -1;
return 0;
}
bool cmp(const point &a, const point &b) {
return (a.y < b.y || (a.y == b.y && a.x < b.x));
}
vector<point> convexPath(const vector<point> &po) {
vector<point> s;
for (int i = (0); i <= (1); ++i) s.push_back(po[i]);
for (int i = (2); i <= ((int)po.size() - 1); ++i) {
while ((int)s.size() > 1 &&
CCW(s[(int)s.size() - 2], s[(int)s.size() - 1], po[i]) <= 0)
s.pop_back();
s.push_back(po[i]);
}
return s;
}
vector<point> convexHull(const vector<point> &po) {
point A = *min_element(po.begin(), po.end(), cmp);
point B = *max_element(po.begin(), po.end(), cmp);
vector<point> P, Q;
for (point p : po) {
int x = CCW(A, p, B);
if (x <= 0) P.push_back(p);
if (x >= 0) Q.push_back(p);
}
sort(P.begin(), P.end(), cmp);
reverse(P.begin(), P.end());
P = convexPath(P);
sort(Q.begin(), Q.end(), cmp);
Q = convexPath(Q);
vector<point> ans;
for (int i = (0); i <= ((int)Q.size() - 2); ++i) ans.push_back(Q[i]);
for (int i = (0); i <= ((int)P.size() - 2); ++i) ans.push_back(P[i]);
if (sq(*ans.begin(), *ans.rbegin()) == 0) ans.pop_back();
return ans;
}
void getEngine(vector<point> &p, int &n) {
for (int i = (1); i <= (n); ++i) {
int x, y;
scanf("%d%d", &x, &y);
p.push_back({x, y});
}
p = convexHull(p);
n = (int)p.size();
}
int n, m;
vector<point> p1, p2;
point rot90(point p) { return {-p.y, p.x}; }
void simplify(vector<point> &p) {
int minX = p[0].x, minY = p[0].y;
for (int i = (0); i <= ((int)p.size() - 1); ++i) {
minX = min(minX, p[i].x);
minY = min(minY, p[i].y);
}
for (int i = (0); i <= ((int)p.size() - 1); ++i) {
p[i].x -= minX;
p[i].y -= minY;
}
}
int getMinimumRotation(const vector<point> &p) {
int n = (int)p.size();
vector<long long> s(n);
for (int i = (0); i <= (n - 1); ++i) s[i] = 1LL * p[i].x * 100000001 + p[i].y;
for (int i = (0); i <= (n - 1); ++i) s.push_back(s[i]);
n *= 2;
vector<int> f(n, -1);
int k = 0;
for (int j = (1); j <= (n - 1); ++j) {
long long sj = s[j];
int i = f[j - k - 1];
while (i != -1 && sj != s[k + i + 1]) {
if (sj < s[k + i + 1]) k = j - i - 1;
i = f[i];
}
if (sj != s[k + i + 1]) {
if (sj < s[k]) k = j;
f[j - k] = -1;
} else
f[j - k] = i + 1;
}
return k;
}
bool isEqual(const vector<point> &p1, const vector<point> &p2) {
int n = (int)p1.size();
int k1 = getMinimumRotation(p1), k2 = getMinimumRotation(p2);
for (int i = (0); i <= (n - 1); ++i)
if (!(p1[(i + k1) % n] == p2[(i + k2) % n])) return false;
return true;
}
int main() {
scanf("%d%d", &n, &m);
getEngine(p1, n);
getEngine(p2, m);
if (n != m) {
puts("NO");
return 0;
}
bool ans = false;
if (n == 2) {
ans = (sq(p1[0], p1[1]) == sq(p2[0], p2[1]));
} else {
simplify(p1);
for (int dir = (0); dir <= (3); ++dir) {
for (int i = (0); i <= (n - 1); ++i) p2[i] = rot90(p2[i]);
simplify(p2);
if (isEqual(p1, p2)) ans = true;
}
}
puts((ans) ? "YES" : "NO");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
inline int inn() {
int x, ch;
while ((ch = getchar()) < '0' || ch > '9')
;
x = ch ^ '0';
while ((ch = getchar()) >= '0' && ch <= '9')
x = (x << 1) + (x << 3) + (ch ^ '0');
return x;
}
inline int sgn(long long x) { return (x < 0) ? -1 : (x > 0); }
inline int sgn(long long x, long long y) { return sgn(x - y); }
inline long long gabs(long long x) { return x < 0 ? -x : x; }
struct P {
long long x, y;
P(long long _x = 0, long long _y = 0) { x = _x, y = _y; }
inline P operator=(const P &p) {
x = p.x, y = p.y;
return *this;
}
inline bool operator<(const P &p) const {
return sgn(x - p.x) ? x < p.x : y < p.y;
}
inline P operator+(const P &p) const { return P(x + p.x, y + p.y); }
inline P operator-(const P &p) const { return P(x - p.x, y - p.y); }
inline int init() { return x = inn(), y = inn(), 0; }
};
inline long long dot(const P &v1, const P &v2) {
return v1.x * v2.x + v1.y * v2.y;
}
inline long long dot(const P &s, const P &p1, const P &p2) {
return dot(p1 - s, p2 - s);
}
inline long long cross(const P &v1, const P &v2) {
return v1.x * v2.y - v1.y * v2.x;
}
inline long long cross(const P &s, const P &p1, const P &p2) {
return cross(p1 - s, p2 - s);
}
inline long long Len2(const P &v) { return dot(v, v); }
inline vector<P> convexHull(vector<P> ps, int need_sort = 1) {
int n = (int)ps.size();
if (n <= 1) return ps;
vector<P> ans(2 * n);
int k = 0;
if (need_sort) sort(ps.begin(), ps.end());
for (int i = 0; i < n; ans[k++] = ps[i++])
while (k > 1 && sgn(cross(ans[k - 2], ans[k - 1], ps[i])) <= 0) k--;
for (int i = n - 1, t = k; i >= 0; ans[k++] = ps[i--])
while (k > t && sgn(cross(ans[k - 2], ans[k - 1], ps[i])) <= 0) k--;
ans.resize(k - 1);
return ans;
}
vector<P> A, B;
inline int getList(vector<P> &ps, long long *S) {
int ss = 0, n = (int)ps.size();
for (int i = 0; i < n; i++) {
P A = ps[(i + 1) % n], B = ps[i], C = ps[(i + 2) % n];
S[++ss] = Len2(B - A), S[++ss] = cross(A, B, C);
}
return ss;
}
int nxt[1000010];
long long S[1000010], T[1000010];
inline int getnxt(long long *s, int n) {
for (int i = 2; i <= n; i++) {
int &j = nxt[i];
j = nxt[i - 1];
while (j && sgn(s[j + 1] - s[i])) j = nxt[j];
if (sgn(s[j + 1] - s[i]) == 0) j++;
}
return 0;
}
inline int ocr(long long *s, int n, long long *t, int m) {
for (int i = 1, j = 0; i <= n; i++) {
while (j && sgn(t[j + 1] - s[i])) j = nxt[j];
if (sgn(t[j + 1] - s[i]) == 0) j++;
if (j == m) return 1;
}
return 0;
}
int main() {
int n = inn(), m = inn();
A.resize(n), B.resize(m);
for (int i = 0; i < n; i++) A[i].init();
for (int i = 0; i < m; i++) B[i].init();
A = convexHull(A), B = convexHull(B);
if (A.size() != B.size()) return !printf("NO\n");
int slen = getList(A, S), tlen = getList(B, T);
for (int i = slen + 1; i < slen * 2; i++) S[i] = S[i - slen];
slen = slen * 2 - 1, getnxt(T, tlen);
if (ocr(S, slen, T, tlen)) return !printf("YES\n");
return !printf("no\n");
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
inline int inn() {
int x, ch;
while ((ch = getchar()) < '0' || ch > '9')
;
x = ch ^ '0';
while ((ch = getchar()) >= '0' && ch <= '9')
x = (x << 1) + (x << 3) + (ch ^ '0');
return x;
}
inline int sgn(long long x) { return (x < 0) ? -1 : (x > 0); }
inline int sgn(long long x, long long y) { return sgn(x - y); }
inline long long gabs(long long x) { return x < 0 ? -x : x; }
struct P {
long long x, y;
P(long long _x = 0, long long _y = 0) { x = _x, y = _y; }
inline P operator=(const P &p) {
x = p.x, y = p.y;
return *this;
}
inline bool operator<(const P &p) const {
return sgn(x - p.x) ? x < p.x : y < p.y;
}
inline P operator+(const P &p) const { return P(x + p.x, y + p.y); }
inline P operator-(const P &p) const { return P(x - p.x, y - p.y); }
inline int init() { return x = inn(), y = inn(), 0; }
};
inline long long dot(const P &v1, const P &v2) {
return v1.x * v2.x + v1.y * v2.y;
}
inline long long dot(const P &s, const P &p1, const P &p2) {
return dot(p1 - s, p2 - s);
}
inline long long cross(const P &v1, const P &v2) {
return v1.x * v2.y - v1.y * v2.x;
}
inline long long cross(const P &s, const P &p1, const P &p2) {
return cross(p1 - s, p2 - s);
}
inline long long Len2(const P &v) { return dot(v, v); }
inline vector<P> convexHull(vector<P> ps, int need_sort = 1) {
int n = (int)ps.size();
if (n <= 1) return ps;
vector<P> ans(2 * n);
int k = 0;
if (need_sort) sort(ps.begin(), ps.end());
for (int i = 0; i < n; ans[k++] = ps[i++])
while (k > 1 && sgn(cross(ans[k - 2], ans[k - 1], ps[i])) <= 0) k--;
for (int i = n - 1, t = k; i >= 0; ans[k++] = ps[i--])
while (k > t && sgn(cross(ans[k - 2], ans[k - 1], ps[i])) <= 0) k--;
ans.resize(k - 1);
return ans;
}
vector<P> A, B;
inline int getList(vector<P> &ps, long long *S) {
int ss = 0, n = (int)ps.size();
for (int i = 0; i < n; i++) {
P A = ps[(i + 1) % n], B = ps[i], C = ps[(i + 2) % n];
S[++ss] = Len2(B - A), S[++ss] = cross(A, B, C);
}
return ss;
}
int nxt[1000010];
long long S[1000010], T[1000010];
inline int getnxt(long long *s, int n) {
for (int i = 2; i <= n; i++) {
int &j = nxt[i];
j = nxt[i - 1];
while (j && sgn(s[j + 1] - s[i])) j = nxt[j];
if (sgn(s[j + 1] - s[i]) == 0) j++;
}
return 0;
}
inline int ocr(long long *s, int n, long long *t, int m) {
for (int i = 1, j = 0; i <= n; i++) {
while (j && sgn(t[j + 1] - s[i])) j = nxt[j];
if (sgn(t[j + 1] - s[i]) == 0) j++;
if (j == m) return 1;
}
return 0;
}
int main() {
int n = inn(), m = inn();
A.resize(n), B.resize(m);
for (int i = 0; i < n; i++) A[i].init();
for (int i = 0; i < m; i++) B[i].init();
A = convexHull(A), B = convexHull(B);
if (A.size() != B.size()) return !printf("NO\n");
int slen = getList(A, S), tlen = getList(B, T);
for (int i = slen + 1; i < slen * 2; i++) S[i] = S[i - slen];
slen = slen * 2 - 1, getnxt(T, tlen);
if (ocr(S, slen, T, tlen)) return !printf("YES\n");
return !printf("no\n");
}
``` |
#include <bits/stdc++.h>
using namespace std;
int myRnd() { return abs(((rand() << 14) ^ rand())); }
int myRnd(int L, int R) {
return abs(((rand() << 15) ^ rand())) % (R - L + 1) + L;
}
void Parr(int *arr, int L, int R) {
for (int i = L; R >= i; i++) {
printf("%d%c", arr[i], " \n"[i == R]);
}
}
void Pvec(vector<int> v) {
for (int i = 0; ((int)(v).size()) > i; i++) {
printf("%d%c", v[i], " \n"[i == ((int)(v).size()) - 1]);
}
}
void Sarr(int *arr, int L, int R) {
for (int i = L; R >= i; i++) {
scanf("%d", &(arr[i]));
;
}
}
const int N = 2e5 + 6;
const int INF = 0x3f3f3f3f;
int mods[10] = {1000000861, 1000004329, 1000025261, 1000075057, 1000099999,
1123465789, 1123564987, 1123586479, 1124396857, 1124638579};
pair<long long, long long> operator-(const pair<long long, long long> &p1,
const pair<long long, long long> &p2) {
return make_pair(p1.first - p2.first, p1.second - p2.second);
}
long long operator^(const pair<long long, long long> &p1,
const pair<long long, long long> &p2) {
return p1.first * p2.second - p1.second * p2.first;
}
long long Val(long long x) {
if (x > 0)
return 1;
else if (x == 0)
return 0;
else if (x < 0)
return -1;
}
void make_tobou(vector<pair<long long, long long> > &v) {
sort(v.begin(), v.end());
vector<pair<long long, long long> > dd;
for (pair<long long, long long> p : v) {
if (((int)(dd).size()) < 2)
dd.push_back(p);
else {
while (((int)(dd).size()) >= 2 && Val((p - dd[((int)(dd).size()) - 2]) ^
(dd[((int)(dd).size()) - 1] -
dd[((int)(dd).size()) - 2])) >= 0)
dd.pop_back();
dd.push_back(p);
}
}
reverse(v.begin(), v.end());
vector<pair<long long, long long> > uu;
for (pair<long long, long long> p : v) {
if (((int)(uu).size()) < 2)
uu.push_back(p);
else {
while (((int)(uu).size()) >= 2 && Val((p - uu[((int)(uu).size()) - 2]) ^
(uu[((int)(uu).size()) - 1] -
uu[((int)(uu).size()) - 2])) >= 0)
uu.pop_back();
uu.push_back(p);
}
}
vector<pair<long long, long long> > vv;
for (pair<long long, long long> p : dd) vv.push_back(p);
vv.pop_back();
for (pair<long long, long long> p : uu) vv.push_back(p);
vv.pop_back();
v = vv;
}
long long get_area(vector<pair<long long, long long> > all) {
long long ans = 0;
for (int i = 1; ((int)(all).size()) - 1 > i; i++) {
ans += ((all[i] - all[0]) ^ (all[i + 1] - all[0]));
}
ans = abs(ans);
return ans;
}
int n, m;
long long ppow[N];
long long pre1[N], pre2[N];
int xx[5] = {880301, 106111601, 991, 47017, 10430579};
pair<long long, long long> c1 = make_pair(0, 0), c2 = make_pair(0, 0);
long long get_dis(long long x1, long long y1, long long x2, long long y2,
long long mod) {
long long x = ((abs(x1 - x2)) % mod + mod) % mod;
long long y = ((abs(y1 - y2)) % mod + mod) % mod;
return (x * x + y * y) % mod;
}
bool solve(vector<pair<long long, long long> > v1,
vector<pair<long long, long long> > v2, long long X, long long mod1,
long long mod2) {
ppow[0] = 1;
for (int i = 1; (N - 1) >= i; ++i) {
ppow[i] = (ppow[i - 1] * X) % mod2;
}
long long hash1 = 0, hash2 = 0;
int n = ((int)(v1).size());
v1.push_back(v1.front());
for (int i = 1; n >= i; i++) {
hash1 += ((get_dis(v1[i].first, v1[i].second, v1[i - 1].first,
v1[i - 1].second, mod1)) *
ppow[i]);
hash1 %= mod2;
hash2 += ((get_dis(c1.first, c1.second, v1[i - 1].first, v1[i - 1].second,
mod1)) *
ppow[i]);
hash2 %= mod2;
}
for (int i = 0; n > i; i++) {
v2.push_back(v2[i]);
}
v2.push_back(v2.front());
for (int i = 1; 2 * n >= i; i++) {
pre1[i] =
pre1[i - 1] + ((get_dis(v2[i].first, v2[i].second, v2[i - 1].first,
v2[i - 1].second, mod1)) *
ppow[i]);
pre1[i] %= mod2;
pre2[i] = pre2[i - 1] + ((get_dis(c2.first, c2.second, v2[i - 1].first,
v2[i - 1].second, mod1)) *
ppow[i]);
pre2[i] %= mod2;
}
for (int i = 1; n >= i; i++) {
int L = i, R = i + n - 1;
long long mod = mod2;
if (((pre1[R] - pre1[L - 1] + mod) % mod) * ppow[N - R] % mod2 ==
hash1 * ppow[N - n] % mod2) {
if (((pre2[R] - pre2[L - 1] + mod) % mod) * ppow[N - R] % mod2 ==
hash2 * ppow[N - n] % mod2)
return true;
}
}
return false;
}
int main() {
srand(clock());
vector<pair<long long, long long> > v1, v2;
scanf("%d %d", &(n), &(m));
;
for (int i = 1; (n) >= i; ++i) {
long long x, y;
scanf("%lld %lld", &(x), &(y));
;
v1.push_back(make_pair(x, y));
}
for (int i = 1; (m) >= i; ++i) {
long long x, y;
scanf("%lld %lld", &(x), &(y));
;
v2.push_back(make_pair(x, y));
}
make_tobou(v1);
make_tobou(v2);
if (((int)(v1).size()) != ((int)(v2).size())) {
puts("NO");
return 0;
}
if (get_area(v1) != get_area(v2)) {
puts("NO");
return 0;
}
int nn = ((int)(v1).size());
for (pair<long long, long long> &p : v1) {
p.first *= nn;
p.second *= nn;
c1.first += p.first;
c1.second += p.second;
}
c1.first /= nn;
c1.second /= nn;
for (pair<long long, long long> &p : v2) {
p.first *= nn;
p.second *= nn;
c2.first += p.first;
c2.second += p.second;
}
c2.first /= nn;
c2.second /= nn;
int ff[5] = {0, 1, 2, 3, 4}, gg[5] = {0, 1, 2, 3, 4}, hh[5] = {0, 1, 2, 3, 4};
random_shuffle(ff, ff + 5);
random_shuffle(gg, gg + 5);
random_shuffle(hh, hh + 5);
for (int i = 0; 5 > i; i++) {
if (solve(v1, v2, xx[ff[i]], mods[gg[i]], mods[hh[i] + 5]) == false) {
puts("NO");
return 0;
}
}
puts("YES");
}
| ### Prompt
In cpp, your task is to solve the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int myRnd() { return abs(((rand() << 14) ^ rand())); }
int myRnd(int L, int R) {
return abs(((rand() << 15) ^ rand())) % (R - L + 1) + L;
}
void Parr(int *arr, int L, int R) {
for (int i = L; R >= i; i++) {
printf("%d%c", arr[i], " \n"[i == R]);
}
}
void Pvec(vector<int> v) {
for (int i = 0; ((int)(v).size()) > i; i++) {
printf("%d%c", v[i], " \n"[i == ((int)(v).size()) - 1]);
}
}
void Sarr(int *arr, int L, int R) {
for (int i = L; R >= i; i++) {
scanf("%d", &(arr[i]));
;
}
}
const int N = 2e5 + 6;
const int INF = 0x3f3f3f3f;
int mods[10] = {1000000861, 1000004329, 1000025261, 1000075057, 1000099999,
1123465789, 1123564987, 1123586479, 1124396857, 1124638579};
pair<long long, long long> operator-(const pair<long long, long long> &p1,
const pair<long long, long long> &p2) {
return make_pair(p1.first - p2.first, p1.second - p2.second);
}
long long operator^(const pair<long long, long long> &p1,
const pair<long long, long long> &p2) {
return p1.first * p2.second - p1.second * p2.first;
}
long long Val(long long x) {
if (x > 0)
return 1;
else if (x == 0)
return 0;
else if (x < 0)
return -1;
}
void make_tobou(vector<pair<long long, long long> > &v) {
sort(v.begin(), v.end());
vector<pair<long long, long long> > dd;
for (pair<long long, long long> p : v) {
if (((int)(dd).size()) < 2)
dd.push_back(p);
else {
while (((int)(dd).size()) >= 2 && Val((p - dd[((int)(dd).size()) - 2]) ^
(dd[((int)(dd).size()) - 1] -
dd[((int)(dd).size()) - 2])) >= 0)
dd.pop_back();
dd.push_back(p);
}
}
reverse(v.begin(), v.end());
vector<pair<long long, long long> > uu;
for (pair<long long, long long> p : v) {
if (((int)(uu).size()) < 2)
uu.push_back(p);
else {
while (((int)(uu).size()) >= 2 && Val((p - uu[((int)(uu).size()) - 2]) ^
(uu[((int)(uu).size()) - 1] -
uu[((int)(uu).size()) - 2])) >= 0)
uu.pop_back();
uu.push_back(p);
}
}
vector<pair<long long, long long> > vv;
for (pair<long long, long long> p : dd) vv.push_back(p);
vv.pop_back();
for (pair<long long, long long> p : uu) vv.push_back(p);
vv.pop_back();
v = vv;
}
long long get_area(vector<pair<long long, long long> > all) {
long long ans = 0;
for (int i = 1; ((int)(all).size()) - 1 > i; i++) {
ans += ((all[i] - all[0]) ^ (all[i + 1] - all[0]));
}
ans = abs(ans);
return ans;
}
int n, m;
long long ppow[N];
long long pre1[N], pre2[N];
int xx[5] = {880301, 106111601, 991, 47017, 10430579};
pair<long long, long long> c1 = make_pair(0, 0), c2 = make_pair(0, 0);
long long get_dis(long long x1, long long y1, long long x2, long long y2,
long long mod) {
long long x = ((abs(x1 - x2)) % mod + mod) % mod;
long long y = ((abs(y1 - y2)) % mod + mod) % mod;
return (x * x + y * y) % mod;
}
bool solve(vector<pair<long long, long long> > v1,
vector<pair<long long, long long> > v2, long long X, long long mod1,
long long mod2) {
ppow[0] = 1;
for (int i = 1; (N - 1) >= i; ++i) {
ppow[i] = (ppow[i - 1] * X) % mod2;
}
long long hash1 = 0, hash2 = 0;
int n = ((int)(v1).size());
v1.push_back(v1.front());
for (int i = 1; n >= i; i++) {
hash1 += ((get_dis(v1[i].first, v1[i].second, v1[i - 1].first,
v1[i - 1].second, mod1)) *
ppow[i]);
hash1 %= mod2;
hash2 += ((get_dis(c1.first, c1.second, v1[i - 1].first, v1[i - 1].second,
mod1)) *
ppow[i]);
hash2 %= mod2;
}
for (int i = 0; n > i; i++) {
v2.push_back(v2[i]);
}
v2.push_back(v2.front());
for (int i = 1; 2 * n >= i; i++) {
pre1[i] =
pre1[i - 1] + ((get_dis(v2[i].first, v2[i].second, v2[i - 1].first,
v2[i - 1].second, mod1)) *
ppow[i]);
pre1[i] %= mod2;
pre2[i] = pre2[i - 1] + ((get_dis(c2.first, c2.second, v2[i - 1].first,
v2[i - 1].second, mod1)) *
ppow[i]);
pre2[i] %= mod2;
}
for (int i = 1; n >= i; i++) {
int L = i, R = i + n - 1;
long long mod = mod2;
if (((pre1[R] - pre1[L - 1] + mod) % mod) * ppow[N - R] % mod2 ==
hash1 * ppow[N - n] % mod2) {
if (((pre2[R] - pre2[L - 1] + mod) % mod) * ppow[N - R] % mod2 ==
hash2 * ppow[N - n] % mod2)
return true;
}
}
return false;
}
int main() {
srand(clock());
vector<pair<long long, long long> > v1, v2;
scanf("%d %d", &(n), &(m));
;
for (int i = 1; (n) >= i; ++i) {
long long x, y;
scanf("%lld %lld", &(x), &(y));
;
v1.push_back(make_pair(x, y));
}
for (int i = 1; (m) >= i; ++i) {
long long x, y;
scanf("%lld %lld", &(x), &(y));
;
v2.push_back(make_pair(x, y));
}
make_tobou(v1);
make_tobou(v2);
if (((int)(v1).size()) != ((int)(v2).size())) {
puts("NO");
return 0;
}
if (get_area(v1) != get_area(v2)) {
puts("NO");
return 0;
}
int nn = ((int)(v1).size());
for (pair<long long, long long> &p : v1) {
p.first *= nn;
p.second *= nn;
c1.first += p.first;
c1.second += p.second;
}
c1.first /= nn;
c1.second /= nn;
for (pair<long long, long long> &p : v2) {
p.first *= nn;
p.second *= nn;
c2.first += p.first;
c2.second += p.second;
}
c2.first /= nn;
c2.second /= nn;
int ff[5] = {0, 1, 2, 3, 4}, gg[5] = {0, 1, 2, 3, 4}, hh[5] = {0, 1, 2, 3, 4};
random_shuffle(ff, ff + 5);
random_shuffle(gg, gg + 5);
random_shuffle(hh, hh + 5);
for (int i = 0; 5 > i; i++) {
if (solve(v1, v2, xx[ff[i]], mods[gg[i]], mods[hh[i] + 5]) == false) {
puts("NO");
return 0;
}
}
puts("YES");
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
typedef struct point {
long long x, y;
point(long long _x = 0, long long _y = 0) : x(_x), y(_y) {}
bool operator<(const point &B) const { return x != B.x ? x < B.x : y < B.y; }
point operator-(const point &B) const { return point(x - B.x, y - B.y); }
long long operator*(const point &B) const { return x * B.y - y * B.x; }
long long len_2() { return x * x + y * y; }
} P;
bool cmp(point &A, point &B) {
return A * B != 0 ? A * B > 0 : A.len_2() < B.len_2();
}
void _sort(point A[], point B[], int len) {
sort(A, A + len);
for (int i = 1; i < len; i++) {
B[i] = A[i] - A[0];
}
sort(B + 1, B + len, cmp);
B[0] = point(0, 0);
}
int ConvexHull(point A[], point B[], int len) {
_sort(A, A, len);
int tot = 0;
for (int i = 0; i < len; i++) {
if (tot <= 1)
B[tot++] = A[i];
else {
while (tot >= 2 && (B[tot - 1] - B[tot - 2]) * (A[i] - B[tot - 2]) <= 0)
tot--;
B[tot++] = A[i];
}
}
return tot;
}
P A[maxn], B[maxn], A1[maxn], B1[maxn];
int len1, len2;
int n, m;
tuple<long long, long long, long long> X[maxn * 2], Y[maxn];
void GetNextval(tuple<long long, long long, long long> p[], int next[],
int pLen) {
next[0] = -1;
int k = -1;
int j = 0;
while (j < pLen) {
if (k == -1 || p[j] == p[k]) {
++j, ++k;
if (p[j] != p[k]) {
next[j] = k;
} else {
next[j] = next[k];
}
} else {
k = next[k];
}
}
}
int nxt[123456];
int KmpSearch(tuple<long long, long long, long long> s[],
tuple<long long, long long, long long> p[], int sLen, int pLen) {
memset(nxt, -1, sizeof(nxt));
GetNextval(p, nxt, pLen);
int i = 0, j = 0;
while (i < sLen && j < pLen) {
if (j == -1 || s[i] == p[j]) {
i++, j++;
} else {
j = nxt[j];
}
}
if (j == pLen) {
return i - j;
} else {
return -1;
}
}
void kmp(point A[], point B[], int len) {
for (int i = 0; i < len; i++) {
X[i] = tuple<long long, long long, long long>(
(A[(i - 1 + len) % len] - A[i]).len_2(),
(A[(i + 1) % len] - A[i]).len_2(),
(A[(i - 1 + len) % len] - A[i]) * (A[(i + 1) % len] - A[i]));
X[i + len] = X[i];
}
for (int i = 0; i < len; i++) {
Y[i] = tuple<long long, long long, long long>(
(B[(i - 1 + len) % len] - B[i]).len_2(),
(B[(i + 1) % len] - B[i]).len_2(),
(B[(i - 1 + len) % len] - B[i]) * (B[(i + 1) % len] - B[i]));
}
if (KmpSearch(X, Y, 2 * len, len) != -1)
puts("YES");
else
puts("NO");
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++) {
scanf("%lld%lld", &A[i].x, &A[i].y);
}
for (int i = 0; i < m; i++) {
scanf("%lld%lld", &B[i].x, &B[i].y);
}
len1 = ConvexHull(A, A1, n);
len2 = ConvexHull(B, B1, m);
if (len1 == len2)
kmp(A1, B1, len1);
else
puts("NO");
return 0;
}
| ### Prompt
In Cpp, your task is to solve the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
typedef struct point {
long long x, y;
point(long long _x = 0, long long _y = 0) : x(_x), y(_y) {}
bool operator<(const point &B) const { return x != B.x ? x < B.x : y < B.y; }
point operator-(const point &B) const { return point(x - B.x, y - B.y); }
long long operator*(const point &B) const { return x * B.y - y * B.x; }
long long len_2() { return x * x + y * y; }
} P;
bool cmp(point &A, point &B) {
return A * B != 0 ? A * B > 0 : A.len_2() < B.len_2();
}
void _sort(point A[], point B[], int len) {
sort(A, A + len);
for (int i = 1; i < len; i++) {
B[i] = A[i] - A[0];
}
sort(B + 1, B + len, cmp);
B[0] = point(0, 0);
}
int ConvexHull(point A[], point B[], int len) {
_sort(A, A, len);
int tot = 0;
for (int i = 0; i < len; i++) {
if (tot <= 1)
B[tot++] = A[i];
else {
while (tot >= 2 && (B[tot - 1] - B[tot - 2]) * (A[i] - B[tot - 2]) <= 0)
tot--;
B[tot++] = A[i];
}
}
return tot;
}
P A[maxn], B[maxn], A1[maxn], B1[maxn];
int len1, len2;
int n, m;
tuple<long long, long long, long long> X[maxn * 2], Y[maxn];
void GetNextval(tuple<long long, long long, long long> p[], int next[],
int pLen) {
next[0] = -1;
int k = -1;
int j = 0;
while (j < pLen) {
if (k == -1 || p[j] == p[k]) {
++j, ++k;
if (p[j] != p[k]) {
next[j] = k;
} else {
next[j] = next[k];
}
} else {
k = next[k];
}
}
}
int nxt[123456];
int KmpSearch(tuple<long long, long long, long long> s[],
tuple<long long, long long, long long> p[], int sLen, int pLen) {
memset(nxt, -1, sizeof(nxt));
GetNextval(p, nxt, pLen);
int i = 0, j = 0;
while (i < sLen && j < pLen) {
if (j == -1 || s[i] == p[j]) {
i++, j++;
} else {
j = nxt[j];
}
}
if (j == pLen) {
return i - j;
} else {
return -1;
}
}
void kmp(point A[], point B[], int len) {
for (int i = 0; i < len; i++) {
X[i] = tuple<long long, long long, long long>(
(A[(i - 1 + len) % len] - A[i]).len_2(),
(A[(i + 1) % len] - A[i]).len_2(),
(A[(i - 1 + len) % len] - A[i]) * (A[(i + 1) % len] - A[i]));
X[i + len] = X[i];
}
for (int i = 0; i < len; i++) {
Y[i] = tuple<long long, long long, long long>(
(B[(i - 1 + len) % len] - B[i]).len_2(),
(B[(i + 1) % len] - B[i]).len_2(),
(B[(i - 1 + len) % len] - B[i]) * (B[(i + 1) % len] - B[i]));
}
if (KmpSearch(X, Y, 2 * len, len) != -1)
puts("YES");
else
puts("NO");
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++) {
scanf("%lld%lld", &A[i].x, &A[i].y);
}
for (int i = 0; i < m; i++) {
scanf("%lld%lld", &B[i].x, &B[i].y);
}
len1 = ConvexHull(A, A1, n);
len2 = ConvexHull(B, B1, m);
if (len1 == len2)
kmp(A1, B1, len1);
else
puts("NO");
return 0;
}
``` |
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int INF = 2147483647;
const int mod = 1000000007;
const int mod2 = 998244353;
const double eps = 1e-9;
const double pi = acos(-1.0);
inline int rint() {
int x;
scanf("%d", &x);
return x;
}
inline long long rll() {
long long x;
scanf("%lld", &x);
return x;
}
inline long long ri64() {
long long x;
scanf("%I64d", &x);
return x;
}
void rvi(vector<int> &a, int n) {
for (int i = 0; i < n; i++) {
a.push_back(rint());
}
}
void rvll(vector<long long> &a, int n) {
for (int i = 0; i < n; i++) {
a.push_back(rll());
}
}
void showvi(vector<int> &a) {
for (int i = 0; i < a.size(); i++) {
printf("%d%c", a[i], (i + 1 == a.size() ? '\n' : ' '));
}
}
void showvll(vector<long long> &a) {
for (int i = 0; i < a.size(); i++) {
printf("%lld%c", a[i], (i + 1 == a.size() ? '\n' : ' '));
}
}
void showviln(vector<int> &a) {
for (int i = 0; i < a.size(); i++) {
printf("%d%c", a[i], '\n');
}
}
void showvllln(vector<long long> &a) {
for (int i = 0; i < a.size(); i++) {
printf("%lld%c", a[i], '\n');
}
}
void showi(int x) { printf("%d", x); }
long long pw(long long a, long long b, long long c) {
long long ans = 1;
long long k = a % c;
while (b > 0) {
if (b % 2 == 1) ans = (ans * k) % c;
b = b / 2;
k = (k * k) % c;
}
return ans;
}
const int maxn = 100011;
struct node {
double x, y;
};
int n;
int tot;
node p[maxn];
node P[maxn];
int tot2;
node P2[maxn];
double X(node A, node B, node C) {
return (B.x - A.x) * (C.y - A.y) - (C.x - A.x) * (B.y - A.y);
}
double dian(node A, node B, node C) {
return (B.x - A.x) * (C.x - A.x) + (C.y - A.y) * (B.y - A.y);
}
double len(node A, node B) {
return sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
}
bool cmp(node A, node B) {
double pp = X(p[0], A, B);
if (pp > 0) return true;
if (pp < 0) return false;
if (pp == 0) return len(p[0], A) < len(p[0], B);
}
void build() {
for (int i = 0; i < n; i++) {
if (p[i].y < p[0].y) swap(p[i], p[0]);
if (p[i].y == p[0].y && p[i].x < p[0].x) swap(p[i], p[0]);
}
sort(p + 1, p + n, cmp);
P[0] = p[0];
P[1] = p[1];
tot = 1;
for (int i = 2; i < n; i++) {
while (tot > 0 && X(P[tot - 1], P[tot], p[i]) <= 0) tot--;
tot++;
P[tot] = p[i];
}
tot++;
}
int x1[100011];
int asdfeawf[100011];
int x2[100011];
int y2[100011];
int a[400044];
int b[400044];
int ha[400044];
int hb[400044];
int pwb[400044];
const int hmod = 1000000007;
const int mmod = 998244353;
const int base = 233;
int dis(int x1, int asdfeawf, int x2, int y2) {
int ans = 1LL * (x1 - x2) * (x1 - x2) % mmod +
1LL * (asdfeawf - y2) * (asdfeawf - y2) % mmod;
if (ans >= mmod) ans -= mmod;
return ans;
}
void work() {
for (int i = 0; i < (tot); i++) {
x1[i] = P[i].x + 0.1;
asdfeawf[i] = P[i].y + 0.1;
}
for (int i = 0; i < (tot2); i++) {
x2[i] = P2[i].x + 0.1;
y2[i] = P2[i].y + 0.1;
}
if (tot != tot2) {
cout << "NO" << '\n';
return;
}
for (int i = 0; i < (tot); i++) {
int X0 = x1[(i - 1 + tot) % tot];
int Y0 = asdfeawf[(i - 1 + tot) % tot];
int X1 = x1[i];
int Y1 = asdfeawf[i];
int X2 = x1[(((i) + 1) % (tot))];
int Y2 = asdfeawf[(((i) + 1) % (tot))];
int len0 = dis(X0, Y0, X1, Y1);
int len1 = dis(X1, Y1, X2, Y2);
a[2 * i + 1] = a[2 * i + 2 * tot + 1] = len1;
int VX0 = X0 - X1;
long long VY0 = Y0 - Y1;
int VX1 = X2 - X1;
long long VY1 = Y2 - Y1;
if (VX0 < 0) VX0 += mmod;
if (VX1 < 0) VX1 += mmod;
if (VY1 < 0) VY1 += mmod;
if (VY0 < 0) VY0 += mmod;
int DOT = (1LL * VX0 * VX1 % mmod + 1LL * VY0 * VY1 % mmod);
if (DOT >= mmod) DOT -= mod;
int MUL = 1LL * len0 * len1 % mmod;
int ANG = 1LL * DOT * pw(MUL, mmod - 2, mmod) % mmod;
a[2 * i] = a[2 * i + 2 * tot] = ANG;
}
for (int i = 0; i < (tot); i++) {
int X0 = x2[(i - 1 + tot) % tot];
int Y0 = y2[(i - 1 + tot) % tot];
int X1 = x2[i];
int Y1 = y2[i];
int X2 = x2[(((i) + 1) % (tot))];
int Y2 = y2[(((i) + 1) % (tot))];
int len0 = dis(X0, Y0, X1, Y1);
int len1 = dis(X1, Y1, X2, Y2);
b[2 * i + 1] = b[2 * i + 2 * tot + 1] = len1;
int VX0 = X0 - X1;
long long VY0 = Y0 - Y1;
int VX1 = X2 - X1;
long long VY1 = Y2 - Y1;
if (VX0 < 0) VX0 += mmod;
if (VX1 < 0) VX1 += mmod;
if (VY1 < 0) VY1 += mmod;
if (VY0 < 0) VY0 += mmod;
int DOT = (1LL * VX0 * VX1 % mmod + 1LL * VY0 * VY1 % mmod);
if (DOT >= mmod) DOT -= mod;
int MUL = 1LL * len0 * len1 % mmod;
int ANG = 1LL * DOT * pw(MUL, mmod - 2, mmod) % mmod;
b[2 * i] = b[2 * i + 2 * tot] = ANG;
}
for (int i = 0; i < (4 * tot); i++) {
ha[i] = 0;
if (i) ha[i] = 1LL * ha[i - 1] * base % hmod;
ha[i] = (ha[i] + a[i]);
if (ha[i] >= hmod) ha[i] -= hmod;
hb[i] = 0;
if (i) hb[i] = 1LL * hb[i - 1] * base % hmod;
hb[i] = (hb[i] + b[i]);
if (hb[i] >= hmod) hb[i] -= hmod;
}
int h1 = ha[2 * tot - 1];
for (int i = 0; i < 2 * tot; i += 2) {
int l = i;
int r = i + 2 * tot - 1;
int h2 = hb[r];
if (l > 0) h2 = h2 - 1LL * hb[l - 1] * pwb[r - l + 1] % hmod;
if (h2 < 0) h2 += hmod;
if (h1 == h2) {
cout << "YES" << '\n';
return;
}
}
cout << "NO" << '\n';
}
int main() {
pwb[0] = 1;
for (int i = 1; i < 400044; i++) pwb[i] = 1LL * pwb[i - 1] * base % hmod;
int n1, n2;
scanf("%d%d", &n1, &n2);
tot = 0;
n = n1;
for (int i = 0; i < (n1); i++) {
p[i].x = rint();
p[i].y = rint();
}
build();
for (int i = 0; i < (tot); i++) {
P2[i].x = P[i].x;
P2[i].y = P[i].y;
}
tot2 = tot;
tot = 0;
n = n2;
for (int i = 0; i < (n2); i++) {
p[i].x = rint();
p[i].y = rint();
}
build();
work();
}
| ### Prompt
In Cpp, your task is to solve the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int INF = 2147483647;
const int mod = 1000000007;
const int mod2 = 998244353;
const double eps = 1e-9;
const double pi = acos(-1.0);
inline int rint() {
int x;
scanf("%d", &x);
return x;
}
inline long long rll() {
long long x;
scanf("%lld", &x);
return x;
}
inline long long ri64() {
long long x;
scanf("%I64d", &x);
return x;
}
void rvi(vector<int> &a, int n) {
for (int i = 0; i < n; i++) {
a.push_back(rint());
}
}
void rvll(vector<long long> &a, int n) {
for (int i = 0; i < n; i++) {
a.push_back(rll());
}
}
void showvi(vector<int> &a) {
for (int i = 0; i < a.size(); i++) {
printf("%d%c", a[i], (i + 1 == a.size() ? '\n' : ' '));
}
}
void showvll(vector<long long> &a) {
for (int i = 0; i < a.size(); i++) {
printf("%lld%c", a[i], (i + 1 == a.size() ? '\n' : ' '));
}
}
void showviln(vector<int> &a) {
for (int i = 0; i < a.size(); i++) {
printf("%d%c", a[i], '\n');
}
}
void showvllln(vector<long long> &a) {
for (int i = 0; i < a.size(); i++) {
printf("%lld%c", a[i], '\n');
}
}
void showi(int x) { printf("%d", x); }
long long pw(long long a, long long b, long long c) {
long long ans = 1;
long long k = a % c;
while (b > 0) {
if (b % 2 == 1) ans = (ans * k) % c;
b = b / 2;
k = (k * k) % c;
}
return ans;
}
const int maxn = 100011;
struct node {
double x, y;
};
int n;
int tot;
node p[maxn];
node P[maxn];
int tot2;
node P2[maxn];
double X(node A, node B, node C) {
return (B.x - A.x) * (C.y - A.y) - (C.x - A.x) * (B.y - A.y);
}
double dian(node A, node B, node C) {
return (B.x - A.x) * (C.x - A.x) + (C.y - A.y) * (B.y - A.y);
}
double len(node A, node B) {
return sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
}
bool cmp(node A, node B) {
double pp = X(p[0], A, B);
if (pp > 0) return true;
if (pp < 0) return false;
if (pp == 0) return len(p[0], A) < len(p[0], B);
}
void build() {
for (int i = 0; i < n; i++) {
if (p[i].y < p[0].y) swap(p[i], p[0]);
if (p[i].y == p[0].y && p[i].x < p[0].x) swap(p[i], p[0]);
}
sort(p + 1, p + n, cmp);
P[0] = p[0];
P[1] = p[1];
tot = 1;
for (int i = 2; i < n; i++) {
while (tot > 0 && X(P[tot - 1], P[tot], p[i]) <= 0) tot--;
tot++;
P[tot] = p[i];
}
tot++;
}
int x1[100011];
int asdfeawf[100011];
int x2[100011];
int y2[100011];
int a[400044];
int b[400044];
int ha[400044];
int hb[400044];
int pwb[400044];
const int hmod = 1000000007;
const int mmod = 998244353;
const int base = 233;
int dis(int x1, int asdfeawf, int x2, int y2) {
int ans = 1LL * (x1 - x2) * (x1 - x2) % mmod +
1LL * (asdfeawf - y2) * (asdfeawf - y2) % mmod;
if (ans >= mmod) ans -= mmod;
return ans;
}
void work() {
for (int i = 0; i < (tot); i++) {
x1[i] = P[i].x + 0.1;
asdfeawf[i] = P[i].y + 0.1;
}
for (int i = 0; i < (tot2); i++) {
x2[i] = P2[i].x + 0.1;
y2[i] = P2[i].y + 0.1;
}
if (tot != tot2) {
cout << "NO" << '\n';
return;
}
for (int i = 0; i < (tot); i++) {
int X0 = x1[(i - 1 + tot) % tot];
int Y0 = asdfeawf[(i - 1 + tot) % tot];
int X1 = x1[i];
int Y1 = asdfeawf[i];
int X2 = x1[(((i) + 1) % (tot))];
int Y2 = asdfeawf[(((i) + 1) % (tot))];
int len0 = dis(X0, Y0, X1, Y1);
int len1 = dis(X1, Y1, X2, Y2);
a[2 * i + 1] = a[2 * i + 2 * tot + 1] = len1;
int VX0 = X0 - X1;
long long VY0 = Y0 - Y1;
int VX1 = X2 - X1;
long long VY1 = Y2 - Y1;
if (VX0 < 0) VX0 += mmod;
if (VX1 < 0) VX1 += mmod;
if (VY1 < 0) VY1 += mmod;
if (VY0 < 0) VY0 += mmod;
int DOT = (1LL * VX0 * VX1 % mmod + 1LL * VY0 * VY1 % mmod);
if (DOT >= mmod) DOT -= mod;
int MUL = 1LL * len0 * len1 % mmod;
int ANG = 1LL * DOT * pw(MUL, mmod - 2, mmod) % mmod;
a[2 * i] = a[2 * i + 2 * tot] = ANG;
}
for (int i = 0; i < (tot); i++) {
int X0 = x2[(i - 1 + tot) % tot];
int Y0 = y2[(i - 1 + tot) % tot];
int X1 = x2[i];
int Y1 = y2[i];
int X2 = x2[(((i) + 1) % (tot))];
int Y2 = y2[(((i) + 1) % (tot))];
int len0 = dis(X0, Y0, X1, Y1);
int len1 = dis(X1, Y1, X2, Y2);
b[2 * i + 1] = b[2 * i + 2 * tot + 1] = len1;
int VX0 = X0 - X1;
long long VY0 = Y0 - Y1;
int VX1 = X2 - X1;
long long VY1 = Y2 - Y1;
if (VX0 < 0) VX0 += mmod;
if (VX1 < 0) VX1 += mmod;
if (VY1 < 0) VY1 += mmod;
if (VY0 < 0) VY0 += mmod;
int DOT = (1LL * VX0 * VX1 % mmod + 1LL * VY0 * VY1 % mmod);
if (DOT >= mmod) DOT -= mod;
int MUL = 1LL * len0 * len1 % mmod;
int ANG = 1LL * DOT * pw(MUL, mmod - 2, mmod) % mmod;
b[2 * i] = b[2 * i + 2 * tot] = ANG;
}
for (int i = 0; i < (4 * tot); i++) {
ha[i] = 0;
if (i) ha[i] = 1LL * ha[i - 1] * base % hmod;
ha[i] = (ha[i] + a[i]);
if (ha[i] >= hmod) ha[i] -= hmod;
hb[i] = 0;
if (i) hb[i] = 1LL * hb[i - 1] * base % hmod;
hb[i] = (hb[i] + b[i]);
if (hb[i] >= hmod) hb[i] -= hmod;
}
int h1 = ha[2 * tot - 1];
for (int i = 0; i < 2 * tot; i += 2) {
int l = i;
int r = i + 2 * tot - 1;
int h2 = hb[r];
if (l > 0) h2 = h2 - 1LL * hb[l - 1] * pwb[r - l + 1] % hmod;
if (h2 < 0) h2 += hmod;
if (h1 == h2) {
cout << "YES" << '\n';
return;
}
}
cout << "NO" << '\n';
}
int main() {
pwb[0] = 1;
for (int i = 1; i < 400044; i++) pwb[i] = 1LL * pwb[i - 1] * base % hmod;
int n1, n2;
scanf("%d%d", &n1, &n2);
tot = 0;
n = n1;
for (int i = 0; i < (n1); i++) {
p[i].x = rint();
p[i].y = rint();
}
build();
for (int i = 0; i < (tot); i++) {
P2[i].x = P[i].x;
P2[i].y = P[i].y;
}
tot2 = tot;
tot = 0;
n = n2;
for (int i = 0; i < (n2); i++) {
p[i].x = rint();
p[i].y = rint();
}
build();
work();
}
``` |
#include <bits/stdc++.h>
using namespace std;
struct Pt {
long long x, y;
Pt() {}
Pt(long long _x, long long _y) : x(_x), y(_y) {}
Pt operator+(const Pt &p) const { return Pt(x + p.x, y + p.y); }
Pt operator-(const Pt &p) const { return Pt(x - p.x, y - p.y); }
Pt operator*(long long c) const { return Pt(x * c, y * c); }
Pt operator/(long long c) const { return Pt(x / c, y / c); }
long long abs2() { return x * x + y * y; }
};
inline long long dist2(Pt a, Pt b) {
long long dx = a.x - b.x;
long long dy = a.y - b.y;
return dx * dx + dy * dy;
}
inline long long cross(Pt a, Pt b) { return a.x * b.y - a.y * b.x; }
long long dot(Pt a, Pt b) { return a.x * b.x + a.y * b.y; }
inline int ccw(Pt p, Pt q, Pt r) {
long long v = cross(p - q, p - r);
if (v < 0) return -1;
if (v > 0) return 1;
return 0;
}
vector<Pt> convexHull(vector<Pt> &P) {
sort(P.begin(), P.end(), [](const Pt &a, const Pt &b) {
if (a.x != b.x) return a.x < b.x;
return a.y < b.y;
});
vector<Pt> up, dn, hull;
for (int i = 0; i < P.size(); ++i) {
while (up.size() > 1 && ccw(up[up.size() - 2], up.back(), P[i]) >= 0)
up.pop_back();
while (dn.size() > 1 && ccw(dn[dn.size() - 2], dn.back(), P[i]) <= 0)
dn.pop_back();
up.push_back(P[i]);
dn.push_back(P[i]);
}
hull = up;
for (int i = (int)dn.size() - 2; i > 0; --i) hull.push_back(dn[i]);
reverse(hull.begin(), hull.end());
return hull;
}
vector<long long> getdata(int n) {
vector<Pt> A(n);
for (int i = 0; i < n; ++i) {
scanf("%lld%lld", &A[i].x, &A[i].y);
}
A = convexHull(A);
if (A.size() == 2) return {(A[1] - A[0]).abs2()};
n = A.size();
vector<long long> data;
for (int i = 0; i < n; ++i) {
int a = (i + n - 1) % n, b = (i + 1) % n;
Pt aa = A[a] - A[i], bb = A[b] - A[i];
long long ang = dot(aa, bb);
data.push_back(aa.abs2());
data.push_back(ang);
}
return data;
}
void no() {
printf("NO\n");
exit(0);
}
void yes() {
printf("YES\n");
exit(0);
}
const int MAXH = 3;
long long P[MAXH] = {998244353, 1000000007, 1000000009};
long long Q[MAXH] = {10000103, 10000079, 10000019};
const int MAXN = 200005;
long long pw[MAXH][MAXN];
int main() {
int n, m;
scanf("%d%d", &n, &m);
auto a = getdata(n);
auto b = getdata(m);
if (a.size() > b.size()) swap(a, b);
if (a.size() == 1) {
if (b.size() > 1) no();
if (a[0] != b[0]) no();
yes();
}
n = a.size();
m = b.size();
if (n != m) no();
for (int h = 0; h < 3; ++h) {
pw[h][0] = 1;
for (int i = 1; i < MAXN; ++i) {
pw[h][i] = pw[h][i - 1] * Q[h] % P[h];
}
}
set<vector<long long>> has;
auto aa = a;
aa.resize(2 * n);
copy_n(aa.begin(), n, aa.begin() + n);
long long val[3] = {};
for (int i = 0; i < 2 * n; i += 2) {
for (int h = 0; h < 3; ++h) {
val[h] = (val[h] * Q[h] + aa[i]) % P[h];
if (i >= n) val[h] = (val[h] - aa[i - n] % P[h] * pw[h][n]) % P[h];
val[h] = (val[h] * Q[h] + aa[i + 1]) % P[h];
if (i >= n) val[h] = (val[h] - aa[i + 1 - n] % P[h] * pw[h][n]) % P[h];
if (val[h] < 0) val[h] += P[h];
}
if (i >= n) {
has.insert({val[0], val[1], val[2]});
}
}
memset(val, 0, sizeof(val));
for (int i = 0; i < n; i += 2) {
for (int h = 0; h < 3; ++h) {
val[h] = (val[h] * Q[h] + b[i]) % P[h];
val[h] = (val[h] * Q[h] + b[i + 1]) % P[h];
if (val[h] < 0) val[h] += P[h];
}
}
if (has.count({val[0], val[1], val[2]}))
yes();
else
no();
}
| ### Prompt
Create a solution in cpp for the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
struct Pt {
long long x, y;
Pt() {}
Pt(long long _x, long long _y) : x(_x), y(_y) {}
Pt operator+(const Pt &p) const { return Pt(x + p.x, y + p.y); }
Pt operator-(const Pt &p) const { return Pt(x - p.x, y - p.y); }
Pt operator*(long long c) const { return Pt(x * c, y * c); }
Pt operator/(long long c) const { return Pt(x / c, y / c); }
long long abs2() { return x * x + y * y; }
};
inline long long dist2(Pt a, Pt b) {
long long dx = a.x - b.x;
long long dy = a.y - b.y;
return dx * dx + dy * dy;
}
inline long long cross(Pt a, Pt b) { return a.x * b.y - a.y * b.x; }
long long dot(Pt a, Pt b) { return a.x * b.x + a.y * b.y; }
inline int ccw(Pt p, Pt q, Pt r) {
long long v = cross(p - q, p - r);
if (v < 0) return -1;
if (v > 0) return 1;
return 0;
}
vector<Pt> convexHull(vector<Pt> &P) {
sort(P.begin(), P.end(), [](const Pt &a, const Pt &b) {
if (a.x != b.x) return a.x < b.x;
return a.y < b.y;
});
vector<Pt> up, dn, hull;
for (int i = 0; i < P.size(); ++i) {
while (up.size() > 1 && ccw(up[up.size() - 2], up.back(), P[i]) >= 0)
up.pop_back();
while (dn.size() > 1 && ccw(dn[dn.size() - 2], dn.back(), P[i]) <= 0)
dn.pop_back();
up.push_back(P[i]);
dn.push_back(P[i]);
}
hull = up;
for (int i = (int)dn.size() - 2; i > 0; --i) hull.push_back(dn[i]);
reverse(hull.begin(), hull.end());
return hull;
}
vector<long long> getdata(int n) {
vector<Pt> A(n);
for (int i = 0; i < n; ++i) {
scanf("%lld%lld", &A[i].x, &A[i].y);
}
A = convexHull(A);
if (A.size() == 2) return {(A[1] - A[0]).abs2()};
n = A.size();
vector<long long> data;
for (int i = 0; i < n; ++i) {
int a = (i + n - 1) % n, b = (i + 1) % n;
Pt aa = A[a] - A[i], bb = A[b] - A[i];
long long ang = dot(aa, bb);
data.push_back(aa.abs2());
data.push_back(ang);
}
return data;
}
void no() {
printf("NO\n");
exit(0);
}
void yes() {
printf("YES\n");
exit(0);
}
const int MAXH = 3;
long long P[MAXH] = {998244353, 1000000007, 1000000009};
long long Q[MAXH] = {10000103, 10000079, 10000019};
const int MAXN = 200005;
long long pw[MAXH][MAXN];
int main() {
int n, m;
scanf("%d%d", &n, &m);
auto a = getdata(n);
auto b = getdata(m);
if (a.size() > b.size()) swap(a, b);
if (a.size() == 1) {
if (b.size() > 1) no();
if (a[0] != b[0]) no();
yes();
}
n = a.size();
m = b.size();
if (n != m) no();
for (int h = 0; h < 3; ++h) {
pw[h][0] = 1;
for (int i = 1; i < MAXN; ++i) {
pw[h][i] = pw[h][i - 1] * Q[h] % P[h];
}
}
set<vector<long long>> has;
auto aa = a;
aa.resize(2 * n);
copy_n(aa.begin(), n, aa.begin() + n);
long long val[3] = {};
for (int i = 0; i < 2 * n; i += 2) {
for (int h = 0; h < 3; ++h) {
val[h] = (val[h] * Q[h] + aa[i]) % P[h];
if (i >= n) val[h] = (val[h] - aa[i - n] % P[h] * pw[h][n]) % P[h];
val[h] = (val[h] * Q[h] + aa[i + 1]) % P[h];
if (i >= n) val[h] = (val[h] - aa[i + 1 - n] % P[h] * pw[h][n]) % P[h];
if (val[h] < 0) val[h] += P[h];
}
if (i >= n) {
has.insert({val[0], val[1], val[2]});
}
}
memset(val, 0, sizeof(val));
for (int i = 0; i < n; i += 2) {
for (int h = 0; h < 3; ++h) {
val[h] = (val[h] * Q[h] + b[i]) % P[h];
val[h] = (val[h] * Q[h] + b[i + 1]) % P[h];
if (val[h] < 0) val[h] += P[h];
}
}
if (has.count({val[0], val[1], val[2]}))
yes();
else
no();
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int Imx = 2147483647;
const long long Lbig = 2e18;
const int mod = 1e9 + 7;
struct fastio {
char s[100000];
int it, len;
fastio() { it = len = 0; }
inline char get() {
if (it < len) return s[it++];
it = 0;
len = fread(s, 1, 100000, stdin);
if (len == 0)
return EOF;
else
return s[it++];
}
bool notend() {
char c = get();
while (c == ' ' || c == '\n') c = get();
if (it > 0) it--;
return c != EOF;
}
} _buff;
inline long long getnum() {
long long r = 0;
bool ng = 0;
char c;
c = _buff.get();
while (c != '-' && (c < '0' || c > '9')) c = _buff.get();
if (c == '-') ng = 1, c = _buff.get();
while (c >= '0' && c <= '9') r = r * 10 + c - '0', c = _buff.get();
return ng ? -r : r;
}
template <class T>
inline void putnum(T x) {
if (x < 0) putchar('-'), x = -x;
register short a[20] = {}, sz = 0;
while (x) a[sz++] = x % 10, x /= 10;
if (sz == 0) putchar('0');
for (int i = sz - 1; i >= 0; i--) putchar('0' + a[i]);
}
inline char getreal() {
char c = _buff.get();
while (c <= 32) c = _buff.get();
return c;
}
long long qpow(long long x, long long k) {
return k == 0 ? 1
: 1ll * qpow(1ll * x * x % mod, k >> 1) * (k & 1 ? x : 1) % mod;
}
struct point {
long long x, y;
point(long long X = 0, long long Y = 0) {
x = X;
y = Y;
}
point operator+(const point &t) const { return point(x + t.x, y + t.y); }
point operator-(const point &t) const { return point(x - t.x, y - t.y); }
point operator*(const point &t) const {
return point(x * t.x - y * t.y, x * t.y + y * t.x);
}
long long det(const point &t) const {
double tmp = 1.0 * x * t.y - 1.0 * y * t.x;
if (tmp > 1e18) return 1e18;
if (tmp < -1e18) return -1e18;
return x * t.y - y * t.x;
}
bool operator<(const point &t) const {
return make_pair(x, y) < make_pair(t.x, t.y);
}
bool operator==(const point &t) const {
return make_pair(x, y) == make_pair(t.x, t.y);
}
long long sqrl() { return x * x + y * y; }
};
point a[100111];
int ch[100111], chn;
int st[100111], stn;
vector<long long> mini(vector<long long> v) {
if (v.size() == 1) return v;
vector<long long> ret;
int i = 0, j = 1, k = 0, n = v.size();
while (i != j && i < n && j < n) {
k = 0;
while (k < n && v[(i + k) % n] == v[(j + k) % n]) k++;
if (v[(i + k) % n] < v[(j + k) % n])
j = max(j + k + 1, i + 1);
else
i = max(i + k + 1, j + 1);
}
int pos = min(i, j);
for (int t = 0; t < n; t++) ret.push_back(v[(pos + t) % n]);
return ret;
}
vector<long long> getv(int n) {
for (int i = 1; i <= n; i++) {
a[i].x = getnum(), a[i].y = getnum();
}
stn = 0;
sort(a + 1, a + n + 1);
chn = 0;
for (int i = 1; i <= n; i++) {
while (chn >= 2 &&
(a[i] - a[ch[chn - 2]]).det(a[ch[chn - 1]] - a[ch[chn - 2]]) >= 0)
chn--;
ch[chn++] = i;
}
for (int i = 0; i < chn; i++) st[stn++] = ch[i];
chn = 0;
for (int i = 1; i <= n; i++) {
while (chn >= 2 &&
(a[i] - a[ch[chn - 2]]).det(a[ch[chn - 1]] - a[ch[chn - 2]]) <= 0)
chn--;
ch[chn++] = i;
}
for (int i = chn - 2; i > 0; i--) st[stn++] = ch[i];
st[stn] = st[0];
vector<point> v;
for (int i = 0; i < stn; i++) {
v.push_back(a[st[i + 1]] - a[st[i]]);
}
v.push_back(v[0]);
vector<long long> ret;
for (int i = 0; i + 1 < v.size(); i++) {
ret.push_back(v[i].sqrl());
ret.push_back((v[i + 1] - v[i]).sqrl());
}
ret = mini(ret);
return ret;
}
vector<long long> A, B;
int n, m;
int main() {
n = getnum(), m = getnum();
A = getv(n);
B = getv(m);
if (A == B)
puts("YES");
else
puts("NO");
return 0;
}
| ### Prompt
Please formulate a cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int Imx = 2147483647;
const long long Lbig = 2e18;
const int mod = 1e9 + 7;
struct fastio {
char s[100000];
int it, len;
fastio() { it = len = 0; }
inline char get() {
if (it < len) return s[it++];
it = 0;
len = fread(s, 1, 100000, stdin);
if (len == 0)
return EOF;
else
return s[it++];
}
bool notend() {
char c = get();
while (c == ' ' || c == '\n') c = get();
if (it > 0) it--;
return c != EOF;
}
} _buff;
inline long long getnum() {
long long r = 0;
bool ng = 0;
char c;
c = _buff.get();
while (c != '-' && (c < '0' || c > '9')) c = _buff.get();
if (c == '-') ng = 1, c = _buff.get();
while (c >= '0' && c <= '9') r = r * 10 + c - '0', c = _buff.get();
return ng ? -r : r;
}
template <class T>
inline void putnum(T x) {
if (x < 0) putchar('-'), x = -x;
register short a[20] = {}, sz = 0;
while (x) a[sz++] = x % 10, x /= 10;
if (sz == 0) putchar('0');
for (int i = sz - 1; i >= 0; i--) putchar('0' + a[i]);
}
inline char getreal() {
char c = _buff.get();
while (c <= 32) c = _buff.get();
return c;
}
long long qpow(long long x, long long k) {
return k == 0 ? 1
: 1ll * qpow(1ll * x * x % mod, k >> 1) * (k & 1 ? x : 1) % mod;
}
struct point {
long long x, y;
point(long long X = 0, long long Y = 0) {
x = X;
y = Y;
}
point operator+(const point &t) const { return point(x + t.x, y + t.y); }
point operator-(const point &t) const { return point(x - t.x, y - t.y); }
point operator*(const point &t) const {
return point(x * t.x - y * t.y, x * t.y + y * t.x);
}
long long det(const point &t) const {
double tmp = 1.0 * x * t.y - 1.0 * y * t.x;
if (tmp > 1e18) return 1e18;
if (tmp < -1e18) return -1e18;
return x * t.y - y * t.x;
}
bool operator<(const point &t) const {
return make_pair(x, y) < make_pair(t.x, t.y);
}
bool operator==(const point &t) const {
return make_pair(x, y) == make_pair(t.x, t.y);
}
long long sqrl() { return x * x + y * y; }
};
point a[100111];
int ch[100111], chn;
int st[100111], stn;
vector<long long> mini(vector<long long> v) {
if (v.size() == 1) return v;
vector<long long> ret;
int i = 0, j = 1, k = 0, n = v.size();
while (i != j && i < n && j < n) {
k = 0;
while (k < n && v[(i + k) % n] == v[(j + k) % n]) k++;
if (v[(i + k) % n] < v[(j + k) % n])
j = max(j + k + 1, i + 1);
else
i = max(i + k + 1, j + 1);
}
int pos = min(i, j);
for (int t = 0; t < n; t++) ret.push_back(v[(pos + t) % n]);
return ret;
}
vector<long long> getv(int n) {
for (int i = 1; i <= n; i++) {
a[i].x = getnum(), a[i].y = getnum();
}
stn = 0;
sort(a + 1, a + n + 1);
chn = 0;
for (int i = 1; i <= n; i++) {
while (chn >= 2 &&
(a[i] - a[ch[chn - 2]]).det(a[ch[chn - 1]] - a[ch[chn - 2]]) >= 0)
chn--;
ch[chn++] = i;
}
for (int i = 0; i < chn; i++) st[stn++] = ch[i];
chn = 0;
for (int i = 1; i <= n; i++) {
while (chn >= 2 &&
(a[i] - a[ch[chn - 2]]).det(a[ch[chn - 1]] - a[ch[chn - 2]]) <= 0)
chn--;
ch[chn++] = i;
}
for (int i = chn - 2; i > 0; i--) st[stn++] = ch[i];
st[stn] = st[0];
vector<point> v;
for (int i = 0; i < stn; i++) {
v.push_back(a[st[i + 1]] - a[st[i]]);
}
v.push_back(v[0]);
vector<long long> ret;
for (int i = 0; i + 1 < v.size(); i++) {
ret.push_back(v[i].sqrl());
ret.push_back((v[i + 1] - v[i]).sqrl());
}
ret = mini(ret);
return ret;
}
vector<long long> A, B;
int n, m;
int main() {
n = getnum(), m = getnum();
A = getv(n);
B = getv(m);
if (A == B)
puts("YES");
else
puts("NO");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = int(1e5) + 100;
bool cmp(pair<long long, long long> b, pair<long long, long long> c);
long long det(pair<long long, long long> b, pair<long long, long long> c,
pair<long long, long long> o);
pair<long long, long long> tmp[maxn];
struct convex {
int n;
pair<long long, long long> p[maxn];
convex() { n = 0; }
void build() {
sort(p + 1, p + 1 + n, cmp);
for (int i = 1; i <= n; ++i) tmp[i] = p[i];
int m = n;
n = 1;
for (int i = 2; i <= m; ++i) {
while (n > 1 && det(tmp[i], p[n], p[n - 1]) >= 0) --n;
p[++n] = tmp[i];
}
int tm = n;
for (int i = m - 1; i; --i) {
while (n > tm && det(tmp[i], p[n], p[n - 1]) >= 0) --n;
p[++n] = tmp[i];
}
--n;
p[0] = p[n];
}
};
convex cx1, cx2;
long long len[maxn], angle[maxn];
int fail[maxn];
long long det(pair<long long, long long> b, pair<long long, long long> c,
pair<long long, long long> o) {
return (b.first - o.first) * (c.second - o.second) -
(b.second - o.second) * (c.first - o.first);
}
long long dot(pair<long long, long long> b, pair<long long, long long> c,
pair<long long, long long> o) {
return (b.first - o.first) * (c.first - o.first) +
(b.second - o.second) * (c.second - o.second);
}
inline long long sqr(long long first) { return first * first; }
long long dis(pair<long long, long long> b, pair<long long, long long> c) {
return sqr(b.first - c.first) + sqr(b.second - c.second);
}
bool cmp(pair<long long, long long> b, pair<long long, long long> c) {
if (b.second != c.second) return b.second < c.second;
return b.first < c.first;
}
void read() {
scanf("%d%d", &cx1.n, &cx2.n);
for (int i = 1; i <= cx1.n; ++i)
scanf("%lld%lld", &cx1.p[i].first, &cx1.p[i].second);
for (int i = 1; i <= cx2.n; ++i)
scanf("%lld%lld", &cx2.p[i].first, &cx2.p[i].second);
}
void solve() {
cx1.build();
cx2.build();
if (cx1.n != cx2.n) {
puts("NO");
return;
}
int n = cx2.n;
for (int i = 1; i <= n; ++i) {
len[i] = dis(cx1.p[i], cx1.p[i + 1]);
angle[i] = dot(cx1.p[i + 1], cx1.p[i - 1], cx1.p[i]);
}
fail[1] = 0;
for (int i = 2, j = 0; i <= n; ++i) {
while (j && !(len[j + 1] == len[i] && angle[j + 1] == angle[i]))
j = fail[j];
if (len[j + 1] == len[i] && angle[j + 1] == angle[i]) ++j;
fail[i] = j;
}
for (int i = n + 1; i <= n * 2 + 1; ++i) cx2.p[i] = cx2.p[i - n];
for (int i = 1, j = 0; i <= n * 2; ++i) {
long long tlen = dis(cx2.p[i], cx2.p[i + 1]);
long long tangle = dot(cx2.p[i + 1], cx2.p[i - 1], cx2.p[i]);
while (j && !(len[j + 1] == tlen && angle[j + 1] == tangle)) j = fail[j];
if (len[j + 1] == tlen && angle[j + 1] == tangle) ++j;
if (j == n) {
puts("YES");
return;
}
}
puts("NO");
}
int main() {
read();
solve();
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = int(1e5) + 100;
bool cmp(pair<long long, long long> b, pair<long long, long long> c);
long long det(pair<long long, long long> b, pair<long long, long long> c,
pair<long long, long long> o);
pair<long long, long long> tmp[maxn];
struct convex {
int n;
pair<long long, long long> p[maxn];
convex() { n = 0; }
void build() {
sort(p + 1, p + 1 + n, cmp);
for (int i = 1; i <= n; ++i) tmp[i] = p[i];
int m = n;
n = 1;
for (int i = 2; i <= m; ++i) {
while (n > 1 && det(tmp[i], p[n], p[n - 1]) >= 0) --n;
p[++n] = tmp[i];
}
int tm = n;
for (int i = m - 1; i; --i) {
while (n > tm && det(tmp[i], p[n], p[n - 1]) >= 0) --n;
p[++n] = tmp[i];
}
--n;
p[0] = p[n];
}
};
convex cx1, cx2;
long long len[maxn], angle[maxn];
int fail[maxn];
long long det(pair<long long, long long> b, pair<long long, long long> c,
pair<long long, long long> o) {
return (b.first - o.first) * (c.second - o.second) -
(b.second - o.second) * (c.first - o.first);
}
long long dot(pair<long long, long long> b, pair<long long, long long> c,
pair<long long, long long> o) {
return (b.first - o.first) * (c.first - o.first) +
(b.second - o.second) * (c.second - o.second);
}
inline long long sqr(long long first) { return first * first; }
long long dis(pair<long long, long long> b, pair<long long, long long> c) {
return sqr(b.first - c.first) + sqr(b.second - c.second);
}
bool cmp(pair<long long, long long> b, pair<long long, long long> c) {
if (b.second != c.second) return b.second < c.second;
return b.first < c.first;
}
void read() {
scanf("%d%d", &cx1.n, &cx2.n);
for (int i = 1; i <= cx1.n; ++i)
scanf("%lld%lld", &cx1.p[i].first, &cx1.p[i].second);
for (int i = 1; i <= cx2.n; ++i)
scanf("%lld%lld", &cx2.p[i].first, &cx2.p[i].second);
}
void solve() {
cx1.build();
cx2.build();
if (cx1.n != cx2.n) {
puts("NO");
return;
}
int n = cx2.n;
for (int i = 1; i <= n; ++i) {
len[i] = dis(cx1.p[i], cx1.p[i + 1]);
angle[i] = dot(cx1.p[i + 1], cx1.p[i - 1], cx1.p[i]);
}
fail[1] = 0;
for (int i = 2, j = 0; i <= n; ++i) {
while (j && !(len[j + 1] == len[i] && angle[j + 1] == angle[i]))
j = fail[j];
if (len[j + 1] == len[i] && angle[j + 1] == angle[i]) ++j;
fail[i] = j;
}
for (int i = n + 1; i <= n * 2 + 1; ++i) cx2.p[i] = cx2.p[i - n];
for (int i = 1, j = 0; i <= n * 2; ++i) {
long long tlen = dis(cx2.p[i], cx2.p[i + 1]);
long long tangle = dot(cx2.p[i + 1], cx2.p[i - 1], cx2.p[i]);
while (j && !(len[j + 1] == tlen && angle[j + 1] == tangle)) j = fail[j];
if (len[j + 1] == tlen && angle[j + 1] == tangle) ++j;
if (j == n) {
puts("YES");
return;
}
}
puts("NO");
}
int main() {
read();
solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
const double eps = 1e-8;
const double pi = acos(-1.0);
const long long INF = 0x3f3f3f3f3f3f3f3f;
int dcmp(double x) {
if (fabs(x) < eps) return 0;
return (x > 0 ? 1 : -1);
}
inline double sqr(double x) { return x * x; }
struct Point {
long long x, y;
Point() { x = 0, y = 0; }
Point(long long _x, long long _y) : x(_x), y(_y) {}
void input() { scanf("%lld%lld", &x, &y); }
void output() { printf("%lld %lld\n", x, y); }
friend istream &operator>>(istream &os, Point &b) {
os >> b.x >> b.y;
return os;
}
friend ostream &operator<<(ostream &os, Point &b) {
os << b.x << ' ' << b.y;
return os;
}
bool operator==(const Point &b) const {
return (dcmp(x - b.x) == 0 && dcmp(y - b.y) == 0);
}
bool operator!=(const Point &b) const {
return !((dcmp(x - b.x) == 0 && dcmp(y - b.y) == 0));
}
bool operator<(const Point &b) const {
return (dcmp(x - b.x) == 0 ? dcmp(y - b.y) < 0 : x < b.x);
}
long long operator^(const Point &b) const { return x * b.y - y * b.x; }
long long operator*(const Point &b) const { return x * b.x + y * b.y; }
Point operator+(const Point &b) const { return Point(x + b.x, y + b.y); }
Point operator-(const Point &b) const { return Point(x - b.x, y - b.y); }
Point operator*(double a) { return Point(x * a, y * a); }
Point operator/(double a) { return Point(x / a, y / a); }
long long len2() { return sqr(x) + sqr(y); }
long long len() { return sqrt(len2()); }
double polar() { return atan2(y, x); }
Point change_len(double r) {
double l = len();
if (dcmp(l) == 0) return *this;
return Point(x * r / l, y * r / l);
}
Point rotate_left() { return Point(-y, x); }
Point rotate_right() { return Point(y, -x); }
Point rotate(Point p, double ang) {
Point v = (*this) - p;
double c = cos(ang), s = sin(ang);
return Point(p.x + v.x * c - v.y * s, p.y + v.x * s + v.y * c);
}
Point normal() { return Point(-y / len(), x / len()); }
};
double cross(Point a, Point b) { return a.x * b.y - a.y * b.x; }
double dot(Point a, Point b) { return a.x * b.x + a.y * b.y; }
double dis(Point a, Point b) {
Point p = b - a;
return p.len();
}
double dis2(Point a, Point b) {
Point p = b - a;
return p.len2();
}
double rad_degree(double rad) { return rad / pi * 180; }
double degree_rad(double deg) { return deg / 180 * pi; }
double rad(Point a, Point b) {
return fabs(atan2(fabs(cross(a, b)), dot(a, b)));
}
bool is_parallel(Point a, Point b) {
double p = rad(a, b);
return dcmp(p) == 0 || dcmp(p - pi) == 0;
}
struct Line {
Point s, e;
Line() {}
Line(Point _s, Point _e) : s(_s), e(_e) {}
Line(Point p, double ang) {
s = p;
if (dcmp(ang - pi / 2) == 0) {
e = s + Point(0, 1);
} else {
e = s + Point(1, tan(ang));
}
}
Line(double a, double b, double c) {
if (dcmp(a) == 0) {
s = Point(0, -c / b);
e = Point(1, -c / b);
} else if (dcmp(b) == 0) {
s = Point(-c / a, 0);
e = Point(-c / a, 1);
} else {
s = Point(0, -c / b);
e = Point(1, (-c - a) / b);
}
}
void input() {
s.input();
e.input();
}
void adjust() {
if (e < s) swap(e, s);
}
double length() { return dis(s, e); }
double polar() { return atan2(e.y - s.y, e.x - s.x); }
double angle() {
double k = atan2(e.y - s.y, e.x - s.x);
if (dcmp(k) < 0) k += pi;
if (dcmp(k - pi) == 0) k -= pi;
return k;
}
Point operator&(const Line &b) const {
Point res = s;
double t = ((s - b.s) ^ (b.s - b.e)) / ((s - e) ^ (b.s - b.e));
res.x += (e.x - s.x) * t;
res.y += (e.y - s.y) * t;
return res;
}
};
int relation(Point p, Line l) {
int c = dcmp(cross(p - l.s, l.e - l.s));
if (c < 0)
return 1;
else if (c > 0)
return 2;
else
return 3;
}
bool point_on_seg(Point p, Line l) {
return dcmp(cross(p - l.s, l.e - l.s)) == 0 &&
dcmp(dot(p - l.s, p - l.e) <= 0);
}
bool is_parallel(Line a, Line b) { return is_parallel(a.e - a.s, b.e - b.s); }
int seg_cross_seg(Line a, Line v) {
int d1 = dcmp(cross(a.e - a.s, v.s - a.s));
int d2 = dcmp(cross(a.e - a.s, v.e - a.s));
int d3 = dcmp(cross(v.e - v.s, a.s - v.s));
int d4 = dcmp(cross(v.e - v.s, a.e - v.s));
if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2;
return (d1 == 0 && dcmp(dot(v.s - a.s, v.s - a.e)) <= 0) ||
(d2 == 0 && dcmp(dot(v.e - a.s, v.e - a.e)) <= 0) ||
(d3 == 0 && dcmp(dot(a.s - v.s, a.s - v.e)) <= 0) ||
(d4 == 0 && dcmp(dot(a.e - v.s, a.e - v.e)) <= 0);
}
int line_cross_seg(Line a, Line v) {
int d1 = dcmp(cross(a.e - a.s, v.s - a.s));
int d2 = dcmp(cross(a.e - a.s, v.e - a.s));
if ((d1 ^ d2) == -2) return 2;
return (d1 == 0 || d2 == 0);
}
int line_cross_line(Line a, Line v) {
if (is_parallel(a, v)) return relation(a.e, v) == 3;
return 2;
}
Point line_intersection(Line a, Line v) {
double a1 = cross(v.e - v.s, a.s - v.s);
double a2 = cross(v.e - v.s, a.e - v.s);
return Point((a.s.x * a2 - a.e.x * a1) / (a2 - a1),
(a.s.y * a2 - a.e.y * a1) / (a2 - a1));
}
Point line_intersection(Line line, double a, double b, double c) {
double u = fabs(a * line.s.x + b * line.s.y + c);
double v = fabs(a * line.e.x + b * line.e.y + c);
return Point((line.s.x * v + line.e.x * u) / (u + v),
(line.s.y * v + line.e.y * u) / (u + v));
}
double point_to_line(Point p, Line a) {
return fabs(cross(p - a.s, a.e - a.s) / a.length());
}
double point_to_seg(Point p, Line a) {
if (dcmp(dot(p - a.s, a.e - a.s)) < 0 || dcmp(dot(p - a.e, a.s - a.e)) < 0)
return min(dis(p, a.e), dis(p, a.s));
return point_to_line(p, a);
}
Point projection(Point p, Line a) {
return a.s + (((a.e - a.s) * dot(a.e - a.s, p - a.s)) / (a.e - a.s).len2());
}
Point symmetry(Point p, Line a) {
Point q = projection(p, a);
return Point(2 * q.x - p.x, 2 * q.y - p.y);
}
Line vector_move_inward(Line line, double d) {
double len = line.length();
Point p((line.s.y - line.e.y) * d / len, (line.e.x - line.s.x) * d / len);
return Line(line.s + p, line.e + p);
}
struct Circle {
Point p;
double r;
Circle() {}
Circle(Point _p, double _r) : p(_p), r(_r) {}
Circle(double a, double b, double _r) {
p = Point(a, b);
r = _r;
}
void input() {
p.input();
scanf("%lf", &r);
}
void output() {
p.output();
printf("%.2f\n", r);
}
bool operator==(const Circle &a) const {
return p == a.p && (dcmp(r - a.r) == 0);
}
double area() { return pi * r * r; }
double circumference() { return 2 * pi * r; }
};
int relation(Point p, Circle a) {
double d = dis(p, a.p);
if (dcmp(d - a.r) == 0) return 1;
return (dcmp(d - a.r) < 0 ? 2 : 0);
}
int relation(Line a, Circle b) {
double p = point_to_line(b.p, a);
if (dcmp(p - b.r) == 0) return 1;
return (dcmp(p - b.r) < 0 ? 2 : 0);
}
int relation(Circle a, Circle v) {
double d = dis(a.p, v.p);
if (dcmp(d - a.r - v.r) > 0) return 5;
if (dcmp(d - a.r - v.r) == 0) return 4;
double l = fabs(a.r - v.r);
if (dcmp(d - l) > 0) return 3;
if (dcmp(d - l) == 0) return 2;
return 1;
}
Circle triangle_out_circle(Point a, Point b, Point c) {
Line u = Line((a + b) / 2, ((a + b) / 2) + (b - a).rotate_left());
Line v = Line((b + c) / 2, ((b + c) / 2) + (c - b).rotate_left());
Point p = line_intersection(u, v);
double r = dis(p, a);
return Circle(p, r);
}
Circle triangle_in_circle(Point a, Point b, Point c) {
Line u, v;
double m = atan2(b.y - a.y, b.x - a.x), n = atan2(c.y - a.y, c.x - a.x);
u.s = a;
u.e = u.s + Point(cos((n + m) / 2), sin((n + m) / 2));
v.s = b;
m = atan2(a.y - b.y, a.x - b.x), n = atan2(c.y - b.y, c.x - b.x);
v.e = v.s + Point(cos((n + m) / 2), sin((n + m) / 2));
Point p = line_intersection(u, v);
double r = point_to_seg(p, Line(a, b));
return Circle(p, r);
}
int circle_intersection(Circle a, Circle v, Point &p1, Point &p2) {
int rel = relation(a, v);
if (rel == 1 || rel == 5) return 0;
double d = dis(a.p, v.p);
double l = (d * d + a.r * a.r - v.r * v.r) / (2 * d);
double h = sqrt(a.r * a.r - l * l);
Point tmp = a.p + (v.p - a.p).change_len(l);
p1 = tmp + ((v.p - a.p).rotate_left().change_len(h));
p2 = tmp + ((v.p - a.p).rotate_right().change_len(h));
if (rel == 2 || rel == 4) return 1;
return 2;
}
int line_circle_intersection(Line v, Circle u, Point &p1, Point &p2) {
if (!relation(v, u)) return 0;
Point a = projection(u.p, v);
double d = point_to_line(u.p, v);
d = sqrt(u.r * u.r - d * d);
if (dcmp(d) == 0) {
p1 = a, p2 = a;
return 1;
}
p1 = a + (v.e - v.s).change_len(d);
p2 = a - (v.e - v.s).change_len(d);
return 2;
}
int get_circle(Point a, Point b, double r1, Circle &c1, Circle &c2) {
Circle x(a, r1), y(b, r1);
int t = circle_intersection(x, y, c1.p, c2.p);
if (!t) return 0;
c1.r = c2.r = r1;
return t;
}
int get_circle(Line u, Point q, double r1, Circle &c1, Circle &c2) {
double d = point_to_line(q, u);
if (dcmp(d - r1 * 2) > 0) return 0;
if (dcmp(d) == 0) {
c1.p = q + ((u.e - u.s).rotate_left().change_len(r1));
c2.p = q + ((u.e - u.s).rotate_right().change_len(r1));
c1.r = c2.r = r1;
return 2;
}
Line u1 = Line(u.s + (u.e - u.s).rotate_left().change_len(r1),
u.e + (u.e - u.s).rotate_left().change_len(r1));
Line u2 = Line(u.s + (u.e - u.s).rotate_right().change_len(r1),
u.e + (u.e - u.s).rotate_right().change_len(r1));
Circle cc = Circle(q, r1);
Point p1, p2;
if (!line_circle_intersection(u1, cc, p1, p2))
line_circle_intersection(u2, cc, p1, p2);
c1 = Circle(p1, r1);
if (p1 == p2) {
c2 = c1;
return 1;
}
c2 = Circle(p2, r1);
return 2;
}
int get_circle(Line u, Line v, double r1, Circle &c1, Circle &c2, Circle &c3,
Circle &c4) {
if (is_parallel(u, v)) return 0;
Line u1 = Line(u.s + (u.e - u.s).rotate_left().change_len(r1),
u.e + (u.e - u.s).rotate_left().change_len(r1));
Line u2 = Line(u.s + (u.e - u.s).rotate_right().change_len(r1),
u.e + (u.e - u.s).rotate_right().change_len(r1));
Line v1 = Line(v.s + (v.e - v.s).rotate_left().change_len(r1),
v.e + (v.e - v.s).rotate_left().change_len(r1));
Line v2 = Line(v.s + (v.e - v.s).rotate_right().change_len(r1),
v.e + (v.e - v.s).rotate_right().change_len(r1));
c1.r = c2.r = c3.r = c4.r = r1;
c1.p = line_intersection(u1, v1);
c2.p = line_intersection(u1, v2);
c3.p = line_intersection(u2, v1);
c4.p = line_intersection(u2, v2);
return 4;
}
int get_circle(Circle cx, Circle cy, double r1, Circle &c1, Circle &c2) {
Circle x(cx.p, r1 + cx.r), y(cy.p, r1 + cy.r);
int t = circle_intersection(x, y, c1.p, c2.p);
if (!t) return 0;
c1.r = c2.r = r1;
return t;
}
Circle get_circle(Point a, Point b, double k) {
double k2 = k * k;
double d = (2 * b.x - 2 * a.x / k2) / ((1 / k2) - 1);
double e = (2 * b.y - 2 * a.y / k2) / ((1 / k2) - 1);
double f = (a.x * a.x / k2 - b.x * b.x + a.y * a.y / k2 - b.y * b.y) /
((1 / k2) - 1);
double cx = -d / 2, cy = -e / 2;
double r = sqrt((d * d + e * e) / 4 - f);
return Circle(Point(cx, cy), r);
}
bool get_circle(Point a, Point b, double r, Point &res1, Point &res2) {
if (dcmp(dis(a, b) - 2 * r) > 0) return 0;
Point mid((a.x + b.x) / 2, (a.y + b.y) / 2);
double angle = atan2(b.y - a.y, b.x - a.x),
d = sqrt(r * r - sqr(dis(a, b) / 2));
res1 = Point(mid.x + d * sin(angle), mid.y - d * cos(angle));
res2 = Point(mid.x - d * sin(angle), mid.y + d * cos(angle));
return 1;
}
Circle get_circle(Point p1, Point p2, Point p3) {
if (dcmp(cross(p2 - p1, p3 - p1)) == 0) {
if (p2 < p1) swap(p1, p2);
if (p3 < p1) swap(p1, p3);
if (p3 < p2) swap(p2, p3);
return Circle((p1 + p3) / 2, dis(p1, p3) / 2);
}
double x1 = p1.x, x2 = p2.x, x3 = p3.x;
double y1 = p1.y, y2 = p2.y, y3 = p3.y;
double a = x1 - x2, b = y1 - y2, c = x1 - x3, d = y1 - y3;
double e = ((x1 * x1 - x2 * x2) + (y1 * y1 - y2 * y2)) / 2;
double f = ((x1 * x1 - x3 * x3) + (y1 * y1 - y3 * y3)) / 2;
double delta = b * c - a * d;
Point pc = Point(-(d * e - b * f) / delta, -(a * f - c * e) / delta);
return Circle(pc, dis(pc, p1));
}
int tan_line(Point q, Circle a, Line &u, Line &v) {
int x = relation(q, a);
if (x == 2) return 0;
if (x == 1) {
u = Line(q, q + (q - a.p).rotate_left());
v = u;
return 1;
}
double d = dis(a.p, q);
double l = a.r * a.r / d;
double h = sqrt(a.r * a.r - l * l);
u = Line(
q, a.p + (q - a.p).change_len(l) + (q - a.p).rotate_left().change_len(h));
v = Line(q, a.p + (q - a.p).change_len(l) +
(q - a.p).rotate_right().change_len(h));
return 2;
}
double area_circle(Circle a, Circle v) {
int rel = relation(a, v);
if (rel >= 4) return 0;
if (rel <= 2) return min(a.area(), v.area());
double d = dis(a.p, v.p);
double hf = (a.r + v.r + d) / 2;
double ss = 2 * sqrt(hf * (hf - a.r) * (hf - v.r) * (hf - d));
double a1 = acos((a.r * a.r + d * d - v.r * v.r) / (2 * a.r * d));
a1 = a1 * a.r * a.r;
double a2 = acos((v.r * v.r + d * d - a.r * a.r) / (2 * v.r * d));
a2 = a2 * v.r * v.r;
return a1 + a2 - ss;
}
double circle_traingle_area(Point a, Point b, Circle c) {
Point p = c.p;
double r = c.r;
if (dcmp(cross(p - a, p - b)) == 0) return 0;
Point q[5];
int len = 0;
q[len++] = a;
Line l(a, b);
Point p1, p2;
if (line_circle_intersection(l, c, q[1], q[2]) == 2) {
if (dcmp(dot(a - q[1], b - q[1])) < 0) q[len++] = q[1];
if (dcmp(dot(a - q[2], b - q[2])) < 0) q[len++] = q[2];
}
q[len++] = b;
if (len == 4 && dcmp(dot(q[0] - q[1], q[2] - q[1])) > 0) swap(q[1], q[2]);
double res = 0;
for (int i = 0; i < len - 1; i++) {
if (relation(q[i], c) == 0 || relation(q[i + 1], c) == 0) {
double arg = rad(q[i] - p, q[i + 1] - p);
res += r * r * arg / 2;
} else
res += fabs(cross(q[i] - p, q[i + 1] - p)) / 2;
}
return res;
}
int circle_cover_point(Point *p, int n, double r) {
int ans = 1;
pair<double, int> angle[MAXN];
for (int i = 0; i < n; i++) {
int m = 0;
double d;
for (int j = 0; j < n; j++)
if (dcmp((d = dis(p[i], p[j])) - 2 * r) <= 0 && i != j) {
double delta = acos(d / (2 * r)), polar = (p[j] - p[i]).polar();
angle[m++] = make_pair(polar - delta, 1);
angle[m++] = make_pair(polar + delta, -1);
}
sort(angle, angle + m);
for (int j = 0, sum = 1; j < m; j++) {
sum += angle[j].second;
ans = max(ans, sum);
}
}
return ans;
}
Circle circle_min_conver(Point *p, int n) {
Point center = p[0];
double r = 0;
random_shuffle(p, p + n);
for (int i = 1; i < n; i++) {
if (dcmp(dis(center, p[i]) - r) > 0) {
center = p[i], r = 0;
for (int j = 0; j < i; j++) {
if (dcmp(dis(center, p[j]) - r) > 0) {
center = (p[i] + p[j]) / 2, r = dis(center, p[j]);
for (int k = 0; k < j; k++) {
if (dcmp(dis(center, p[k]) - r) > 0) {
center = get_circle(p[i], p[j], p[k]).p;
r = dis(center, p[i]);
}
}
}
}
}
}
return Circle(center, r);
}
bool counter_wise(Point *p, int n) {
for (int i = 1; i < n - 1; i++)
if (dcmp(cross(p[i] - p[i - 1], p[i + 1] - p[i - 1])) > 0)
return 0;
else if (dcmp(cross(p[i] - p[i - 1], p[i + 1] - p[i - 1])) < 0) {
reverse(p, p + n);
return 1;
}
return 1;
}
bool clock_wise(Point *p, int n) {
for (int i = 1; i < n - 1; i++)
if (dcmp(cross(p[i] - p[i - 1], p[i + 1] - p[i - 1])) < 0)
return 0;
else if (dcmp(cross(p[i] - p[i - 1], p[i + 1] - p[i - 1])) > 0) {
reverse(p, p + n);
return 1;
}
return 1;
}
bool is_convex(Point *p, int n) {
for (int i = 0; i < n; i++) {
int j = (i + 1) % n;
int k = (j + 1) % n;
if (dcmp(cross(p[j] - p[i], p[k] - p[j])) == -1) return 0;
}
return 1;
}
double polygon_area(Point *p, int n) {
double area = 0;
for (int i = 1; i < n - 1; i++) area += cross(p[i] - p[0], p[i + 1] - p[0]);
return area / 2;
}
double convex_diameter(Point *p, int n) {
if (n == 1) return 0;
if (n == 2) return dis(p[0], p[1]);
if (n == 3)
return max(dis(p[0], p[1]), max(dis(p[1], p[2]), dis(p[2], p[0])));
counter_wise(p, n);
double ans = 0;
p[n] = p[0];
for (int i = 0, j = 2; i < n; i++) {
while (dcmp(cross(p[i + 1] - p[i], p[j] - p[i]) -
cross(p[i + 1] - p[i], p[j + 1] - p[i])) < 0)
j = (j + 1) % n;
ans = max(ans, max(dis(p[i], p[j]), dis(p[i + 1], p[j])));
}
return ans;
}
double convex_min_distance(Point *a, int n, Point *b, int m) {
int u = 0, v = 0, tmp;
double ans = INF;
counter_wise(a, n);
a[n] = a[0];
counter_wise(b, m);
b[m] = b[0];
for (int i = 0; i < n; i++)
if (dcmp(a[u].y - a[i].y) > 0) u = i;
for (int i = 0; i < m; i++)
if (dcmp(b[v].y - b[i].y) < 0) v = i;
for (int i = 0; i < n; i++) {
while ((tmp = dcmp(cross(a[u + 1] - a[u], b[v] - a[u]) -
cross(a[u + 1] - a[u], b[v + 1] - a[u]))) < 0)
v = (v + 1) % m;
if (tmp == 1)
ans = min(ans, point_to_seg(b[v], Line(a[u], a[u + 1])));
else
ans = min(ans, min(min(point_to_seg(a[u], Line(b[v], b[v + 1])),
point_to_seg(a[u + 1], Line(b[v], b[v + 1]))),
min(point_to_seg(b[v], Line(a[u], a[u + 1])),
point_to_seg(b[v + 1], Line(a[u], a[u + 1])))));
u = (u + 1) % n;
}
return ans;
}
double convex_rectangle_area(Point *p, int n) {
if (n <= 2) return 0;
counter_wise(p, n);
p[n] = p[0];
int l, r = 1, t = 1;
double ans = INF;
for (int i = 0; i < n; i++) {
while (dcmp(cross(p[i + 1] - p[i], p[t] - p[i]) -
cross(p[i + 1] - p[i], p[t + 1] - p[i])) < 0)
t = (t + 1) % n;
while (dcmp(dot(p[i + 1] - p[i], p[r] - p[i]) -
dot(p[i + 1] - p[i], p[r + 1] - p[i])) < 0)
r = (r + 1) % n;
if (!i) l = r;
while (dcmp(dot(p[i + 1] - p[i], p[l] - p[i]) -
dot(p[i + 1] - p[i], p[l + 1] - p[i])) >= 0)
l = (l + 1) % n;
double d = dis(p[i + 1], p[i]);
ans = min(ans, cross(p[i + 1] - p[i], p[t] - p[i]) *
(dot(p[i + 1] - p[i], p[r] - p[i]) -
dot(p[i + 1] - p[i], p[l] - p[i])) /
d / d);
}
return ans;
}
int relation(Point a, Point *b, int n) {
int w = 0;
b[n] = b[0];
for (int i = 0; i < n; i++) {
if (a == b[i] || a == b[i + 1]) return 3;
if (point_on_seg(a, Line(b[i + 1], b[i]))) return 2;
int k = dcmp(cross(b[i + 1] - b[i], a - b[i]));
int d1 = dcmp(b[i].y - a.y);
int d2 = dcmp(b[i + 1].y - a.y);
if (k > 0 && d1 <= 0 && d2 > 0) w++;
if (k < 0 && d2 <= 0 && d1 > 0) w--;
}
if (w != 0) return 1;
return 0;
}
void polygon_cut(Point *p, int &n, Line u) {
int top = 0;
Point tmp[MAXN];
p[n] = p[0];
for (int i = 0; i < n; i++) {
int d1 = dcmp(cross(u.e - u.s, p[i] - u.s));
int d2 = dcmp(cross(u.e - u.s, p[i + 1] - u.s));
if (d1 >= 0) tmp[top++] = p[i];
if (d1 * d2 < 0) tmp[top++] = line_intersection(u, Line(p[i], p[i + 1]));
}
for (int i = 0; i < top; i++) p[i] = tmp[i];
n = top;
}
void polygon_cut(Point *p, int &n, double a, double b, double c) {
int top = 0;
Point tmp[MAXN];
p[n] = p[0], p[n + 1] = p[1];
for (int i = 1; i <= n; i++) {
if (a * p[i].x + b * p[i].y + c > -eps)
tmp[top++] = p[i];
else {
if (a * p[i - 1].x + b * p[i - 1].y + c > eps)
tmp[top++] = line_intersection(Line(p[i - 1], p[i]), Line(a, b, c));
if (a * p[i + 1].x + b * p[i + 1].y + c > eps)
tmp[top++] = line_intersection(Line(p[i], p[i + 1]), Line(a, b, c));
}
}
for (int i = 0; i < top; i++) p[i] = tmp[i];
n = top;
}
double polygon_circumference(Point *p, int n) {
double ans = 0;
p[n] = p[0];
for (int i = 0; i < n; i++) ans += dis(p[i], p[i + 1]);
return ans;
}
double area_polygon_circle(Circle c, Point *p, int n) {
double ans = 0;
for (int i = 0; i < n; i++) {
int j = (i + 1) % n;
if (dcmp(cross(p[j] - c.p, p[i] - c.p)) >= 0)
ans += circle_traingle_area(p[i], p[j], c);
else
ans -= circle_traingle_area(p[i], p[j], c);
}
return fabs(ans);
}
Point centre_of_polygon(Point *p, int n) {
double sum = 0, sumx = 0, sumy = 0;
Point p1 = p[0], p2 = p[1], p3;
for (int i = 2; i < n; i++) {
p3 = p[i];
double area = cross(p2 - p1, p3 - p2) / 2;
sum += area;
sumx += (p1.x + p2.x + p3.x) * area;
sumy += (p1.y + p2.y + p3.y) * area;
p2 = p3;
}
return Point(sumx / (3 * sum), sumy / (3 * sum));
}
Point tmp[MAXN];
int convex_hull(Point *p, int n, Point *ch) {
int m = 0;
sort(p, p + n);
for (int i = 0; i < n; i++) {
while (m > 1 && cross(tmp[m - 1] - tmp[m - 2], p[i] - tmp[m - 1]) <= 0) m--;
tmp[m++] = p[i];
}
int k = m;
for (int i = n - 2; i >= 0; i--) {
while (m > k && cross(tmp[m - 1] - tmp[m - 2], p[i] - tmp[m - 1]) <= 0) m--;
tmp[m++] = p[i];
}
if (n > 1) m--;
for (int i = 0; i < m; i++) ch[i] = tmp[i];
return m;
}
bool polar_cmp(Line a, Line b) {
double x = a.polar(), y = b.polar();
if (dcmp(x - y) == 0) return dcmp(cross(a.e - a.s, b.e - a.s)) < 0;
return dcmp(x - y) < 0;
}
int half_plane_intersection(Line *line, int n, Point *ch) {
Line L[MAXN], que[MAXN];
int head = 0, tail = 1, cnt = 1, num = 0;
for (int i = 0; i < n; i++) L[i] = line[i];
sort(L, L + n, polar_cmp);
for (int i = 1; i < n; i++)
if (dcmp(L[i].polar() - L[cnt - 1].polar()) != 0) L[cnt++] = L[i];
que[0] = L[0], que[1] = L[1];
for (int i = 2; i < cnt; i++) {
while (tail > head && relation(que[tail] & que[tail - 1], L[i]) == 2)
tail--;
while (tail > head && relation(que[head] & que[head + 1], L[i]) == 2)
head++;
que[++tail] = L[i];
}
while (tail > head && relation(que[tail] & que[tail - 1], que[head]) == 2)
tail--;
while (tail > head && relation(que[head] & que[head + 1], que[tail]) == 2)
head++;
if (tail - head <= 1) return -1;
for (int i = head; i <= tail; i++) {
int j = (i == tail ? head : i + 1);
ch[num++] = que[i] & que[j];
}
return num;
}
double polygon_fermat_point(Point *p, int n, Point &res) {
double T = 1e10, ans = INF, dt = 0.999;
Point before(-1, -1);
while (T > eps) {
double k = 2.0 * pi * (rand() * 1.0 / RAND_MAX), sum1 = 0, sum2 = 0;
Point cur(before.x + T * cos(k), before.y + T * sin(k));
for (int i = 0; i < n; i++) sum1 += dis(p[i], before);
for (int i = 0; i < n; i++) sum2 += dis(p[i], cur);
if (dcmp(ans - sum1) >= 0) ans = sum1, res = before;
if (dcmp(ans - sum2) >= 0) ans = sum2, res = cur;
if (dcmp(sum1 - sum2) >= 0 ||
exp((sum1 - sum2) / T) >= (rand() * 1.0 / RAND_MAX))
before = cur;
T *= dt;
}
return ans;
}
double sphere_distance(double r, double x1, double y1, double x2, double y2) {
return r * acos(sin(y1) * sin(y2) + cos(y1) * cos(y2) * cos(x1 - x2));
}
double f(double x) { return x * x + x; }
double cal(double l, double r) {
return (r - l) * (f(l) + f(r) + 4 * f((l + r) / 2)) / 6;
}
double simpson(double l, double r, double ans) {
double mid = (l + r) / 2, ans1 = cal(l, mid), ans2 = cal(mid, r);
if (fabs(ans1 + ans2 - ans) <= eps) return ans;
return simpson(l, mid, ans1) + simpson(mid, r, ans2);
}
int __next[MAXN];
Point pa[MAXN], push_back[MAXN];
struct node {
long long len;
double delta;
node() {}
node(long long a, double b) { len = a, delta = b; }
bool operator==(const node &n) const {
if (len == n.len && (dcmp(delta - n.delta) == 0 ||
dcmp(fabs(delta - n.delta) - 2 * pi) == 0))
return 1;
return 0;
}
bool operator!=(const node &n) const {
if (len == n.len && (dcmp(delta - n.delta) == 0 ||
dcmp(fabs(delta - n.delta) - 2 * pi) == 0))
return 0;
return 1;
}
} infa[MAXN], infb[MAXN];
void get_next(int n) {
memset(__next, 0, sizeof(__next));
__next[0] = -1;
for (int i = 1, k = -1; i < n; i++) {
while (k != -1 && infa[k + 1] != infa[i]) k = __next[k];
if (infa[k + 1] == infa[i]) k++;
__next[i] = k;
}
}
bool kmp(int n) {
for (int i = 0, k = -1; i < 2 * n; i++) {
while (k != -1 && infa[k + 1] != infb[i % n]) k = __next[k];
if (infa[k + 1] == infb[i % n]) k++;
if (k == n - 1) return 1;
}
return 0;
}
int main(void) {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++) pa[i].input();
for (int i = 0; i < m; i++) push_back[i].input();
n = convex_hull(pa, n, pa);
m = convex_hull(push_back, m, push_back);
if (n != m) return puts("NO"), 0;
for (int i = 0; i < n; i++)
infa[i] = node((pa[(i + 1) % n] - pa[i]).len2(),
(pa[(i + 1) % n] - pa[i]).polar() -
(pa[i] - pa[(i - 1 + n) % n]).polar()),
infb[i] = node((push_back[(i + 1) % n] - push_back[i]).len2(),
(push_back[(i + 1) % n] - push_back[i]).polar() -
(push_back[i] - push_back[(i - 1 + n) % n]).polar());
get_next(n);
if (kmp(n))
puts("YES");
else
puts("NO");
return 0;
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
const double eps = 1e-8;
const double pi = acos(-1.0);
const long long INF = 0x3f3f3f3f3f3f3f3f;
int dcmp(double x) {
if (fabs(x) < eps) return 0;
return (x > 0 ? 1 : -1);
}
inline double sqr(double x) { return x * x; }
struct Point {
long long x, y;
Point() { x = 0, y = 0; }
Point(long long _x, long long _y) : x(_x), y(_y) {}
void input() { scanf("%lld%lld", &x, &y); }
void output() { printf("%lld %lld\n", x, y); }
friend istream &operator>>(istream &os, Point &b) {
os >> b.x >> b.y;
return os;
}
friend ostream &operator<<(ostream &os, Point &b) {
os << b.x << ' ' << b.y;
return os;
}
bool operator==(const Point &b) const {
return (dcmp(x - b.x) == 0 && dcmp(y - b.y) == 0);
}
bool operator!=(const Point &b) const {
return !((dcmp(x - b.x) == 0 && dcmp(y - b.y) == 0));
}
bool operator<(const Point &b) const {
return (dcmp(x - b.x) == 0 ? dcmp(y - b.y) < 0 : x < b.x);
}
long long operator^(const Point &b) const { return x * b.y - y * b.x; }
long long operator*(const Point &b) const { return x * b.x + y * b.y; }
Point operator+(const Point &b) const { return Point(x + b.x, y + b.y); }
Point operator-(const Point &b) const { return Point(x - b.x, y - b.y); }
Point operator*(double a) { return Point(x * a, y * a); }
Point operator/(double a) { return Point(x / a, y / a); }
long long len2() { return sqr(x) + sqr(y); }
long long len() { return sqrt(len2()); }
double polar() { return atan2(y, x); }
Point change_len(double r) {
double l = len();
if (dcmp(l) == 0) return *this;
return Point(x * r / l, y * r / l);
}
Point rotate_left() { return Point(-y, x); }
Point rotate_right() { return Point(y, -x); }
Point rotate(Point p, double ang) {
Point v = (*this) - p;
double c = cos(ang), s = sin(ang);
return Point(p.x + v.x * c - v.y * s, p.y + v.x * s + v.y * c);
}
Point normal() { return Point(-y / len(), x / len()); }
};
double cross(Point a, Point b) { return a.x * b.y - a.y * b.x; }
double dot(Point a, Point b) { return a.x * b.x + a.y * b.y; }
double dis(Point a, Point b) {
Point p = b - a;
return p.len();
}
double dis2(Point a, Point b) {
Point p = b - a;
return p.len2();
}
double rad_degree(double rad) { return rad / pi * 180; }
double degree_rad(double deg) { return deg / 180 * pi; }
double rad(Point a, Point b) {
return fabs(atan2(fabs(cross(a, b)), dot(a, b)));
}
bool is_parallel(Point a, Point b) {
double p = rad(a, b);
return dcmp(p) == 0 || dcmp(p - pi) == 0;
}
struct Line {
Point s, e;
Line() {}
Line(Point _s, Point _e) : s(_s), e(_e) {}
Line(Point p, double ang) {
s = p;
if (dcmp(ang - pi / 2) == 0) {
e = s + Point(0, 1);
} else {
e = s + Point(1, tan(ang));
}
}
Line(double a, double b, double c) {
if (dcmp(a) == 0) {
s = Point(0, -c / b);
e = Point(1, -c / b);
} else if (dcmp(b) == 0) {
s = Point(-c / a, 0);
e = Point(-c / a, 1);
} else {
s = Point(0, -c / b);
e = Point(1, (-c - a) / b);
}
}
void input() {
s.input();
e.input();
}
void adjust() {
if (e < s) swap(e, s);
}
double length() { return dis(s, e); }
double polar() { return atan2(e.y - s.y, e.x - s.x); }
double angle() {
double k = atan2(e.y - s.y, e.x - s.x);
if (dcmp(k) < 0) k += pi;
if (dcmp(k - pi) == 0) k -= pi;
return k;
}
Point operator&(const Line &b) const {
Point res = s;
double t = ((s - b.s) ^ (b.s - b.e)) / ((s - e) ^ (b.s - b.e));
res.x += (e.x - s.x) * t;
res.y += (e.y - s.y) * t;
return res;
}
};
int relation(Point p, Line l) {
int c = dcmp(cross(p - l.s, l.e - l.s));
if (c < 0)
return 1;
else if (c > 0)
return 2;
else
return 3;
}
bool point_on_seg(Point p, Line l) {
return dcmp(cross(p - l.s, l.e - l.s)) == 0 &&
dcmp(dot(p - l.s, p - l.e) <= 0);
}
bool is_parallel(Line a, Line b) { return is_parallel(a.e - a.s, b.e - b.s); }
int seg_cross_seg(Line a, Line v) {
int d1 = dcmp(cross(a.e - a.s, v.s - a.s));
int d2 = dcmp(cross(a.e - a.s, v.e - a.s));
int d3 = dcmp(cross(v.e - v.s, a.s - v.s));
int d4 = dcmp(cross(v.e - v.s, a.e - v.s));
if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2;
return (d1 == 0 && dcmp(dot(v.s - a.s, v.s - a.e)) <= 0) ||
(d2 == 0 && dcmp(dot(v.e - a.s, v.e - a.e)) <= 0) ||
(d3 == 0 && dcmp(dot(a.s - v.s, a.s - v.e)) <= 0) ||
(d4 == 0 && dcmp(dot(a.e - v.s, a.e - v.e)) <= 0);
}
int line_cross_seg(Line a, Line v) {
int d1 = dcmp(cross(a.e - a.s, v.s - a.s));
int d2 = dcmp(cross(a.e - a.s, v.e - a.s));
if ((d1 ^ d2) == -2) return 2;
return (d1 == 0 || d2 == 0);
}
int line_cross_line(Line a, Line v) {
if (is_parallel(a, v)) return relation(a.e, v) == 3;
return 2;
}
Point line_intersection(Line a, Line v) {
double a1 = cross(v.e - v.s, a.s - v.s);
double a2 = cross(v.e - v.s, a.e - v.s);
return Point((a.s.x * a2 - a.e.x * a1) / (a2 - a1),
(a.s.y * a2 - a.e.y * a1) / (a2 - a1));
}
Point line_intersection(Line line, double a, double b, double c) {
double u = fabs(a * line.s.x + b * line.s.y + c);
double v = fabs(a * line.e.x + b * line.e.y + c);
return Point((line.s.x * v + line.e.x * u) / (u + v),
(line.s.y * v + line.e.y * u) / (u + v));
}
double point_to_line(Point p, Line a) {
return fabs(cross(p - a.s, a.e - a.s) / a.length());
}
double point_to_seg(Point p, Line a) {
if (dcmp(dot(p - a.s, a.e - a.s)) < 0 || dcmp(dot(p - a.e, a.s - a.e)) < 0)
return min(dis(p, a.e), dis(p, a.s));
return point_to_line(p, a);
}
Point projection(Point p, Line a) {
return a.s + (((a.e - a.s) * dot(a.e - a.s, p - a.s)) / (a.e - a.s).len2());
}
Point symmetry(Point p, Line a) {
Point q = projection(p, a);
return Point(2 * q.x - p.x, 2 * q.y - p.y);
}
Line vector_move_inward(Line line, double d) {
double len = line.length();
Point p((line.s.y - line.e.y) * d / len, (line.e.x - line.s.x) * d / len);
return Line(line.s + p, line.e + p);
}
struct Circle {
Point p;
double r;
Circle() {}
Circle(Point _p, double _r) : p(_p), r(_r) {}
Circle(double a, double b, double _r) {
p = Point(a, b);
r = _r;
}
void input() {
p.input();
scanf("%lf", &r);
}
void output() {
p.output();
printf("%.2f\n", r);
}
bool operator==(const Circle &a) const {
return p == a.p && (dcmp(r - a.r) == 0);
}
double area() { return pi * r * r; }
double circumference() { return 2 * pi * r; }
};
int relation(Point p, Circle a) {
double d = dis(p, a.p);
if (dcmp(d - a.r) == 0) return 1;
return (dcmp(d - a.r) < 0 ? 2 : 0);
}
int relation(Line a, Circle b) {
double p = point_to_line(b.p, a);
if (dcmp(p - b.r) == 0) return 1;
return (dcmp(p - b.r) < 0 ? 2 : 0);
}
int relation(Circle a, Circle v) {
double d = dis(a.p, v.p);
if (dcmp(d - a.r - v.r) > 0) return 5;
if (dcmp(d - a.r - v.r) == 0) return 4;
double l = fabs(a.r - v.r);
if (dcmp(d - l) > 0) return 3;
if (dcmp(d - l) == 0) return 2;
return 1;
}
Circle triangle_out_circle(Point a, Point b, Point c) {
Line u = Line((a + b) / 2, ((a + b) / 2) + (b - a).rotate_left());
Line v = Line((b + c) / 2, ((b + c) / 2) + (c - b).rotate_left());
Point p = line_intersection(u, v);
double r = dis(p, a);
return Circle(p, r);
}
Circle triangle_in_circle(Point a, Point b, Point c) {
Line u, v;
double m = atan2(b.y - a.y, b.x - a.x), n = atan2(c.y - a.y, c.x - a.x);
u.s = a;
u.e = u.s + Point(cos((n + m) / 2), sin((n + m) / 2));
v.s = b;
m = atan2(a.y - b.y, a.x - b.x), n = atan2(c.y - b.y, c.x - b.x);
v.e = v.s + Point(cos((n + m) / 2), sin((n + m) / 2));
Point p = line_intersection(u, v);
double r = point_to_seg(p, Line(a, b));
return Circle(p, r);
}
int circle_intersection(Circle a, Circle v, Point &p1, Point &p2) {
int rel = relation(a, v);
if (rel == 1 || rel == 5) return 0;
double d = dis(a.p, v.p);
double l = (d * d + a.r * a.r - v.r * v.r) / (2 * d);
double h = sqrt(a.r * a.r - l * l);
Point tmp = a.p + (v.p - a.p).change_len(l);
p1 = tmp + ((v.p - a.p).rotate_left().change_len(h));
p2 = tmp + ((v.p - a.p).rotate_right().change_len(h));
if (rel == 2 || rel == 4) return 1;
return 2;
}
int line_circle_intersection(Line v, Circle u, Point &p1, Point &p2) {
if (!relation(v, u)) return 0;
Point a = projection(u.p, v);
double d = point_to_line(u.p, v);
d = sqrt(u.r * u.r - d * d);
if (dcmp(d) == 0) {
p1 = a, p2 = a;
return 1;
}
p1 = a + (v.e - v.s).change_len(d);
p2 = a - (v.e - v.s).change_len(d);
return 2;
}
int get_circle(Point a, Point b, double r1, Circle &c1, Circle &c2) {
Circle x(a, r1), y(b, r1);
int t = circle_intersection(x, y, c1.p, c2.p);
if (!t) return 0;
c1.r = c2.r = r1;
return t;
}
int get_circle(Line u, Point q, double r1, Circle &c1, Circle &c2) {
double d = point_to_line(q, u);
if (dcmp(d - r1 * 2) > 0) return 0;
if (dcmp(d) == 0) {
c1.p = q + ((u.e - u.s).rotate_left().change_len(r1));
c2.p = q + ((u.e - u.s).rotate_right().change_len(r1));
c1.r = c2.r = r1;
return 2;
}
Line u1 = Line(u.s + (u.e - u.s).rotate_left().change_len(r1),
u.e + (u.e - u.s).rotate_left().change_len(r1));
Line u2 = Line(u.s + (u.e - u.s).rotate_right().change_len(r1),
u.e + (u.e - u.s).rotate_right().change_len(r1));
Circle cc = Circle(q, r1);
Point p1, p2;
if (!line_circle_intersection(u1, cc, p1, p2))
line_circle_intersection(u2, cc, p1, p2);
c1 = Circle(p1, r1);
if (p1 == p2) {
c2 = c1;
return 1;
}
c2 = Circle(p2, r1);
return 2;
}
int get_circle(Line u, Line v, double r1, Circle &c1, Circle &c2, Circle &c3,
Circle &c4) {
if (is_parallel(u, v)) return 0;
Line u1 = Line(u.s + (u.e - u.s).rotate_left().change_len(r1),
u.e + (u.e - u.s).rotate_left().change_len(r1));
Line u2 = Line(u.s + (u.e - u.s).rotate_right().change_len(r1),
u.e + (u.e - u.s).rotate_right().change_len(r1));
Line v1 = Line(v.s + (v.e - v.s).rotate_left().change_len(r1),
v.e + (v.e - v.s).rotate_left().change_len(r1));
Line v2 = Line(v.s + (v.e - v.s).rotate_right().change_len(r1),
v.e + (v.e - v.s).rotate_right().change_len(r1));
c1.r = c2.r = c3.r = c4.r = r1;
c1.p = line_intersection(u1, v1);
c2.p = line_intersection(u1, v2);
c3.p = line_intersection(u2, v1);
c4.p = line_intersection(u2, v2);
return 4;
}
int get_circle(Circle cx, Circle cy, double r1, Circle &c1, Circle &c2) {
Circle x(cx.p, r1 + cx.r), y(cy.p, r1 + cy.r);
int t = circle_intersection(x, y, c1.p, c2.p);
if (!t) return 0;
c1.r = c2.r = r1;
return t;
}
Circle get_circle(Point a, Point b, double k) {
double k2 = k * k;
double d = (2 * b.x - 2 * a.x / k2) / ((1 / k2) - 1);
double e = (2 * b.y - 2 * a.y / k2) / ((1 / k2) - 1);
double f = (a.x * a.x / k2 - b.x * b.x + a.y * a.y / k2 - b.y * b.y) /
((1 / k2) - 1);
double cx = -d / 2, cy = -e / 2;
double r = sqrt((d * d + e * e) / 4 - f);
return Circle(Point(cx, cy), r);
}
bool get_circle(Point a, Point b, double r, Point &res1, Point &res2) {
if (dcmp(dis(a, b) - 2 * r) > 0) return 0;
Point mid((a.x + b.x) / 2, (a.y + b.y) / 2);
double angle = atan2(b.y - a.y, b.x - a.x),
d = sqrt(r * r - sqr(dis(a, b) / 2));
res1 = Point(mid.x + d * sin(angle), mid.y - d * cos(angle));
res2 = Point(mid.x - d * sin(angle), mid.y + d * cos(angle));
return 1;
}
Circle get_circle(Point p1, Point p2, Point p3) {
if (dcmp(cross(p2 - p1, p3 - p1)) == 0) {
if (p2 < p1) swap(p1, p2);
if (p3 < p1) swap(p1, p3);
if (p3 < p2) swap(p2, p3);
return Circle((p1 + p3) / 2, dis(p1, p3) / 2);
}
double x1 = p1.x, x2 = p2.x, x3 = p3.x;
double y1 = p1.y, y2 = p2.y, y3 = p3.y;
double a = x1 - x2, b = y1 - y2, c = x1 - x3, d = y1 - y3;
double e = ((x1 * x1 - x2 * x2) + (y1 * y1 - y2 * y2)) / 2;
double f = ((x1 * x1 - x3 * x3) + (y1 * y1 - y3 * y3)) / 2;
double delta = b * c - a * d;
Point pc = Point(-(d * e - b * f) / delta, -(a * f - c * e) / delta);
return Circle(pc, dis(pc, p1));
}
int tan_line(Point q, Circle a, Line &u, Line &v) {
int x = relation(q, a);
if (x == 2) return 0;
if (x == 1) {
u = Line(q, q + (q - a.p).rotate_left());
v = u;
return 1;
}
double d = dis(a.p, q);
double l = a.r * a.r / d;
double h = sqrt(a.r * a.r - l * l);
u = Line(
q, a.p + (q - a.p).change_len(l) + (q - a.p).rotate_left().change_len(h));
v = Line(q, a.p + (q - a.p).change_len(l) +
(q - a.p).rotate_right().change_len(h));
return 2;
}
double area_circle(Circle a, Circle v) {
int rel = relation(a, v);
if (rel >= 4) return 0;
if (rel <= 2) return min(a.area(), v.area());
double d = dis(a.p, v.p);
double hf = (a.r + v.r + d) / 2;
double ss = 2 * sqrt(hf * (hf - a.r) * (hf - v.r) * (hf - d));
double a1 = acos((a.r * a.r + d * d - v.r * v.r) / (2 * a.r * d));
a1 = a1 * a.r * a.r;
double a2 = acos((v.r * v.r + d * d - a.r * a.r) / (2 * v.r * d));
a2 = a2 * v.r * v.r;
return a1 + a2 - ss;
}
double circle_traingle_area(Point a, Point b, Circle c) {
Point p = c.p;
double r = c.r;
if (dcmp(cross(p - a, p - b)) == 0) return 0;
Point q[5];
int len = 0;
q[len++] = a;
Line l(a, b);
Point p1, p2;
if (line_circle_intersection(l, c, q[1], q[2]) == 2) {
if (dcmp(dot(a - q[1], b - q[1])) < 0) q[len++] = q[1];
if (dcmp(dot(a - q[2], b - q[2])) < 0) q[len++] = q[2];
}
q[len++] = b;
if (len == 4 && dcmp(dot(q[0] - q[1], q[2] - q[1])) > 0) swap(q[1], q[2]);
double res = 0;
for (int i = 0; i < len - 1; i++) {
if (relation(q[i], c) == 0 || relation(q[i + 1], c) == 0) {
double arg = rad(q[i] - p, q[i + 1] - p);
res += r * r * arg / 2;
} else
res += fabs(cross(q[i] - p, q[i + 1] - p)) / 2;
}
return res;
}
int circle_cover_point(Point *p, int n, double r) {
int ans = 1;
pair<double, int> angle[MAXN];
for (int i = 0; i < n; i++) {
int m = 0;
double d;
for (int j = 0; j < n; j++)
if (dcmp((d = dis(p[i], p[j])) - 2 * r) <= 0 && i != j) {
double delta = acos(d / (2 * r)), polar = (p[j] - p[i]).polar();
angle[m++] = make_pair(polar - delta, 1);
angle[m++] = make_pair(polar + delta, -1);
}
sort(angle, angle + m);
for (int j = 0, sum = 1; j < m; j++) {
sum += angle[j].second;
ans = max(ans, sum);
}
}
return ans;
}
Circle circle_min_conver(Point *p, int n) {
Point center = p[0];
double r = 0;
random_shuffle(p, p + n);
for (int i = 1; i < n; i++) {
if (dcmp(dis(center, p[i]) - r) > 0) {
center = p[i], r = 0;
for (int j = 0; j < i; j++) {
if (dcmp(dis(center, p[j]) - r) > 0) {
center = (p[i] + p[j]) / 2, r = dis(center, p[j]);
for (int k = 0; k < j; k++) {
if (dcmp(dis(center, p[k]) - r) > 0) {
center = get_circle(p[i], p[j], p[k]).p;
r = dis(center, p[i]);
}
}
}
}
}
}
return Circle(center, r);
}
bool counter_wise(Point *p, int n) {
for (int i = 1; i < n - 1; i++)
if (dcmp(cross(p[i] - p[i - 1], p[i + 1] - p[i - 1])) > 0)
return 0;
else if (dcmp(cross(p[i] - p[i - 1], p[i + 1] - p[i - 1])) < 0) {
reverse(p, p + n);
return 1;
}
return 1;
}
bool clock_wise(Point *p, int n) {
for (int i = 1; i < n - 1; i++)
if (dcmp(cross(p[i] - p[i - 1], p[i + 1] - p[i - 1])) < 0)
return 0;
else if (dcmp(cross(p[i] - p[i - 1], p[i + 1] - p[i - 1])) > 0) {
reverse(p, p + n);
return 1;
}
return 1;
}
bool is_convex(Point *p, int n) {
for (int i = 0; i < n; i++) {
int j = (i + 1) % n;
int k = (j + 1) % n;
if (dcmp(cross(p[j] - p[i], p[k] - p[j])) == -1) return 0;
}
return 1;
}
double polygon_area(Point *p, int n) {
double area = 0;
for (int i = 1; i < n - 1; i++) area += cross(p[i] - p[0], p[i + 1] - p[0]);
return area / 2;
}
double convex_diameter(Point *p, int n) {
if (n == 1) return 0;
if (n == 2) return dis(p[0], p[1]);
if (n == 3)
return max(dis(p[0], p[1]), max(dis(p[1], p[2]), dis(p[2], p[0])));
counter_wise(p, n);
double ans = 0;
p[n] = p[0];
for (int i = 0, j = 2; i < n; i++) {
while (dcmp(cross(p[i + 1] - p[i], p[j] - p[i]) -
cross(p[i + 1] - p[i], p[j + 1] - p[i])) < 0)
j = (j + 1) % n;
ans = max(ans, max(dis(p[i], p[j]), dis(p[i + 1], p[j])));
}
return ans;
}
double convex_min_distance(Point *a, int n, Point *b, int m) {
int u = 0, v = 0, tmp;
double ans = INF;
counter_wise(a, n);
a[n] = a[0];
counter_wise(b, m);
b[m] = b[0];
for (int i = 0; i < n; i++)
if (dcmp(a[u].y - a[i].y) > 0) u = i;
for (int i = 0; i < m; i++)
if (dcmp(b[v].y - b[i].y) < 0) v = i;
for (int i = 0; i < n; i++) {
while ((tmp = dcmp(cross(a[u + 1] - a[u], b[v] - a[u]) -
cross(a[u + 1] - a[u], b[v + 1] - a[u]))) < 0)
v = (v + 1) % m;
if (tmp == 1)
ans = min(ans, point_to_seg(b[v], Line(a[u], a[u + 1])));
else
ans = min(ans, min(min(point_to_seg(a[u], Line(b[v], b[v + 1])),
point_to_seg(a[u + 1], Line(b[v], b[v + 1]))),
min(point_to_seg(b[v], Line(a[u], a[u + 1])),
point_to_seg(b[v + 1], Line(a[u], a[u + 1])))));
u = (u + 1) % n;
}
return ans;
}
double convex_rectangle_area(Point *p, int n) {
if (n <= 2) return 0;
counter_wise(p, n);
p[n] = p[0];
int l, r = 1, t = 1;
double ans = INF;
for (int i = 0; i < n; i++) {
while (dcmp(cross(p[i + 1] - p[i], p[t] - p[i]) -
cross(p[i + 1] - p[i], p[t + 1] - p[i])) < 0)
t = (t + 1) % n;
while (dcmp(dot(p[i + 1] - p[i], p[r] - p[i]) -
dot(p[i + 1] - p[i], p[r + 1] - p[i])) < 0)
r = (r + 1) % n;
if (!i) l = r;
while (dcmp(dot(p[i + 1] - p[i], p[l] - p[i]) -
dot(p[i + 1] - p[i], p[l + 1] - p[i])) >= 0)
l = (l + 1) % n;
double d = dis(p[i + 1], p[i]);
ans = min(ans, cross(p[i + 1] - p[i], p[t] - p[i]) *
(dot(p[i + 1] - p[i], p[r] - p[i]) -
dot(p[i + 1] - p[i], p[l] - p[i])) /
d / d);
}
return ans;
}
int relation(Point a, Point *b, int n) {
int w = 0;
b[n] = b[0];
for (int i = 0; i < n; i++) {
if (a == b[i] || a == b[i + 1]) return 3;
if (point_on_seg(a, Line(b[i + 1], b[i]))) return 2;
int k = dcmp(cross(b[i + 1] - b[i], a - b[i]));
int d1 = dcmp(b[i].y - a.y);
int d2 = dcmp(b[i + 1].y - a.y);
if (k > 0 && d1 <= 0 && d2 > 0) w++;
if (k < 0 && d2 <= 0 && d1 > 0) w--;
}
if (w != 0) return 1;
return 0;
}
void polygon_cut(Point *p, int &n, Line u) {
int top = 0;
Point tmp[MAXN];
p[n] = p[0];
for (int i = 0; i < n; i++) {
int d1 = dcmp(cross(u.e - u.s, p[i] - u.s));
int d2 = dcmp(cross(u.e - u.s, p[i + 1] - u.s));
if (d1 >= 0) tmp[top++] = p[i];
if (d1 * d2 < 0) tmp[top++] = line_intersection(u, Line(p[i], p[i + 1]));
}
for (int i = 0; i < top; i++) p[i] = tmp[i];
n = top;
}
void polygon_cut(Point *p, int &n, double a, double b, double c) {
int top = 0;
Point tmp[MAXN];
p[n] = p[0], p[n + 1] = p[1];
for (int i = 1; i <= n; i++) {
if (a * p[i].x + b * p[i].y + c > -eps)
tmp[top++] = p[i];
else {
if (a * p[i - 1].x + b * p[i - 1].y + c > eps)
tmp[top++] = line_intersection(Line(p[i - 1], p[i]), Line(a, b, c));
if (a * p[i + 1].x + b * p[i + 1].y + c > eps)
tmp[top++] = line_intersection(Line(p[i], p[i + 1]), Line(a, b, c));
}
}
for (int i = 0; i < top; i++) p[i] = tmp[i];
n = top;
}
double polygon_circumference(Point *p, int n) {
double ans = 0;
p[n] = p[0];
for (int i = 0; i < n; i++) ans += dis(p[i], p[i + 1]);
return ans;
}
double area_polygon_circle(Circle c, Point *p, int n) {
double ans = 0;
for (int i = 0; i < n; i++) {
int j = (i + 1) % n;
if (dcmp(cross(p[j] - c.p, p[i] - c.p)) >= 0)
ans += circle_traingle_area(p[i], p[j], c);
else
ans -= circle_traingle_area(p[i], p[j], c);
}
return fabs(ans);
}
Point centre_of_polygon(Point *p, int n) {
double sum = 0, sumx = 0, sumy = 0;
Point p1 = p[0], p2 = p[1], p3;
for (int i = 2; i < n; i++) {
p3 = p[i];
double area = cross(p2 - p1, p3 - p2) / 2;
sum += area;
sumx += (p1.x + p2.x + p3.x) * area;
sumy += (p1.y + p2.y + p3.y) * area;
p2 = p3;
}
return Point(sumx / (3 * sum), sumy / (3 * sum));
}
Point tmp[MAXN];
int convex_hull(Point *p, int n, Point *ch) {
int m = 0;
sort(p, p + n);
for (int i = 0; i < n; i++) {
while (m > 1 && cross(tmp[m - 1] - tmp[m - 2], p[i] - tmp[m - 1]) <= 0) m--;
tmp[m++] = p[i];
}
int k = m;
for (int i = n - 2; i >= 0; i--) {
while (m > k && cross(tmp[m - 1] - tmp[m - 2], p[i] - tmp[m - 1]) <= 0) m--;
tmp[m++] = p[i];
}
if (n > 1) m--;
for (int i = 0; i < m; i++) ch[i] = tmp[i];
return m;
}
bool polar_cmp(Line a, Line b) {
double x = a.polar(), y = b.polar();
if (dcmp(x - y) == 0) return dcmp(cross(a.e - a.s, b.e - a.s)) < 0;
return dcmp(x - y) < 0;
}
int half_plane_intersection(Line *line, int n, Point *ch) {
Line L[MAXN], que[MAXN];
int head = 0, tail = 1, cnt = 1, num = 0;
for (int i = 0; i < n; i++) L[i] = line[i];
sort(L, L + n, polar_cmp);
for (int i = 1; i < n; i++)
if (dcmp(L[i].polar() - L[cnt - 1].polar()) != 0) L[cnt++] = L[i];
que[0] = L[0], que[1] = L[1];
for (int i = 2; i < cnt; i++) {
while (tail > head && relation(que[tail] & que[tail - 1], L[i]) == 2)
tail--;
while (tail > head && relation(que[head] & que[head + 1], L[i]) == 2)
head++;
que[++tail] = L[i];
}
while (tail > head && relation(que[tail] & que[tail - 1], que[head]) == 2)
tail--;
while (tail > head && relation(que[head] & que[head + 1], que[tail]) == 2)
head++;
if (tail - head <= 1) return -1;
for (int i = head; i <= tail; i++) {
int j = (i == tail ? head : i + 1);
ch[num++] = que[i] & que[j];
}
return num;
}
double polygon_fermat_point(Point *p, int n, Point &res) {
double T = 1e10, ans = INF, dt = 0.999;
Point before(-1, -1);
while (T > eps) {
double k = 2.0 * pi * (rand() * 1.0 / RAND_MAX), sum1 = 0, sum2 = 0;
Point cur(before.x + T * cos(k), before.y + T * sin(k));
for (int i = 0; i < n; i++) sum1 += dis(p[i], before);
for (int i = 0; i < n; i++) sum2 += dis(p[i], cur);
if (dcmp(ans - sum1) >= 0) ans = sum1, res = before;
if (dcmp(ans - sum2) >= 0) ans = sum2, res = cur;
if (dcmp(sum1 - sum2) >= 0 ||
exp((sum1 - sum2) / T) >= (rand() * 1.0 / RAND_MAX))
before = cur;
T *= dt;
}
return ans;
}
double sphere_distance(double r, double x1, double y1, double x2, double y2) {
return r * acos(sin(y1) * sin(y2) + cos(y1) * cos(y2) * cos(x1 - x2));
}
double f(double x) { return x * x + x; }
double cal(double l, double r) {
return (r - l) * (f(l) + f(r) + 4 * f((l + r) / 2)) / 6;
}
double simpson(double l, double r, double ans) {
double mid = (l + r) / 2, ans1 = cal(l, mid), ans2 = cal(mid, r);
if (fabs(ans1 + ans2 - ans) <= eps) return ans;
return simpson(l, mid, ans1) + simpson(mid, r, ans2);
}
int __next[MAXN];
Point pa[MAXN], push_back[MAXN];
struct node {
long long len;
double delta;
node() {}
node(long long a, double b) { len = a, delta = b; }
bool operator==(const node &n) const {
if (len == n.len && (dcmp(delta - n.delta) == 0 ||
dcmp(fabs(delta - n.delta) - 2 * pi) == 0))
return 1;
return 0;
}
bool operator!=(const node &n) const {
if (len == n.len && (dcmp(delta - n.delta) == 0 ||
dcmp(fabs(delta - n.delta) - 2 * pi) == 0))
return 0;
return 1;
}
} infa[MAXN], infb[MAXN];
void get_next(int n) {
memset(__next, 0, sizeof(__next));
__next[0] = -1;
for (int i = 1, k = -1; i < n; i++) {
while (k != -1 && infa[k + 1] != infa[i]) k = __next[k];
if (infa[k + 1] == infa[i]) k++;
__next[i] = k;
}
}
bool kmp(int n) {
for (int i = 0, k = -1; i < 2 * n; i++) {
while (k != -1 && infa[k + 1] != infb[i % n]) k = __next[k];
if (infa[k + 1] == infb[i % n]) k++;
if (k == n - 1) return 1;
}
return 0;
}
int main(void) {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++) pa[i].input();
for (int i = 0; i < m; i++) push_back[i].input();
n = convex_hull(pa, n, pa);
m = convex_hull(push_back, m, push_back);
if (n != m) return puts("NO"), 0;
for (int i = 0; i < n; i++)
infa[i] = node((pa[(i + 1) % n] - pa[i]).len2(),
(pa[(i + 1) % n] - pa[i]).polar() -
(pa[i] - pa[(i - 1 + n) % n]).polar()),
infb[i] = node((push_back[(i + 1) % n] - push_back[i]).len2(),
(push_back[(i + 1) % n] - push_back[i]).polar() -
(push_back[i] - push_back[(i - 1 + n) % n]).polar());
get_next(n);
if (kmp(n))
puts("YES");
else
puts("NO");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
inline int read() {
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
namespace Mymath {
long long qp(long long x, long long p, long long mod) {
long long ans = 1;
while (p) {
if (p & 1) ans = ans * x % mod;
x = x * x % mod;
p >>= 1;
}
return ans;
}
long long inv(long long x, long long mod) { return qp(x, mod - 2, mod); }
long long C(long long N, long long K, long long fact[], long long mod) {
return fact[N] * inv(fact[K], mod) % mod * inv(fact[N - K], mod) % mod;
}
template <typename Tp>
Tp gcd(Tp A, Tp B) {
if (B == 0) return A;
return gcd(B, A % B);
}
template <typename Tp>
Tp lcm(Tp A, Tp B) {
return A * B / gcd(A, B);
}
}; // namespace Mymath
namespace fwt {
using namespace Mymath;
void FWT(int a[], int n, long long mod) {
for (int d = 1; d < n; d <<= 1)
for (int m = d << 1, i = 0; i < n; i += m)
for (int j = 0; j < d; j++) {
int x = a[i + j], y = a[i + j + d];
a[i + j] = (x + y) % mod, a[i + j + d] = (x - y + mod) % mod;
}
}
void UFWT(int a[], int n, long long mod) {
long long rev = inv(2, mod);
for (int d = 1; d < n; d <<= 1)
for (int m = d << 1, i = 0; i < n; i += m)
for (int j = 0; j < d; j++) {
int x = a[i + j], y = a[i + j + d];
a[i + j] = 1LL * (x + y) * rev % mod,
a[i + j + d] = (1LL * (x - y) * rev % mod + mod) % mod;
}
}
void solve(int a[], int b[], int n, long long mod) {
FWT(a, n, mod);
FWT(b, n, mod);
for (int i = 0; i < n; i++) a[i] = 1LL * a[i] * b[i] % mod;
UFWT(a, n, mod);
}
}; // namespace fwt
const int Maxn = 100005;
struct P {
long long x, y;
P(long long X, long long Y) { x = X, y = Y; }
P() { x = y = 0; }
P operator+(P p) { return P(x + p.x, y + p.y); }
P operator-(P p) { return P(x - p.x, y - p.y); }
P operator*(double k) { return P(x * k, y * k); }
long long det(P p) { return x * p.y - y * p.x; }
long long dot(P p) { return x * p.x + y * p.y; }
bool operator<(const P& p) const {
if (x != p.x) return x < p.x;
return y < p.y;
}
double dist() { return sqrt((double)x * x + y * y); }
};
P A[Maxn], B[Maxn];
int n, m;
vector<P> xp, yp;
vector<P> build(P X[], int l) {
vector<P> ret;
sort(X + 1, X + 1 + l);
ret.push_back(X[1]);
for (int i = 2; i <= l; i++) {
while (ret.size() >= 2 &&
(ret[ret.size() - 2] - X[i]).det(ret[ret.size() - 1] - X[i]) >= 0) {
ret.pop_back();
}
ret.push_back(X[i]);
}
int rs = ret.size() + 1;
for (int i = l - 1; i >= 1; i--) {
while (ret.size() >= rs &&
(ret[ret.size() - 2] - X[i]).det(ret[ret.size() - 1] - X[i]) >= 0) {
ret.pop_back();
}
ret.push_back(X[i]);
}
return ret;
}
using namespace Mymath;
const long long mod1 = 998244353;
const long long mod2 = 1000000007;
long long aa[Maxn], ab[Maxn];
long long pa1[Maxn], pa2[Maxn];
long long pb1[Maxn], pb2[Maxn];
long long Fa1[Maxn], Fa2[Maxn];
long long ifa1[Maxn], ifa2[Maxn];
int main() {
n = read();
m = read();
for (int i = 1; i <= n; i++) {
A[i].x = read();
A[i].y = read();
}
for (int i = 1; i <= m; i++) {
B[i].x = read();
B[i].y = read();
}
xp = build(A, n);
yp = build(B, m);
if (xp.size() != yp.size()) {
printf("NO\n");
return 0;
}
for (int i = 1; i < xp.size() - 1; i++) {
P ta = xp[i - 1] - xp[i], tb = xp[i + 1] - xp[i];
long long X = ta.det(tb);
aa[i] = X;
}
{
P ta = xp[1] - xp[0], tb = xp[xp.size() - 2] - xp[0];
long long X = ta.det(tb);
X = -(X);
aa[0] = aa[xp.size() - 1] = X;
}
for (int i = 1; i < yp.size() - 1; i++) {
P ta = yp[i - 1] - yp[i], tb = yp[i + 1] - yp[i];
long long X = ta.det(tb);
ab[i] = X;
}
{
P ta = yp[1] - yp[0], tb = yp[yp.size() - 2] - yp[0];
long long X = ta.det(tb);
X = -X;
ab[0] = ab[yp.size() - 1] = X;
}
vector<pair<long long, long long> > P1, P2;
P1.push_back(make_pair(0, 0));
for (int i = 0; i < xp.size() - 1; i++) {
P ed = xp[i + 1] - xp[i];
long long dd = ed.x * ed.x + ed.y * ed.y;
long long h1 = dd % mod1 * 239429834 + aa[i] % mod1 * 834728734 +
aa[i + 1] % mod1 * 349828333;
h1 %= mod1;
long long h2 = dd % mod2 * 239429834 + aa[i] % mod2 * 834728734 +
aa[i + 1] % mod2 * 349828333;
h2 %= mod2;
P1.push_back(make_pair(h1, h2));
}
P2.push_back(make_pair(0, 0));
for (int i = 0; i < yp.size() - 1; i++) {
P ed = yp[i + 1] - yp[i];
long long dd = ed.x * ed.x + ed.y * ed.y;
long long h1 = dd % mod1 * 239429834 + ab[i] % mod1 * 834728734 +
ab[i + 1] % mod1 * 349828333;
h1 %= mod1;
long long h2 = dd % mod2 * 239429834 + ab[i] % mod2 * 834728734 +
ab[i + 1] % mod2 * 349828333;
h2 %= mod2;
P2.push_back(make_pair(h1, h2));
}
int L = P1.size() - 1;
for (int i = 1; i <= L; i++) {
pa1[i] = pa1[i - 1] * 93482874 + P1[i].first;
pa1[i] %= mod1;
pa2[i] = pa2[i - 1] * 93482874 + P1[i].second;
pa2[i] %= mod2;
}
for (int i = 1; i <= L; i++) {
pb1[i] = pb1[i - 1] * 93482874 + P2[i].first;
pb1[i] %= mod1;
pb2[i] = pb2[i - 1] * 93482874 + P2[i].second;
pb2[i] %= mod2;
}
Fa1[0] = Fa2[0] = 1;
for (int i = 1; i < Maxn; i++) {
Fa1[i] = Fa1[i - 1] * 93482874 % mod1;
Fa2[i] = Fa2[i - 1] * 93482874 % mod2;
}
for (int i = 1; i <= L; i++) {
long long h1 = pa1[L] - pa1[i - 1] * Fa1[L - i + 1];
h1 = (h1 % mod1 + mod1) % mod1;
h1 = h1 * Fa1[i - 1] + pa1[i - 1];
h1 %= mod1;
long long h2 = pa2[L] - pa2[i - 1] * Fa2[L - i + 1];
h2 = (h2 % mod2 + mod2) % mod2;
h2 = h2 * Fa2[i - 1] + pa2[i - 1];
h2 %= mod2;
if (h1 == pb1[L] && h2 == pb2[L]) {
printf("YES\n");
return 0;
}
}
printf("NO\n");
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
inline int read() {
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
namespace Mymath {
long long qp(long long x, long long p, long long mod) {
long long ans = 1;
while (p) {
if (p & 1) ans = ans * x % mod;
x = x * x % mod;
p >>= 1;
}
return ans;
}
long long inv(long long x, long long mod) { return qp(x, mod - 2, mod); }
long long C(long long N, long long K, long long fact[], long long mod) {
return fact[N] * inv(fact[K], mod) % mod * inv(fact[N - K], mod) % mod;
}
template <typename Tp>
Tp gcd(Tp A, Tp B) {
if (B == 0) return A;
return gcd(B, A % B);
}
template <typename Tp>
Tp lcm(Tp A, Tp B) {
return A * B / gcd(A, B);
}
}; // namespace Mymath
namespace fwt {
using namespace Mymath;
void FWT(int a[], int n, long long mod) {
for (int d = 1; d < n; d <<= 1)
for (int m = d << 1, i = 0; i < n; i += m)
for (int j = 0; j < d; j++) {
int x = a[i + j], y = a[i + j + d];
a[i + j] = (x + y) % mod, a[i + j + d] = (x - y + mod) % mod;
}
}
void UFWT(int a[], int n, long long mod) {
long long rev = inv(2, mod);
for (int d = 1; d < n; d <<= 1)
for (int m = d << 1, i = 0; i < n; i += m)
for (int j = 0; j < d; j++) {
int x = a[i + j], y = a[i + j + d];
a[i + j] = 1LL * (x + y) * rev % mod,
a[i + j + d] = (1LL * (x - y) * rev % mod + mod) % mod;
}
}
void solve(int a[], int b[], int n, long long mod) {
FWT(a, n, mod);
FWT(b, n, mod);
for (int i = 0; i < n; i++) a[i] = 1LL * a[i] * b[i] % mod;
UFWT(a, n, mod);
}
}; // namespace fwt
const int Maxn = 100005;
struct P {
long long x, y;
P(long long X, long long Y) { x = X, y = Y; }
P() { x = y = 0; }
P operator+(P p) { return P(x + p.x, y + p.y); }
P operator-(P p) { return P(x - p.x, y - p.y); }
P operator*(double k) { return P(x * k, y * k); }
long long det(P p) { return x * p.y - y * p.x; }
long long dot(P p) { return x * p.x + y * p.y; }
bool operator<(const P& p) const {
if (x != p.x) return x < p.x;
return y < p.y;
}
double dist() { return sqrt((double)x * x + y * y); }
};
P A[Maxn], B[Maxn];
int n, m;
vector<P> xp, yp;
vector<P> build(P X[], int l) {
vector<P> ret;
sort(X + 1, X + 1 + l);
ret.push_back(X[1]);
for (int i = 2; i <= l; i++) {
while (ret.size() >= 2 &&
(ret[ret.size() - 2] - X[i]).det(ret[ret.size() - 1] - X[i]) >= 0) {
ret.pop_back();
}
ret.push_back(X[i]);
}
int rs = ret.size() + 1;
for (int i = l - 1; i >= 1; i--) {
while (ret.size() >= rs &&
(ret[ret.size() - 2] - X[i]).det(ret[ret.size() - 1] - X[i]) >= 0) {
ret.pop_back();
}
ret.push_back(X[i]);
}
return ret;
}
using namespace Mymath;
const long long mod1 = 998244353;
const long long mod2 = 1000000007;
long long aa[Maxn], ab[Maxn];
long long pa1[Maxn], pa2[Maxn];
long long pb1[Maxn], pb2[Maxn];
long long Fa1[Maxn], Fa2[Maxn];
long long ifa1[Maxn], ifa2[Maxn];
int main() {
n = read();
m = read();
for (int i = 1; i <= n; i++) {
A[i].x = read();
A[i].y = read();
}
for (int i = 1; i <= m; i++) {
B[i].x = read();
B[i].y = read();
}
xp = build(A, n);
yp = build(B, m);
if (xp.size() != yp.size()) {
printf("NO\n");
return 0;
}
for (int i = 1; i < xp.size() - 1; i++) {
P ta = xp[i - 1] - xp[i], tb = xp[i + 1] - xp[i];
long long X = ta.det(tb);
aa[i] = X;
}
{
P ta = xp[1] - xp[0], tb = xp[xp.size() - 2] - xp[0];
long long X = ta.det(tb);
X = -(X);
aa[0] = aa[xp.size() - 1] = X;
}
for (int i = 1; i < yp.size() - 1; i++) {
P ta = yp[i - 1] - yp[i], tb = yp[i + 1] - yp[i];
long long X = ta.det(tb);
ab[i] = X;
}
{
P ta = yp[1] - yp[0], tb = yp[yp.size() - 2] - yp[0];
long long X = ta.det(tb);
X = -X;
ab[0] = ab[yp.size() - 1] = X;
}
vector<pair<long long, long long> > P1, P2;
P1.push_back(make_pair(0, 0));
for (int i = 0; i < xp.size() - 1; i++) {
P ed = xp[i + 1] - xp[i];
long long dd = ed.x * ed.x + ed.y * ed.y;
long long h1 = dd % mod1 * 239429834 + aa[i] % mod1 * 834728734 +
aa[i + 1] % mod1 * 349828333;
h1 %= mod1;
long long h2 = dd % mod2 * 239429834 + aa[i] % mod2 * 834728734 +
aa[i + 1] % mod2 * 349828333;
h2 %= mod2;
P1.push_back(make_pair(h1, h2));
}
P2.push_back(make_pair(0, 0));
for (int i = 0; i < yp.size() - 1; i++) {
P ed = yp[i + 1] - yp[i];
long long dd = ed.x * ed.x + ed.y * ed.y;
long long h1 = dd % mod1 * 239429834 + ab[i] % mod1 * 834728734 +
ab[i + 1] % mod1 * 349828333;
h1 %= mod1;
long long h2 = dd % mod2 * 239429834 + ab[i] % mod2 * 834728734 +
ab[i + 1] % mod2 * 349828333;
h2 %= mod2;
P2.push_back(make_pair(h1, h2));
}
int L = P1.size() - 1;
for (int i = 1; i <= L; i++) {
pa1[i] = pa1[i - 1] * 93482874 + P1[i].first;
pa1[i] %= mod1;
pa2[i] = pa2[i - 1] * 93482874 + P1[i].second;
pa2[i] %= mod2;
}
for (int i = 1; i <= L; i++) {
pb1[i] = pb1[i - 1] * 93482874 + P2[i].first;
pb1[i] %= mod1;
pb2[i] = pb2[i - 1] * 93482874 + P2[i].second;
pb2[i] %= mod2;
}
Fa1[0] = Fa2[0] = 1;
for (int i = 1; i < Maxn; i++) {
Fa1[i] = Fa1[i - 1] * 93482874 % mod1;
Fa2[i] = Fa2[i - 1] * 93482874 % mod2;
}
for (int i = 1; i <= L; i++) {
long long h1 = pa1[L] - pa1[i - 1] * Fa1[L - i + 1];
h1 = (h1 % mod1 + mod1) % mod1;
h1 = h1 * Fa1[i - 1] + pa1[i - 1];
h1 %= mod1;
long long h2 = pa2[L] - pa2[i - 1] * Fa2[L - i + 1];
h2 = (h2 % mod2 + mod2) % mod2;
h2 = h2 * Fa2[i - 1] + pa2[i - 1];
h2 %= mod2;
if (h1 == pb1[L] && h2 == pb2[L]) {
printf("YES\n");
return 0;
}
}
printf("NO\n");
return 0;
}
``` |
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int INF = 2147483647;
const int mod = 1000000007;
const int mod2 = 998244353;
const double eps = 1e-9;
const double pi = acos(-1.0);
inline int rint() {
int x;
scanf("%d", &x);
return x;
}
inline long long rll() {
long long x;
scanf("%lld", &x);
return x;
}
inline long long ri64() {
long long x;
scanf("%I64d", &x);
return x;
}
void rvi(vector<int> &a, int n) {
for (int i = 0; i < n; i++) {
a.push_back(rint());
}
}
void rvll(vector<long long> &a, int n) {
for (int i = 0; i < n; i++) {
a.push_back(rll());
}
}
void showvi(vector<int> &a) {
for (int i = 0; i < a.size(); i++) {
printf("%d%c", a[i], (i + 1 == a.size() ? '\n' : ' '));
}
}
void showvll(vector<long long> &a) {
for (int i = 0; i < a.size(); i++) {
printf("%lld%c", a[i], (i + 1 == a.size() ? '\n' : ' '));
}
}
void showviln(vector<int> &a) {
for (int i = 0; i < a.size(); i++) {
printf("%d%c", a[i], '\n');
}
}
void showvllln(vector<long long> &a) {
for (int i = 0; i < a.size(); i++) {
printf("%lld%c", a[i], '\n');
}
}
void showi(int x) { printf("%d", x); }
long long pw(long long a, long long b, long long c) {
long long ans = 1;
long long k = a % c;
while (b > 0) {
if (b % 2 == 1) ans = (ans * k) % c;
b = b / 2;
k = (k * k) % c;
}
return ans;
}
const int maxn = 100011;
struct node {
double x, y;
};
int n;
int tot;
node p[maxn];
node P[maxn];
int tot2;
node P2[maxn];
double X(node A, node B, node C) {
return (B.x - A.x) * (C.y - A.y) - (C.x - A.x) * (B.y - A.y);
}
double dian(node A, node B, node C) {
return (B.x - A.x) * (C.x - A.x) + (C.y - A.y) * (B.y - A.y);
}
double len(node A, node B) {
return sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
}
bool cmp(node A, node B) {
double pp = X(p[0], A, B);
if (pp > 0) return true;
if (pp < 0) return false;
if (pp == 0) return len(p[0], A) < len(p[0], B);
}
void build() {
for (int i = 0; i < n; i++) {
if (p[i].y < p[0].y) swap(p[i], p[0]);
if (p[i].y == p[0].y && p[i].x < p[0].x) swap(p[i], p[0]);
}
sort(p + 1, p + n, cmp);
P[0] = p[0];
P[1] = p[1];
tot = 1;
for (int i = 2; i < n; i++) {
while (tot > 0 && X(P[tot - 1], P[tot], p[i]) <= 0) tot--;
tot++;
P[tot] = p[i];
}
tot++;
}
int x1[100011];
int asdfeawf[100011];
int x2[100011];
int y2[100011];
int a[400044];
int b[400044];
int ha[400044];
int hb[400044];
int pwb[400044];
const int hmod = 1000000007;
const int mmod = 998244353;
const int base = 233;
int dis(int x1, int asdfeawf, int x2, int y2) {
int ans = 1LL * (x1 - x2) * (x1 - x2) % mmod +
1LL * (asdfeawf - y2) * (asdfeawf - y2) % mmod;
if (ans >= mmod) ans -= mmod;
return ans;
}
void work() {
for (int i = 0; i < (tot); i++) {
x1[i] = P[i].x + 0.1;
asdfeawf[i] = P[i].y + 0.1;
}
for (int i = 0; i < (tot2); i++) {
x2[i] = P2[i].x + 0.1;
y2[i] = P2[i].y + 0.1;
}
if (tot != tot2) {
cout << "NO" << '\n';
return;
}
for (int i = 0; i < (tot); i++) {
int X0 = x1[(i - 1 + tot) % tot];
int Y0 = asdfeawf[(i - 1 + tot) % tot];
int X1 = x1[i];
int Y1 = asdfeawf[i];
int X2 = x1[(((i) + 1) % (tot))];
int Y2 = asdfeawf[(((i) + 1) % (tot))];
int len0 = dis(X0, Y0, X1, Y1);
int len1 = dis(X1, Y1, X2, Y2);
a[2 * i + 1] = a[2 * i + 2 * tot + 1] = len1;
int VX0 = X0 - X1;
long long VY0 = Y0 - Y1;
int VX1 = X2 - X1;
long long VY1 = Y2 - Y1;
if (VX0 < 0) VX0 += mmod;
if (VX1 < 0) VX1 += mmod;
if (VY1 < 0) VY1 += mmod;
if (VY0 < 0) VY0 += mmod;
int DOT = (1LL * VX0 * VX1 % mmod + 1LL * VY0 * VY1 % mmod);
if (DOT >= mmod) DOT -= mod;
int MUL = 1LL * len0 * len1 % mmod;
int ANG = 1LL * DOT * pw(MUL, mmod - 2, mmod) % mmod;
a[2 * i] = a[2 * i + 2 * tot] = ANG;
}
for (int i = 0; i < (tot); i++) {
int X0 = x2[(i - 1 + tot) % tot];
int Y0 = y2[(i - 1 + tot) % tot];
int X1 = x2[i];
int Y1 = y2[i];
int X2 = x2[(((i) + 1) % (tot))];
int Y2 = y2[(((i) + 1) % (tot))];
int len0 = dis(X0, Y0, X1, Y1);
int len1 = dis(X1, Y1, X2, Y2);
b[2 * i + 1] = b[2 * i + 2 * tot + 1] = len1;
int VX0 = X0 - X1;
long long VY0 = Y0 - Y1;
int VX1 = X2 - X1;
long long VY1 = Y2 - Y1;
if (VX0 < 0) VX0 += mmod;
if (VX1 < 0) VX1 += mmod;
if (VY1 < 0) VY1 += mmod;
if (VY0 < 0) VY0 += mmod;
int DOT = (1LL * VX0 * VX1 % mmod + 1LL * VY0 * VY1 % mmod);
if (DOT >= mmod) DOT -= mod;
int MUL = 1LL * len0 * len1 % mmod;
int ANG = 1LL * DOT * pw(MUL, mmod - 2, mmod) % mmod;
b[2 * i] = b[2 * i + 2 * tot] = ANG;
}
for (int i = 0; i < (4 * tot); i++) {
ha[i] = 0;
if (i) ha[i] = 1LL * ha[i - 1] * base % hmod;
ha[i] = (ha[i] + a[i]);
if (ha[i] >= hmod) ha[i] -= hmod;
hb[i] = 0;
if (i) hb[i] = 1LL * hb[i - 1] * base % hmod;
hb[i] = (hb[i] + b[i]);
if (hb[i] >= hmod) hb[i] -= hmod;
}
int h1 = ha[2 * tot - 1];
for (int i = 0; i < 2 * tot; i += 2) {
int l = i;
int r = i + 2 * tot - 1;
int h2 = hb[r];
if (l > 0) h2 = h2 - 1LL * hb[l - 1] * pwb[r - l + 1] % hmod;
if (h2 < 0) h2 += hmod;
if (h1 == h2) {
cout << "YES" << '\n';
return;
}
}
cout << "NO" << '\n';
}
int main() {
pwb[0] = 1;
for (int i = 1; i < 400044; i++) pwb[i] = 1LL * pwb[i - 1] * base % hmod;
int n1, n2;
scanf("%d%d", &n1, &n2);
tot = 0;
n = n1;
for (int i = 0; i < (n1); i++) {
p[i].x = rint();
p[i].y = rint();
}
build();
for (int i = 0; i < (tot); i++) {
P2[i].x = P[i].x;
P2[i].y = P[i].y;
}
tot2 = tot;
tot = 0;
n = n2;
for (int i = 0; i < (n2); i++) {
p[i].x = rint();
p[i].y = rint();
}
build();
work();
}
| ### Prompt
Generate a cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int INF = 2147483647;
const int mod = 1000000007;
const int mod2 = 998244353;
const double eps = 1e-9;
const double pi = acos(-1.0);
inline int rint() {
int x;
scanf("%d", &x);
return x;
}
inline long long rll() {
long long x;
scanf("%lld", &x);
return x;
}
inline long long ri64() {
long long x;
scanf("%I64d", &x);
return x;
}
void rvi(vector<int> &a, int n) {
for (int i = 0; i < n; i++) {
a.push_back(rint());
}
}
void rvll(vector<long long> &a, int n) {
for (int i = 0; i < n; i++) {
a.push_back(rll());
}
}
void showvi(vector<int> &a) {
for (int i = 0; i < a.size(); i++) {
printf("%d%c", a[i], (i + 1 == a.size() ? '\n' : ' '));
}
}
void showvll(vector<long long> &a) {
for (int i = 0; i < a.size(); i++) {
printf("%lld%c", a[i], (i + 1 == a.size() ? '\n' : ' '));
}
}
void showviln(vector<int> &a) {
for (int i = 0; i < a.size(); i++) {
printf("%d%c", a[i], '\n');
}
}
void showvllln(vector<long long> &a) {
for (int i = 0; i < a.size(); i++) {
printf("%lld%c", a[i], '\n');
}
}
void showi(int x) { printf("%d", x); }
long long pw(long long a, long long b, long long c) {
long long ans = 1;
long long k = a % c;
while (b > 0) {
if (b % 2 == 1) ans = (ans * k) % c;
b = b / 2;
k = (k * k) % c;
}
return ans;
}
const int maxn = 100011;
struct node {
double x, y;
};
int n;
int tot;
node p[maxn];
node P[maxn];
int tot2;
node P2[maxn];
double X(node A, node B, node C) {
return (B.x - A.x) * (C.y - A.y) - (C.x - A.x) * (B.y - A.y);
}
double dian(node A, node B, node C) {
return (B.x - A.x) * (C.x - A.x) + (C.y - A.y) * (B.y - A.y);
}
double len(node A, node B) {
return sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
}
bool cmp(node A, node B) {
double pp = X(p[0], A, B);
if (pp > 0) return true;
if (pp < 0) return false;
if (pp == 0) return len(p[0], A) < len(p[0], B);
}
void build() {
for (int i = 0; i < n; i++) {
if (p[i].y < p[0].y) swap(p[i], p[0]);
if (p[i].y == p[0].y && p[i].x < p[0].x) swap(p[i], p[0]);
}
sort(p + 1, p + n, cmp);
P[0] = p[0];
P[1] = p[1];
tot = 1;
for (int i = 2; i < n; i++) {
while (tot > 0 && X(P[tot - 1], P[tot], p[i]) <= 0) tot--;
tot++;
P[tot] = p[i];
}
tot++;
}
int x1[100011];
int asdfeawf[100011];
int x2[100011];
int y2[100011];
int a[400044];
int b[400044];
int ha[400044];
int hb[400044];
int pwb[400044];
const int hmod = 1000000007;
const int mmod = 998244353;
const int base = 233;
int dis(int x1, int asdfeawf, int x2, int y2) {
int ans = 1LL * (x1 - x2) * (x1 - x2) % mmod +
1LL * (asdfeawf - y2) * (asdfeawf - y2) % mmod;
if (ans >= mmod) ans -= mmod;
return ans;
}
void work() {
for (int i = 0; i < (tot); i++) {
x1[i] = P[i].x + 0.1;
asdfeawf[i] = P[i].y + 0.1;
}
for (int i = 0; i < (tot2); i++) {
x2[i] = P2[i].x + 0.1;
y2[i] = P2[i].y + 0.1;
}
if (tot != tot2) {
cout << "NO" << '\n';
return;
}
for (int i = 0; i < (tot); i++) {
int X0 = x1[(i - 1 + tot) % tot];
int Y0 = asdfeawf[(i - 1 + tot) % tot];
int X1 = x1[i];
int Y1 = asdfeawf[i];
int X2 = x1[(((i) + 1) % (tot))];
int Y2 = asdfeawf[(((i) + 1) % (tot))];
int len0 = dis(X0, Y0, X1, Y1);
int len1 = dis(X1, Y1, X2, Y2);
a[2 * i + 1] = a[2 * i + 2 * tot + 1] = len1;
int VX0 = X0 - X1;
long long VY0 = Y0 - Y1;
int VX1 = X2 - X1;
long long VY1 = Y2 - Y1;
if (VX0 < 0) VX0 += mmod;
if (VX1 < 0) VX1 += mmod;
if (VY1 < 0) VY1 += mmod;
if (VY0 < 0) VY0 += mmod;
int DOT = (1LL * VX0 * VX1 % mmod + 1LL * VY0 * VY1 % mmod);
if (DOT >= mmod) DOT -= mod;
int MUL = 1LL * len0 * len1 % mmod;
int ANG = 1LL * DOT * pw(MUL, mmod - 2, mmod) % mmod;
a[2 * i] = a[2 * i + 2 * tot] = ANG;
}
for (int i = 0; i < (tot); i++) {
int X0 = x2[(i - 1 + tot) % tot];
int Y0 = y2[(i - 1 + tot) % tot];
int X1 = x2[i];
int Y1 = y2[i];
int X2 = x2[(((i) + 1) % (tot))];
int Y2 = y2[(((i) + 1) % (tot))];
int len0 = dis(X0, Y0, X1, Y1);
int len1 = dis(X1, Y1, X2, Y2);
b[2 * i + 1] = b[2 * i + 2 * tot + 1] = len1;
int VX0 = X0 - X1;
long long VY0 = Y0 - Y1;
int VX1 = X2 - X1;
long long VY1 = Y2 - Y1;
if (VX0 < 0) VX0 += mmod;
if (VX1 < 0) VX1 += mmod;
if (VY1 < 0) VY1 += mmod;
if (VY0 < 0) VY0 += mmod;
int DOT = (1LL * VX0 * VX1 % mmod + 1LL * VY0 * VY1 % mmod);
if (DOT >= mmod) DOT -= mod;
int MUL = 1LL * len0 * len1 % mmod;
int ANG = 1LL * DOT * pw(MUL, mmod - 2, mmod) % mmod;
b[2 * i] = b[2 * i + 2 * tot] = ANG;
}
for (int i = 0; i < (4 * tot); i++) {
ha[i] = 0;
if (i) ha[i] = 1LL * ha[i - 1] * base % hmod;
ha[i] = (ha[i] + a[i]);
if (ha[i] >= hmod) ha[i] -= hmod;
hb[i] = 0;
if (i) hb[i] = 1LL * hb[i - 1] * base % hmod;
hb[i] = (hb[i] + b[i]);
if (hb[i] >= hmod) hb[i] -= hmod;
}
int h1 = ha[2 * tot - 1];
for (int i = 0; i < 2 * tot; i += 2) {
int l = i;
int r = i + 2 * tot - 1;
int h2 = hb[r];
if (l > 0) h2 = h2 - 1LL * hb[l - 1] * pwb[r - l + 1] % hmod;
if (h2 < 0) h2 += hmod;
if (h1 == h2) {
cout << "YES" << '\n';
return;
}
}
cout << "NO" << '\n';
}
int main() {
pwb[0] = 1;
for (int i = 1; i < 400044; i++) pwb[i] = 1LL * pwb[i - 1] * base % hmod;
int n1, n2;
scanf("%d%d", &n1, &n2);
tot = 0;
n = n1;
for (int i = 0; i < (n1); i++) {
p[i].x = rint();
p[i].y = rint();
}
build();
for (int i = 0; i < (tot); i++) {
P2[i].x = P[i].x;
P2[i].y = P[i].y;
}
tot2 = tot;
tot = 0;
n = n2;
for (int i = 0; i < (n2); i++) {
p[i].x = rint();
p[i].y = rint();
}
build();
work();
}
``` |
#include <bits/stdc++.h>
using namespace std;
struct vec {
long long first, second;
vec() {}
vec(long long first, long long second) : first(first), second(second) {}
vec operator+(const vec& T) const {
return {first + T.first, second + T.second};
}
vec operator-(const vec& T) const {
return {first - T.first, second - T.second};
}
vec operator*(const long long& T) const { return {first * T, second * T}; }
long long dot(const vec& T) const {
return first * T.first + second * T.second;
}
long long cross(const vec& T) const {
return first * T.second - second * T.first;
}
long long len() { return first * first + second * second; }
bool operator==(const vec& T) const {
return first == T.first && second == T.second;
}
};
bool CHcmp(vec u, vec v) {
if (u.first != v.first) return u.first < v.first;
return u.second < v.second;
}
vector<vec> convex_hull(vector<vec> p) {
sort(p.begin(), p.end(), CHcmp);
vector<vec> hull;
for (int i = 0; i < p.size(); i++) {
while (hull.size() > 1 && (hull.back() - hull[hull.size() - 2])
.cross(p[i] - hull[hull.size() - 2]) <= 0)
hull.pop_back();
hull.push_back(p[i]);
}
int tmp = (int)hull.size();
for (int i = p.size() - 2; i >= 0; i--) {
while (hull.size() > tmp &&
(hull.back() - hull[hull.size() - 2])
.cross(p[i] - hull[hull.size() - 2]) <= 0)
hull.pop_back();
hull.push_back(p[i]);
}
if (p.size() > 1) hull.pop_back();
return hull;
}
int n, m, sz;
vector<vec> ina, inb, ha, hb;
vector<pair<long long, long long> > sa, sb;
long long val[200055], ps[200055];
bool solve(long long p1, long long p2, long long md) {
long long tar = 0;
for (int i = 0; i < sz; i++) {
long long num = sb[i].first * p1 + sb[i].second * p2;
num %= md;
tar = (tar + num * (i + 1)) % md;
}
long long cur = 0;
for (int i = 0; i < sz + sz; i++) {
val[i] = sa[i].first * p1 + sa[i].second * p2;
val[i] %= md;
if (i < sz) cur = (cur + val[i] * (i + 1)) % md;
if (i > 0)
ps[i] = (ps[i - 1] + val[i]) % md;
else
ps[i] = val[i];
}
if (cur == tar) return true;
for (int i = sz; i < sz + sz; i++) {
cur -= ps[i - 1] - (i - 1 - sz >= 0 ? ps[i - 1 - sz] : 0);
cur %= md;
cur = (cur + md) % md;
cur = (cur + val[i] * sz) % md;
if (cur == tar) return true;
}
return false;
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++) {
long long first, second;
scanf("%lld%lld", &first, &second);
ina.push_back(vec(first, second));
}
for (int i = 0; i < m; i++) {
long long first, second;
scanf("%lld%lld", &first, &second);
inb.push_back(vec(first, second));
}
ha = convex_hull(ina);
hb = convex_hull(inb);
if (ha.size() != hb.size()) {
printf("no\n");
return 0;
}
sz = ha.size();
ha.push_back(ha[0]);
ha.push_back(ha[1]);
hb.push_back(hb[0]);
hb.push_back(hb[1]);
for (int i = 0; i < sz; i++) {
long long len = (ha[i] - ha[i + 1]).len();
long long ang = (ha[i] - ha[i + 1]).dot(ha[i] - ha[i + 2]);
sa.push_back({len, ang});
}
for (int i = 0; i < sz; i++) {
long long len = (hb[i] - hb[i + 1]).len();
long long ang = (hb[i] - hb[i + 1]).dot(hb[i] - hb[i + 2]);
sb.push_back({len, ang});
}
for (int i = 0; i < sz; i++) sa.push_back(sa[i]);
if (solve(37, 97, 1610612741)) {
printf("YES\n");
} else {
printf("NO\n");
}
return 0;
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
struct vec {
long long first, second;
vec() {}
vec(long long first, long long second) : first(first), second(second) {}
vec operator+(const vec& T) const {
return {first + T.first, second + T.second};
}
vec operator-(const vec& T) const {
return {first - T.first, second - T.second};
}
vec operator*(const long long& T) const { return {first * T, second * T}; }
long long dot(const vec& T) const {
return first * T.first + second * T.second;
}
long long cross(const vec& T) const {
return first * T.second - second * T.first;
}
long long len() { return first * first + second * second; }
bool operator==(const vec& T) const {
return first == T.first && second == T.second;
}
};
bool CHcmp(vec u, vec v) {
if (u.first != v.first) return u.first < v.first;
return u.second < v.second;
}
vector<vec> convex_hull(vector<vec> p) {
sort(p.begin(), p.end(), CHcmp);
vector<vec> hull;
for (int i = 0; i < p.size(); i++) {
while (hull.size() > 1 && (hull.back() - hull[hull.size() - 2])
.cross(p[i] - hull[hull.size() - 2]) <= 0)
hull.pop_back();
hull.push_back(p[i]);
}
int tmp = (int)hull.size();
for (int i = p.size() - 2; i >= 0; i--) {
while (hull.size() > tmp &&
(hull.back() - hull[hull.size() - 2])
.cross(p[i] - hull[hull.size() - 2]) <= 0)
hull.pop_back();
hull.push_back(p[i]);
}
if (p.size() > 1) hull.pop_back();
return hull;
}
int n, m, sz;
vector<vec> ina, inb, ha, hb;
vector<pair<long long, long long> > sa, sb;
long long val[200055], ps[200055];
bool solve(long long p1, long long p2, long long md) {
long long tar = 0;
for (int i = 0; i < sz; i++) {
long long num = sb[i].first * p1 + sb[i].second * p2;
num %= md;
tar = (tar + num * (i + 1)) % md;
}
long long cur = 0;
for (int i = 0; i < sz + sz; i++) {
val[i] = sa[i].first * p1 + sa[i].second * p2;
val[i] %= md;
if (i < sz) cur = (cur + val[i] * (i + 1)) % md;
if (i > 0)
ps[i] = (ps[i - 1] + val[i]) % md;
else
ps[i] = val[i];
}
if (cur == tar) return true;
for (int i = sz; i < sz + sz; i++) {
cur -= ps[i - 1] - (i - 1 - sz >= 0 ? ps[i - 1 - sz] : 0);
cur %= md;
cur = (cur + md) % md;
cur = (cur + val[i] * sz) % md;
if (cur == tar) return true;
}
return false;
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++) {
long long first, second;
scanf("%lld%lld", &first, &second);
ina.push_back(vec(first, second));
}
for (int i = 0; i < m; i++) {
long long first, second;
scanf("%lld%lld", &first, &second);
inb.push_back(vec(first, second));
}
ha = convex_hull(ina);
hb = convex_hull(inb);
if (ha.size() != hb.size()) {
printf("no\n");
return 0;
}
sz = ha.size();
ha.push_back(ha[0]);
ha.push_back(ha[1]);
hb.push_back(hb[0]);
hb.push_back(hb[1]);
for (int i = 0; i < sz; i++) {
long long len = (ha[i] - ha[i + 1]).len();
long long ang = (ha[i] - ha[i + 1]).dot(ha[i] - ha[i + 2]);
sa.push_back({len, ang});
}
for (int i = 0; i < sz; i++) {
long long len = (hb[i] - hb[i + 1]).len();
long long ang = (hb[i] - hb[i + 1]).dot(hb[i] - hb[i + 2]);
sb.push_back({len, ang});
}
for (int i = 0; i < sz; i++) sa.push_back(sa[i]);
if (solve(37, 97, 1610612741)) {
printf("YES\n");
} else {
printf("NO\n");
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int nr, N, M, pi[200009], ap[200009], st[200009];
long long pat[200009];
vector<long long> v, g, h;
long long det(pair<int, int> a, pair<int, int> b, pair<int, int> c) {
return 1LL * a.first * b.second + 1LL * b.first * c.second +
1LL * c.first * a.second - 1LL * a.first * c.second -
1LL * b.first * a.second - 1LL * c.first * b.second;
}
void add(int i, vector<pair<int, int> > &p) {
while (nr >= 2 && det(p[st[nr - 1]], p[st[nr]], p[i]) >= 0)
ap[st[nr]] = 0, nr--;
ap[i] = 1, st[++nr] = i;
}
long long modul(long long x) {
if (x < 0) return -x;
return x;
}
void takeConvexHull(vector<pair<int, int> > &points,
vector<pair<int, int> > &hull) {
hull.clear();
sort(points.begin(), points.end());
for (int i = 0; i < points.size(); i++) ap[i] = 0;
nr = 2, st[1] = 0, st[2] = 1, ap[1] = 1;
for (int i = 2; i < points.size(); i++)
if (ap[i] == 0) add(i, points);
for (int i = points.size() - 1; i > 0; i--)
if (ap[i] == 0) add(i, points);
add(0, points), nr--;
for (int i = 1; i <= nr; i++) hull.push_back(points[st[i]]);
if (hull.size() == 1) hull.push_back(points.back());
}
long long d2(pair<int, int> a, pair<int, int> b) {
return 1LL * (a.first - b.first) * (a.first - b.first) +
1LL * (a.second - b.second) * (a.second - b.second);
}
void build(vector<pair<int, int> > &hull, vector<long long> &v) {
v.clear();
int L = hull.size();
hull.push_back(hull[0]);
hull.push_back(hull[1]);
for (int i = 0; i < L; i++)
v.push_back(d2(hull[i], hull[i + 1])),
v.push_back(modul(det(hull[i], hull[i + 1], hull[i + 2])));
hull.pop_back();
hull.pop_back();
}
void solve(int N, vector<long long> &v, vector<long long> &v2) {
vector<pair<int, int> > points, hull;
for (int i = 1; i <= N; i++) {
int x, y;
scanf("%d %d", &x, &y);
points.push_back({x, y});
}
takeConvexHull(points, hull);
build(hull, v);
reverse(hull.begin(), hull.end());
build(hull, v2);
}
void tryMatch(vector<long long> &h) {
N = v.size();
for (int i = 1; i <= N; i++) pat[i] = h[i - 1];
for (int i = 0; i < N; i++) v.push_back(v[i]);
int k = 0;
for (int i = 2; i <= N; i++) {
while (k != 0 && pat[k + 1] != pat[i]) k = pi[k];
k += (pat[k + 1] == pat[i]);
pi[i] = k;
}
k = 0;
int currPos = 0;
for (auto e : v) {
currPos++;
while (k != 0 && pat[k + 1] != e) k = pi[k];
k += (pat[k + 1] == e);
if (k == N && currPos % 2 == N % 2) {
printf("YES\n");
exit(0);
}
if (k == N) k = pi[k];
}
}
int main() {
scanf("%d %d", &N, &M);
solve(N, v, h);
solve(M, h, g);
if (v.size() != h.size()) {
printf("NO\n");
return 0;
}
tryMatch(h);
printf("NO\n");
return 0;
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int nr, N, M, pi[200009], ap[200009], st[200009];
long long pat[200009];
vector<long long> v, g, h;
long long det(pair<int, int> a, pair<int, int> b, pair<int, int> c) {
return 1LL * a.first * b.second + 1LL * b.first * c.second +
1LL * c.first * a.second - 1LL * a.first * c.second -
1LL * b.first * a.second - 1LL * c.first * b.second;
}
void add(int i, vector<pair<int, int> > &p) {
while (nr >= 2 && det(p[st[nr - 1]], p[st[nr]], p[i]) >= 0)
ap[st[nr]] = 0, nr--;
ap[i] = 1, st[++nr] = i;
}
long long modul(long long x) {
if (x < 0) return -x;
return x;
}
void takeConvexHull(vector<pair<int, int> > &points,
vector<pair<int, int> > &hull) {
hull.clear();
sort(points.begin(), points.end());
for (int i = 0; i < points.size(); i++) ap[i] = 0;
nr = 2, st[1] = 0, st[2] = 1, ap[1] = 1;
for (int i = 2; i < points.size(); i++)
if (ap[i] == 0) add(i, points);
for (int i = points.size() - 1; i > 0; i--)
if (ap[i] == 0) add(i, points);
add(0, points), nr--;
for (int i = 1; i <= nr; i++) hull.push_back(points[st[i]]);
if (hull.size() == 1) hull.push_back(points.back());
}
long long d2(pair<int, int> a, pair<int, int> b) {
return 1LL * (a.first - b.first) * (a.first - b.first) +
1LL * (a.second - b.second) * (a.second - b.second);
}
void build(vector<pair<int, int> > &hull, vector<long long> &v) {
v.clear();
int L = hull.size();
hull.push_back(hull[0]);
hull.push_back(hull[1]);
for (int i = 0; i < L; i++)
v.push_back(d2(hull[i], hull[i + 1])),
v.push_back(modul(det(hull[i], hull[i + 1], hull[i + 2])));
hull.pop_back();
hull.pop_back();
}
void solve(int N, vector<long long> &v, vector<long long> &v2) {
vector<pair<int, int> > points, hull;
for (int i = 1; i <= N; i++) {
int x, y;
scanf("%d %d", &x, &y);
points.push_back({x, y});
}
takeConvexHull(points, hull);
build(hull, v);
reverse(hull.begin(), hull.end());
build(hull, v2);
}
void tryMatch(vector<long long> &h) {
N = v.size();
for (int i = 1; i <= N; i++) pat[i] = h[i - 1];
for (int i = 0; i < N; i++) v.push_back(v[i]);
int k = 0;
for (int i = 2; i <= N; i++) {
while (k != 0 && pat[k + 1] != pat[i]) k = pi[k];
k += (pat[k + 1] == pat[i]);
pi[i] = k;
}
k = 0;
int currPos = 0;
for (auto e : v) {
currPos++;
while (k != 0 && pat[k + 1] != e) k = pi[k];
k += (pat[k + 1] == e);
if (k == N && currPos % 2 == N % 2) {
printf("YES\n");
exit(0);
}
if (k == N) k = pi[k];
}
}
int main() {
scanf("%d %d", &N, &M);
solve(N, v, h);
solve(M, h, g);
if (v.size() != h.size()) {
printf("NO\n");
return 0;
}
tryMatch(h);
printf("NO\n");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, m;
struct point {
int x, y;
point() {}
point(int x, int y) : x(x), y(y) {}
point operator-(point a) { return point(x - a.x, y - a.y); }
bool operator<(const point& a) const { return x == a.x ? y < a.y : x < a.x; }
} a[N], b[N];
long long dot(point a, point b) {
return (long long)a.x * b.x + (long long)a.y * b.y;
}
long long cross(point a, point b) {
return (long long)a.x * b.y - (long long)a.y * b.x;
}
long long dist2(point a, point b) { return dot(a - b, a - b); }
vector<point> convex_hull(point* p, int n) {
sort(p + 1, p + 1 + n);
vector<point> convex;
for (int i = 1; i <= n; ++i) {
for (int j; (j = convex.size()) > 1 &&
cross(convex[j - 1] - convex[j - 2], p[i] - convex[j - 2]) <= 0;
convex.pop_back())
;
convex.push_back(p[i]);
}
int old = convex.size();
for (int i = n - 1; i; --i) {
for (int j; (j = convex.size()) > old &&
cross(convex[j - 1] - convex[j - 2], p[i] - convex[j - 2]) <= 0;
convex.pop_back())
;
convex.push_back(p[i]);
}
convex.pop_back();
return convex;
}
struct p3 {
long long x, y, z;
p3() {}
p3(long long x, long long y, long long z) : x(x), y(y), z(z) {}
bool operator<(const p3& a) const {
return x == a.x ? y == a.y ? z < a.z : y < a.y : x < a.x;
}
};
long long f(point a, point b) { return dot(a, b) + cross(a, b); }
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i) {
scanf("%d%d", &a[i].x, &a[i].y);
}
for (int i = 1; i <= m; ++i) {
scanf("%d%d", &b[i].x, &b[i].y);
}
vector<point> convex_a = convex_hull(a, n);
vector<point> convex_b = convex_hull(b, m);
if (convex_a.size() != convex_b.size()) {
return puts("NO"), 0;
} else {
map<p3, int> appeared;
for (int i = 0; i < convex_a.size(); ++i) {
int j = i + 1 == convex_a.size() ? 0 : i + 1;
int k = i == 0 ? convex_a.size() - 1 : i - 1;
++appeared[p3(dist2(convex_a[k], convex_a[i]),
f(convex_a[k] - convex_a[i], convex_a[j] - convex_a[i]),
dist2(convex_a[j], convex_a[i]))];
}
for (int i = 0; i < convex_b.size(); ++i) {
int j = i + 1 == convex_b.size() ? 0 : i + 1;
int k = i == 0 ? convex_b.size() - 1 : i - 1;
--appeared[p3(dist2(convex_b[k], convex_b[i]),
f(convex_b[k] - convex_b[i], convex_b[j] - convex_b[i]),
dist2(convex_b[j], convex_b[i]))];
}
bool same = true;
for (map<p3, int>::iterator it = appeared.begin(); it != appeared.end();
++it) {
if (it->second != 0) {
same = false;
}
}
return puts(same ? "YES" : "NO"), 0;
}
return 0;
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, m;
struct point {
int x, y;
point() {}
point(int x, int y) : x(x), y(y) {}
point operator-(point a) { return point(x - a.x, y - a.y); }
bool operator<(const point& a) const { return x == a.x ? y < a.y : x < a.x; }
} a[N], b[N];
long long dot(point a, point b) {
return (long long)a.x * b.x + (long long)a.y * b.y;
}
long long cross(point a, point b) {
return (long long)a.x * b.y - (long long)a.y * b.x;
}
long long dist2(point a, point b) { return dot(a - b, a - b); }
vector<point> convex_hull(point* p, int n) {
sort(p + 1, p + 1 + n);
vector<point> convex;
for (int i = 1; i <= n; ++i) {
for (int j; (j = convex.size()) > 1 &&
cross(convex[j - 1] - convex[j - 2], p[i] - convex[j - 2]) <= 0;
convex.pop_back())
;
convex.push_back(p[i]);
}
int old = convex.size();
for (int i = n - 1; i; --i) {
for (int j; (j = convex.size()) > old &&
cross(convex[j - 1] - convex[j - 2], p[i] - convex[j - 2]) <= 0;
convex.pop_back())
;
convex.push_back(p[i]);
}
convex.pop_back();
return convex;
}
struct p3 {
long long x, y, z;
p3() {}
p3(long long x, long long y, long long z) : x(x), y(y), z(z) {}
bool operator<(const p3& a) const {
return x == a.x ? y == a.y ? z < a.z : y < a.y : x < a.x;
}
};
long long f(point a, point b) { return dot(a, b) + cross(a, b); }
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i) {
scanf("%d%d", &a[i].x, &a[i].y);
}
for (int i = 1; i <= m; ++i) {
scanf("%d%d", &b[i].x, &b[i].y);
}
vector<point> convex_a = convex_hull(a, n);
vector<point> convex_b = convex_hull(b, m);
if (convex_a.size() != convex_b.size()) {
return puts("NO"), 0;
} else {
map<p3, int> appeared;
for (int i = 0; i < convex_a.size(); ++i) {
int j = i + 1 == convex_a.size() ? 0 : i + 1;
int k = i == 0 ? convex_a.size() - 1 : i - 1;
++appeared[p3(dist2(convex_a[k], convex_a[i]),
f(convex_a[k] - convex_a[i], convex_a[j] - convex_a[i]),
dist2(convex_a[j], convex_a[i]))];
}
for (int i = 0; i < convex_b.size(); ++i) {
int j = i + 1 == convex_b.size() ? 0 : i + 1;
int k = i == 0 ? convex_b.size() - 1 : i - 1;
--appeared[p3(dist2(convex_b[k], convex_b[i]),
f(convex_b[k] - convex_b[i], convex_b[j] - convex_b[i]),
dist2(convex_b[j], convex_b[i]))];
}
bool same = true;
for (map<p3, int>::iterator it = appeared.begin(); it != appeared.end();
++it) {
if (it->second != 0) {
same = false;
}
}
return puts(same ? "YES" : "NO"), 0;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1);
int read() {
char c = getchar();
int res = 0;
while (c < '0' || c > '9') c = getchar();
while (48 <= c && c <= 57) {
res = res * 10 + c - '0';
c = getchar();
}
return res;
}
bool bad(pair<double, double> A, pair<double, double> B, pair<double, double> C,
bool t) {
return B.first * (C.second - A.second) + C.first * (A.second - B.second) +
A.first * (B.second - C.second) >=
0;
}
void ConvexHull(vector<pair<double, double> > &p) {
sort(p.begin(), p.end());
vector<pair<double, double> > upper;
upper.clear();
vector<pair<double, double> > lower;
lower.clear();
for (int i = 0; i < p.size(); ++i) {
while (upper.size() > 1) {
int j = upper.size() - 2;
if (bad(upper[j], upper[j + 1], p[i], 1))
upper.pop_back();
else
break;
}
upper.push_back(p[i]);
}
reverse(p.begin(), p.end());
for (int i = 0; i < p.size(); ++i) {
while (lower.size() > 1) {
int j = lower.size() - 2;
if (bad(lower[j], lower[j + 1], p[i], 0))
lower.pop_back();
else
break;
}
lower.push_back(p[i]);
}
p.clear();
for (int i = 1; i < upper.size(); ++i) p.push_back(upper[i]);
for (int i = 1; i < lower.size(); ++i) p.push_back(lower[i]);
}
double len(pair<double, double> u) {
return sqrt(u.first * u.first + u.second * u.second);
}
double ang(pair<double, double> u, pair<double, double> v) {
return acos((u.first * v.first + u.second * v.second) / len(u) / len(v));
}
void TransForm(vector<pair<double, double> > &P) {
vector<pair<double, double> > C;
C.clear();
int n = P.size();
for (int i = 0; i < n; ++i) {
int j = (i + 2) % n;
pair<double, double> u = {P[i].first - P[(i + 1) % n].first,
P[i].second - P[(i + 1) % n].second};
pair<double, double> v = {P[j].first - P[(i + 1) % n].first,
P[j].second - P[(i + 1) % n].second};
C.push_back({len(u), ang(u, v)});
}
P = C;
}
int n, m;
vector<pair<double, double> > A, B;
int KMP[N];
int main() {
A.resize(read());
B.resize(read());
for (int i = 0; i < A.size(); ++i) A[i].first = read(), A[i].second = read();
for (int i = 0; i < B.size(); ++i) B[i].first = read(), B[i].second = read();
ConvexHull(A);
TransForm(A);
ConvexHull(B);
TransForm(B);
if (A.size() != B.size()) return printf("NO"), 0;
int n = A.size();
for (int i = 0; i < n; ++i) A.push_back(A[i]);
KMP[0] = -1;
for (int i = 1; i < n; ++i) {
int j = KMP[i - 1];
while (B[j + 1] != B[i]) {
if (j >= 0)
j = KMP[j];
else {
j = -2;
break;
}
}
KMP[i] = j + 1;
}
int len = -1;
for (int i = 0; i < A.size(); ++i) {
if (len == n - 1) {
printf("YES");
exit(0);
}
if (A[i] == B[len + 1])
++len;
else if (len >= 0)
len = KMP[len];
}
printf("NO");
}
| ### Prompt
Generate a cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1);
int read() {
char c = getchar();
int res = 0;
while (c < '0' || c > '9') c = getchar();
while (48 <= c && c <= 57) {
res = res * 10 + c - '0';
c = getchar();
}
return res;
}
bool bad(pair<double, double> A, pair<double, double> B, pair<double, double> C,
bool t) {
return B.first * (C.second - A.second) + C.first * (A.second - B.second) +
A.first * (B.second - C.second) >=
0;
}
void ConvexHull(vector<pair<double, double> > &p) {
sort(p.begin(), p.end());
vector<pair<double, double> > upper;
upper.clear();
vector<pair<double, double> > lower;
lower.clear();
for (int i = 0; i < p.size(); ++i) {
while (upper.size() > 1) {
int j = upper.size() - 2;
if (bad(upper[j], upper[j + 1], p[i], 1))
upper.pop_back();
else
break;
}
upper.push_back(p[i]);
}
reverse(p.begin(), p.end());
for (int i = 0; i < p.size(); ++i) {
while (lower.size() > 1) {
int j = lower.size() - 2;
if (bad(lower[j], lower[j + 1], p[i], 0))
lower.pop_back();
else
break;
}
lower.push_back(p[i]);
}
p.clear();
for (int i = 1; i < upper.size(); ++i) p.push_back(upper[i]);
for (int i = 1; i < lower.size(); ++i) p.push_back(lower[i]);
}
double len(pair<double, double> u) {
return sqrt(u.first * u.first + u.second * u.second);
}
double ang(pair<double, double> u, pair<double, double> v) {
return acos((u.first * v.first + u.second * v.second) / len(u) / len(v));
}
void TransForm(vector<pair<double, double> > &P) {
vector<pair<double, double> > C;
C.clear();
int n = P.size();
for (int i = 0; i < n; ++i) {
int j = (i + 2) % n;
pair<double, double> u = {P[i].first - P[(i + 1) % n].first,
P[i].second - P[(i + 1) % n].second};
pair<double, double> v = {P[j].first - P[(i + 1) % n].first,
P[j].second - P[(i + 1) % n].second};
C.push_back({len(u), ang(u, v)});
}
P = C;
}
int n, m;
vector<pair<double, double> > A, B;
int KMP[N];
int main() {
A.resize(read());
B.resize(read());
for (int i = 0; i < A.size(); ++i) A[i].first = read(), A[i].second = read();
for (int i = 0; i < B.size(); ++i) B[i].first = read(), B[i].second = read();
ConvexHull(A);
TransForm(A);
ConvexHull(B);
TransForm(B);
if (A.size() != B.size()) return printf("NO"), 0;
int n = A.size();
for (int i = 0; i < n; ++i) A.push_back(A[i]);
KMP[0] = -1;
for (int i = 1; i < n; ++i) {
int j = KMP[i - 1];
while (B[j + 1] != B[i]) {
if (j >= 0)
j = KMP[j];
else {
j = -2;
break;
}
}
KMP[i] = j + 1;
}
int len = -1;
for (int i = 0; i < A.size(); ++i) {
if (len == n - 1) {
printf("YES");
exit(0);
}
if (A[i] == B[len + 1])
++len;
else if (len >= 0)
len = KMP[len];
}
printf("NO");
}
``` |
#include <bits/stdc++.h>
using namespace std;
using namespace std;
struct pt {
long long x, y;
};
bool cmp(pt a, pt b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
bool cw(pt a, pt b, pt c) {
return a.x * (b.y - c.y) + b.x * (c.y - a.y) + c.x * (a.y - b.y) < 0;
}
bool ccw(pt a, pt b, pt c) {
return a.x * (b.y - c.y) + b.x * (c.y - a.y) + c.x * (a.y - b.y) > 0;
}
void convex_hull(vector<pt>& a) {
if (a.size() == 1) return;
sort(a.begin(), a.end(), &cmp);
pt p1 = a[0], p2 = a.back();
vector<pt> up, down;
up.push_back(p1);
down.push_back(p1);
for (size_t i = 1; i < a.size(); ++i) {
if (i == a.size() - 1 || cw(p1, a[i], p2)) {
while (up.size() >= 2 && !cw(up[up.size() - 2], up[up.size() - 1], a[i]))
up.pop_back();
up.push_back(a[i]);
}
if (i == a.size() - 1 || ccw(p1, a[i], p2)) {
while (down.size() >= 2 &&
!ccw(down[down.size() - 2], down[down.size() - 1], a[i]))
down.pop_back();
down.push_back(a[i]);
}
}
a.clear();
for (size_t i = 0; i < up.size(); ++i) a.push_back(up[i]);
for (size_t i = down.size() - 2; i > 0; --i) a.push_back(down[i]);
}
long long d2(pt a, pt b) {
long long dx = a.x - b.x;
long long dy = a.y - b.y;
return dx * dx + dy * dy;
}
long long t2(pt a, pt b, pt c) {
return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
}
pair<long long, long long> mk(pt a, pt b, pt c) {
auto d = d2(a, b);
auto t = t2(a, b, c);
return make_pair(d, t);
}
int pi[3 * 100000 + 7];
bool check(vector<pt>& f, vector<pt>& s) {
if (f.size() != s.size()) return false;
vector<pair<long long, long long>> af;
vector<pair<long long, long long>> bf;
int n = f.size();
for (int i = 2; i < n; ++i) {
af.push_back(mk(f[i], f[i - 1], f[i - 2]));
bf.push_back(mk(s[i], s[i - 1], s[i - 2]));
}
af.push_back(mk(f[0], f[n - 1], f[n - 2]));
af.push_back(mk(f[1], f[0], f[n - 1]));
bf.push_back(mk(s[0], s[n - 1], s[n - 2]));
bf.push_back(mk(s[1], s[0], s[n - 1]));
bf.push_back(make_pair(-1, -1));
for (int u = 0; u < 2; ++u)
for (int i = 0; i < n; ++i) bf.push_back(af[i]);
pi[0] = 0;
for (int i = 1; i < bf.size(); ++i) {
int j = pi[i - 1];
while (j > 0 && bf[i] != bf[j]) j = pi[j - 1];
if (bf[i] == bf[j]) ++j;
pi[i] = j;
if (pi[i] >= n) return true;
}
return false;
}
class TaskE {
public:
void solve(std::istream& in, std::ostream& out) {
int n, m;
in >> n >> m;
vector<pt> f(n);
vector<pt> s(m);
for (int i = 0; i < n; ++i) {
in >> f[i].x >> f[i].y;
}
for (int i = 0; i < m; ++i) {
in >> s[i].x >> s[i].y;
}
convex_hull(f);
convex_hull(s);
if (check(f, s))
out << "YES\n";
else
out << "NO\n";
}
};
int main() {
TaskE solver;
std::istream& in(std::cin);
std::ostream& out(std::cout);
solver.solve(in, out);
return 0;
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
using namespace std;
struct pt {
long long x, y;
};
bool cmp(pt a, pt b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
bool cw(pt a, pt b, pt c) {
return a.x * (b.y - c.y) + b.x * (c.y - a.y) + c.x * (a.y - b.y) < 0;
}
bool ccw(pt a, pt b, pt c) {
return a.x * (b.y - c.y) + b.x * (c.y - a.y) + c.x * (a.y - b.y) > 0;
}
void convex_hull(vector<pt>& a) {
if (a.size() == 1) return;
sort(a.begin(), a.end(), &cmp);
pt p1 = a[0], p2 = a.back();
vector<pt> up, down;
up.push_back(p1);
down.push_back(p1);
for (size_t i = 1; i < a.size(); ++i) {
if (i == a.size() - 1 || cw(p1, a[i], p2)) {
while (up.size() >= 2 && !cw(up[up.size() - 2], up[up.size() - 1], a[i]))
up.pop_back();
up.push_back(a[i]);
}
if (i == a.size() - 1 || ccw(p1, a[i], p2)) {
while (down.size() >= 2 &&
!ccw(down[down.size() - 2], down[down.size() - 1], a[i]))
down.pop_back();
down.push_back(a[i]);
}
}
a.clear();
for (size_t i = 0; i < up.size(); ++i) a.push_back(up[i]);
for (size_t i = down.size() - 2; i > 0; --i) a.push_back(down[i]);
}
long long d2(pt a, pt b) {
long long dx = a.x - b.x;
long long dy = a.y - b.y;
return dx * dx + dy * dy;
}
long long t2(pt a, pt b, pt c) {
return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
}
pair<long long, long long> mk(pt a, pt b, pt c) {
auto d = d2(a, b);
auto t = t2(a, b, c);
return make_pair(d, t);
}
int pi[3 * 100000 + 7];
bool check(vector<pt>& f, vector<pt>& s) {
if (f.size() != s.size()) return false;
vector<pair<long long, long long>> af;
vector<pair<long long, long long>> bf;
int n = f.size();
for (int i = 2; i < n; ++i) {
af.push_back(mk(f[i], f[i - 1], f[i - 2]));
bf.push_back(mk(s[i], s[i - 1], s[i - 2]));
}
af.push_back(mk(f[0], f[n - 1], f[n - 2]));
af.push_back(mk(f[1], f[0], f[n - 1]));
bf.push_back(mk(s[0], s[n - 1], s[n - 2]));
bf.push_back(mk(s[1], s[0], s[n - 1]));
bf.push_back(make_pair(-1, -1));
for (int u = 0; u < 2; ++u)
for (int i = 0; i < n; ++i) bf.push_back(af[i]);
pi[0] = 0;
for (int i = 1; i < bf.size(); ++i) {
int j = pi[i - 1];
while (j > 0 && bf[i] != bf[j]) j = pi[j - 1];
if (bf[i] == bf[j]) ++j;
pi[i] = j;
if (pi[i] >= n) return true;
}
return false;
}
class TaskE {
public:
void solve(std::istream& in, std::ostream& out) {
int n, m;
in >> n >> m;
vector<pt> f(n);
vector<pt> s(m);
for (int i = 0; i < n; ++i) {
in >> f[i].x >> f[i].y;
}
for (int i = 0; i < m; ++i) {
in >> s[i].x >> s[i].y;
}
convex_hull(f);
convex_hull(s);
if (check(f, s))
out << "YES\n";
else
out << "NO\n";
}
};
int main() {
TaskE solver;
std::istream& in(std::cin);
std::ostream& out(std::cout);
solver.solve(in, out);
return 0;
}
``` |
#include <bits/stdc++.h>
const int moder = 998244353;
int powermod(int a, int exp) {
int ret = 1;
for (; exp > 0; exp >>= 1) {
if (exp & 1) {
ret = 1ll * ret * a % moder;
}
a = 1ll * a * a % moder;
}
return ret;
}
struct KMP {
int m;
std::pair<long long, int> p[(1000010)];
int fail[(1000010)];
void makeFail(std::pair<long long, int> *p, int mm) {
memset(this->p, 0, sizeof(this->p));
m = mm;
for (int i = 0; i < m; ++i) {
this->p[i] = p[i];
}
fail[0] = -1;
for (int i = 1, j = -1; i <= m; ++i) {
while (j >= 0 && p[j] != p[i - 1]) j = fail[j];
fail[i] = ++j;
}
}
bool searchForm(std::pair<long long, int> *t, int n) {
for (int i = 0, j = 0; i < n; ++i) {
while (j >= 0 && p[j] != t[i]) j = fail[j];
if (++j == m) {
return true;
}
}
return false;
}
} kmp;
struct P {
int x, y;
P(int x = 0, int y = 0) : x(x), y(y) {}
P operator+(const P &p) const { return P(x + p.x, y + p.y); }
P operator-(const P &p) const { return P(x - p.x, y - p.y); }
long long operator^(const P &p) const {
return 1ll * x * p.y - 1ll * y * p.x;
}
long long operator%(const P &p) const {
return 1ll * x * p.x + 1ll * y * p.y;
}
long long abs2() const { return *this % *this; }
bool operator<(const P &p) const { return x != p.x ? x < p.x : y < p.y; }
void read() { scanf("%d%d", &x, &y); }
};
struct L {
P p, v;
L() {}
L(P a, P b) : p(a), v(b - a) {}
};
long long onLeft(L l, P p) { return l.v ^ (p - l.p); }
const int N = 1000010;
std::vector<P> convexHull(std::vector<P> &ps) {
static P stack[N];
std::sort(ps.begin(), ps.end());
int n = ps.size(), top = 0;
for (int i = 0; i < n; ++i) {
while (top > 1 && onLeft(L(stack[top - 2], stack[top - 1]), ps[i]) <= 0) {
--top;
}
stack[top++] = ps[i];
}
int tmp = top;
for (int i = n - 2; i >= 0; --i) {
while (top > tmp && onLeft(L(stack[top - 2], stack[top - 1]), ps[i]) <= 0) {
--top;
}
stack[top++] = ps[i];
}
if (n > 1) --top;
std::vector<P> ret;
for (int i = 0; i < top; ++i) {
ret.push_back(stack[i]);
}
return ret;
}
std::pair<long long, int> s[N], t[N];
int main() {
int n, m;
scanf("%d%d", &n, &m);
std::vector<P> ps, qs;
for (int i = 0; i < n; ++i) {
P p;
p.read();
ps.push_back(p);
}
for (int i = 0; i < m; ++i) {
P p;
p.read();
qs.push_back(p);
}
ps = convexHull(ps);
qs = convexHull(qs);
if (ps.size() != qs.size()) {
puts("NO");
return 0;
}
n = ps.size();
for (int i = 0; i < n; ++i) {
long long x = (ps[i] - ps[(i + 1) % n]).abs2();
long long y = (qs[i] - qs[(i + 1) % n]).abs2();
int u = ((ps[(i + 1) % n] - ps[i]) % (ps[(i + 1) % n] - ps[(i + 2) % n])) %
moder;
u = 1ll * u * u % moder;
u = 1ll * u * powermod(x, moder - 2) % moder *
powermod((ps[(i + 1) % n] - ps[(i + 2) % n]).abs2(), moder - 2) % moder;
int v = ((qs[(i + 1) % n] - qs[i]) % (qs[(i + 1) % n] - qs[(i + 2) % n])) %
moder;
v = 1ll * v * v % moder;
v = 1ll * v * powermod(y, moder - 2) % moder *
powermod((qs[(i + 1) % n] - qs[(i + 2) % n]).abs2(), moder - 2) % moder;
s[i] = {x, u};
t[i] = {y, v};
}
for (int i = n; i < n << 1; ++i) {
s[i] = s[i - n];
}
kmp.makeFail(t, n);
if (kmp.searchForm(s, n << 1)) {
puts("YES");
} else {
puts("NO");
}
return 0;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
const int moder = 998244353;
int powermod(int a, int exp) {
int ret = 1;
for (; exp > 0; exp >>= 1) {
if (exp & 1) {
ret = 1ll * ret * a % moder;
}
a = 1ll * a * a % moder;
}
return ret;
}
struct KMP {
int m;
std::pair<long long, int> p[(1000010)];
int fail[(1000010)];
void makeFail(std::pair<long long, int> *p, int mm) {
memset(this->p, 0, sizeof(this->p));
m = mm;
for (int i = 0; i < m; ++i) {
this->p[i] = p[i];
}
fail[0] = -1;
for (int i = 1, j = -1; i <= m; ++i) {
while (j >= 0 && p[j] != p[i - 1]) j = fail[j];
fail[i] = ++j;
}
}
bool searchForm(std::pair<long long, int> *t, int n) {
for (int i = 0, j = 0; i < n; ++i) {
while (j >= 0 && p[j] != t[i]) j = fail[j];
if (++j == m) {
return true;
}
}
return false;
}
} kmp;
struct P {
int x, y;
P(int x = 0, int y = 0) : x(x), y(y) {}
P operator+(const P &p) const { return P(x + p.x, y + p.y); }
P operator-(const P &p) const { return P(x - p.x, y - p.y); }
long long operator^(const P &p) const {
return 1ll * x * p.y - 1ll * y * p.x;
}
long long operator%(const P &p) const {
return 1ll * x * p.x + 1ll * y * p.y;
}
long long abs2() const { return *this % *this; }
bool operator<(const P &p) const { return x != p.x ? x < p.x : y < p.y; }
void read() { scanf("%d%d", &x, &y); }
};
struct L {
P p, v;
L() {}
L(P a, P b) : p(a), v(b - a) {}
};
long long onLeft(L l, P p) { return l.v ^ (p - l.p); }
const int N = 1000010;
std::vector<P> convexHull(std::vector<P> &ps) {
static P stack[N];
std::sort(ps.begin(), ps.end());
int n = ps.size(), top = 0;
for (int i = 0; i < n; ++i) {
while (top > 1 && onLeft(L(stack[top - 2], stack[top - 1]), ps[i]) <= 0) {
--top;
}
stack[top++] = ps[i];
}
int tmp = top;
for (int i = n - 2; i >= 0; --i) {
while (top > tmp && onLeft(L(stack[top - 2], stack[top - 1]), ps[i]) <= 0) {
--top;
}
stack[top++] = ps[i];
}
if (n > 1) --top;
std::vector<P> ret;
for (int i = 0; i < top; ++i) {
ret.push_back(stack[i]);
}
return ret;
}
std::pair<long long, int> s[N], t[N];
int main() {
int n, m;
scanf("%d%d", &n, &m);
std::vector<P> ps, qs;
for (int i = 0; i < n; ++i) {
P p;
p.read();
ps.push_back(p);
}
for (int i = 0; i < m; ++i) {
P p;
p.read();
qs.push_back(p);
}
ps = convexHull(ps);
qs = convexHull(qs);
if (ps.size() != qs.size()) {
puts("NO");
return 0;
}
n = ps.size();
for (int i = 0; i < n; ++i) {
long long x = (ps[i] - ps[(i + 1) % n]).abs2();
long long y = (qs[i] - qs[(i + 1) % n]).abs2();
int u = ((ps[(i + 1) % n] - ps[i]) % (ps[(i + 1) % n] - ps[(i + 2) % n])) %
moder;
u = 1ll * u * u % moder;
u = 1ll * u * powermod(x, moder - 2) % moder *
powermod((ps[(i + 1) % n] - ps[(i + 2) % n]).abs2(), moder - 2) % moder;
int v = ((qs[(i + 1) % n] - qs[i]) % (qs[(i + 1) % n] - qs[(i + 2) % n])) %
moder;
v = 1ll * v * v % moder;
v = 1ll * v * powermod(y, moder - 2) % moder *
powermod((qs[(i + 1) % n] - qs[(i + 2) % n]).abs2(), moder - 2) % moder;
s[i] = {x, u};
t[i] = {y, v};
}
for (int i = n; i < n << 1; ++i) {
s[i] = s[i - n];
}
kmp.makeFail(t, n);
if (kmp.searchForm(s, n << 1)) {
puts("YES");
} else {
puts("NO");
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long add(long long a, long long b) { return a + b; }
struct P {
long long x, y;
P(long long x = 0, long long y = 0) : x(x), y(y) {}
P operator+(const P &b) const { return P(add(x, b.x), add(y, b.y)); }
P operator-(const P &b) const { return P(add(x, -b.x), add(y, -b.y)); }
P operator*(long long d) const { return P(x * d, y * d); }
long long operator*(const P &b) const { return add(x * b.x, y * b.y); }
long long operator^(const P &b) const { return add(x * b.y, -y * b.x); }
bool operator<(const P &b) const {
if (x != b.x) return x < b.x;
return y < b.y;
}
bool operator==(const P &b) const { return x == b.x && y == b.y; }
};
vector<P> convex_hull(P *ps, int n) {
sort(ps, ps + n);
int k = 0;
vector<P> qs(2 * n);
for (int i = 0; i < n; i++) {
while (k > 1 && ((qs[k - 2] - qs[k - 1]) ^ (ps[i] - qs[k - 1])) >= 0) --k;
qs[k++] = ps[i];
}
for (int i = n - 2, t = k; i >= 0; i--) {
while (k > t && ((qs[k - 2] - qs[k - 1]) ^ (ps[i] - qs[k - 1])) >= 0) --k;
qs[k++] = ps[i];
}
qs.resize(k - 1);
return qs;
}
long long dist(P p, P q) { return (p - q) * (p - q); }
const int N = 2e5 + 10;
P a[N], b[2][N];
int len[2];
void calc(vector<P> cv, int ci) {
len[ci] = cv.size();
cv.push_back(cv[0]);
cv.push_back(cv[1]);
for (int i = 0; i < len[ci]; i++) {
b[ci][i] = P(dist(cv[i], cv[i + 1]),
(cv[i + 1] - cv[i]) * (cv[i + 2] - cv[i + 1]));
}
}
int getmin(P *s, int n) {
for (int i = 0; i < n; i++) s[i + n] = s[i];
int i = 0, j = 1, k = 0;
while (i < n && j < n && k < n) {
if (s[i + k] == s[j + k])
k++;
else {
if (s[j + k] < s[i + k])
i += k + 1;
else
j += k + 1;
if (i == j) j++;
k = 0;
}
}
return min(i, j);
}
int main() {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++) {
int x, y;
scanf("%d%d", &x, &y);
a[i] = P(x, y);
}
calc(convex_hull(a, n), 0);
for (int i = 0; i < m; i++) {
int x, y;
scanf("%d%d", &x, &y);
a[i] = P(x, y);
}
calc(convex_hull(a, m), 1);
if (len[0] != len[1])
puts("NO");
else {
int x = getmin(b[0], len[0]), y = getmin(b[1], len[1]);
for (int i = 0; i < n; i++) {
if (!(b[0][x + i] == b[1][y + i])) {
puts("NO");
return 0;
}
}
puts("YES");
}
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long add(long long a, long long b) { return a + b; }
struct P {
long long x, y;
P(long long x = 0, long long y = 0) : x(x), y(y) {}
P operator+(const P &b) const { return P(add(x, b.x), add(y, b.y)); }
P operator-(const P &b) const { return P(add(x, -b.x), add(y, -b.y)); }
P operator*(long long d) const { return P(x * d, y * d); }
long long operator*(const P &b) const { return add(x * b.x, y * b.y); }
long long operator^(const P &b) const { return add(x * b.y, -y * b.x); }
bool operator<(const P &b) const {
if (x != b.x) return x < b.x;
return y < b.y;
}
bool operator==(const P &b) const { return x == b.x && y == b.y; }
};
vector<P> convex_hull(P *ps, int n) {
sort(ps, ps + n);
int k = 0;
vector<P> qs(2 * n);
for (int i = 0; i < n; i++) {
while (k > 1 && ((qs[k - 2] - qs[k - 1]) ^ (ps[i] - qs[k - 1])) >= 0) --k;
qs[k++] = ps[i];
}
for (int i = n - 2, t = k; i >= 0; i--) {
while (k > t && ((qs[k - 2] - qs[k - 1]) ^ (ps[i] - qs[k - 1])) >= 0) --k;
qs[k++] = ps[i];
}
qs.resize(k - 1);
return qs;
}
long long dist(P p, P q) { return (p - q) * (p - q); }
const int N = 2e5 + 10;
P a[N], b[2][N];
int len[2];
void calc(vector<P> cv, int ci) {
len[ci] = cv.size();
cv.push_back(cv[0]);
cv.push_back(cv[1]);
for (int i = 0; i < len[ci]; i++) {
b[ci][i] = P(dist(cv[i], cv[i + 1]),
(cv[i + 1] - cv[i]) * (cv[i + 2] - cv[i + 1]));
}
}
int getmin(P *s, int n) {
for (int i = 0; i < n; i++) s[i + n] = s[i];
int i = 0, j = 1, k = 0;
while (i < n && j < n && k < n) {
if (s[i + k] == s[j + k])
k++;
else {
if (s[j + k] < s[i + k])
i += k + 1;
else
j += k + 1;
if (i == j) j++;
k = 0;
}
}
return min(i, j);
}
int main() {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++) {
int x, y;
scanf("%d%d", &x, &y);
a[i] = P(x, y);
}
calc(convex_hull(a, n), 0);
for (int i = 0; i < m; i++) {
int x, y;
scanf("%d%d", &x, &y);
a[i] = P(x, y);
}
calc(convex_hull(a, m), 1);
if (len[0] != len[1])
puts("NO");
else {
int x = getmin(b[0], len[0]), y = getmin(b[1], len[1]);
for (int i = 0; i < n; i++) {
if (!(b[0][x + i] == b[1][y + i])) {
puts("NO");
return 0;
}
}
puts("YES");
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
struct Point {
long long first, second;
};
Point p0;
Point nextToTop(stack<Point> &S) {
Point p = S.top();
S.pop();
Point res = S.top();
S.push(p);
return res;
}
int distSq(Point p1, Point p2) {
return (p1.first - p2.first) * (p1.first - p2.first) +
(p1.second - p2.second) * (p1.second - p2.second);
}
int orientation(Point p, Point q, Point r) {
long long val = (q.second - p.second) * (r.first - q.first);
val -= (q.first - p.first) * (r.second - q.second);
if (val == 0) return 0;
return (val > 0) ? 1 : 2;
}
long long orientation1(Point p, Point q, Point r) {
long long val = (q.first - p.first) * (r.first - q.first);
val += (q.second - p.second) * (r.second - q.second);
return val;
}
bool compare(Point p1, Point p2) {
long long o = orientation(p0, p1, p2);
if (o == 0) return (distSq(p0, p2) >= distSq(p0, p1)) ? true : false;
return (o == 2) ? true : false;
}
bool equal(Point p1, Point p2) {
return (p1.first == p2.first && p1.second == p2.second);
}
vector<Point> convexHull(vector<Point> a) {
long long n = a.size();
p0 = {INT_MAX, INT_MAX};
for (int i = 0; i < n; i++) {
if (p0.second > a[i].second) {
p0 = a[i];
} else if (p0.second == a[i].second) {
p0.first = min(p0.first, a[i].first);
}
}
stack<Point> s;
s.push(p0);
vector<Point> v;
for (int i = 0; i < n; i++) {
if (!equal(a[i], p0)) v.push_back(a[i]);
}
sort(v.begin(), v.end(), compare);
long long pos = 0;
n = v.size();
for (int i = 0; i < n; i++) {
while (i < n - 1 && orientation(p0, v[i], v[i + 1]) == 0) i++;
v[pos++] = v[i];
}
if (pos < 2) {
vector<Point> t;
t.push_back(p0);
t.push_back(v[0]);
return t;
}
s.push(v[0]);
s.push(v[1]);
for (int i = 2; i < pos; i++) {
while (orientation(nextToTop(s), s.top(), v[i]) != 2) s.pop();
s.push(v[i]);
}
vector<Point> t;
while (!s.empty()) {
Point p = s.top();
t.push_back(p);
s.pop();
}
return t;
}
vector<long long> reduce(vector<Point> a) {
int n = a.size();
a.push_back(a[0]);
a.push_back(a[1]);
vector<long long> ans;
for (int i = 1; i < (int)a.size() - 1; i++) {
ans.push_back(distSq(a[i], a[i - 1]));
ans.push_back(orientation1(a[i - 1], a[i], a[i + 1]));
}
return ans;
}
int kmp(vector<long long> &s, vector<long long> &pat) {
int n = pat.size();
long long lps[n];
lps[0] = 0;
int pos = 1;
int len = 0;
while (pos < n) {
if (pat[pos] == pat[len]) {
lps[pos] = len + 1;
pos++;
len++;
} else {
if (len == 0) {
lps[pos] = 0;
pos++;
} else {
len = lps[len - 1];
}
}
}
int m = s.size();
len = 0;
for (int i = 0; i < m; i++) {
if (s[i] == pat[len]) {
len++;
if (len == n) return i;
} else {
if (len != 0) {
i--;
len = lps[len - 1];
}
}
}
return -1;
}
bool sameConvexHull(vector<Point> a, vector<Point> b) {
if (a.size() != b.size()) return false;
vector<long long> first = reduce(a);
vector<long long> second = reduce(b);
int l = first.size();
for (int i = 0; i < l; i++) {
first.push_back(first[i]);
}
return kmp(first, second) == -1 ? false : true;
}
int main() {
std::ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n, m;
cin >> n >> m;
vector<Point> a, b;
for (int i = 0; i < n; i++) {
Point t;
cin >> t.first >> t.second;
a.push_back(t);
}
for (int i = 0; i < m; i++) {
Point t;
cin >> t.first >> t.second;
b.push_back(t);
}
a = convexHull(a);
b = convexHull(b);
if (sameConvexHull(a, b)) {
cout << "YES\n";
} else {
cout << "NO\n";
}
return 0;
}
| ### Prompt
Develop a solution in cpp to the problem described below:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
struct Point {
long long first, second;
};
Point p0;
Point nextToTop(stack<Point> &S) {
Point p = S.top();
S.pop();
Point res = S.top();
S.push(p);
return res;
}
int distSq(Point p1, Point p2) {
return (p1.first - p2.first) * (p1.first - p2.first) +
(p1.second - p2.second) * (p1.second - p2.second);
}
int orientation(Point p, Point q, Point r) {
long long val = (q.second - p.second) * (r.first - q.first);
val -= (q.first - p.first) * (r.second - q.second);
if (val == 0) return 0;
return (val > 0) ? 1 : 2;
}
long long orientation1(Point p, Point q, Point r) {
long long val = (q.first - p.first) * (r.first - q.first);
val += (q.second - p.second) * (r.second - q.second);
return val;
}
bool compare(Point p1, Point p2) {
long long o = orientation(p0, p1, p2);
if (o == 0) return (distSq(p0, p2) >= distSq(p0, p1)) ? true : false;
return (o == 2) ? true : false;
}
bool equal(Point p1, Point p2) {
return (p1.first == p2.first && p1.second == p2.second);
}
vector<Point> convexHull(vector<Point> a) {
long long n = a.size();
p0 = {INT_MAX, INT_MAX};
for (int i = 0; i < n; i++) {
if (p0.second > a[i].second) {
p0 = a[i];
} else if (p0.second == a[i].second) {
p0.first = min(p0.first, a[i].first);
}
}
stack<Point> s;
s.push(p0);
vector<Point> v;
for (int i = 0; i < n; i++) {
if (!equal(a[i], p0)) v.push_back(a[i]);
}
sort(v.begin(), v.end(), compare);
long long pos = 0;
n = v.size();
for (int i = 0; i < n; i++) {
while (i < n - 1 && orientation(p0, v[i], v[i + 1]) == 0) i++;
v[pos++] = v[i];
}
if (pos < 2) {
vector<Point> t;
t.push_back(p0);
t.push_back(v[0]);
return t;
}
s.push(v[0]);
s.push(v[1]);
for (int i = 2; i < pos; i++) {
while (orientation(nextToTop(s), s.top(), v[i]) != 2) s.pop();
s.push(v[i]);
}
vector<Point> t;
while (!s.empty()) {
Point p = s.top();
t.push_back(p);
s.pop();
}
return t;
}
vector<long long> reduce(vector<Point> a) {
int n = a.size();
a.push_back(a[0]);
a.push_back(a[1]);
vector<long long> ans;
for (int i = 1; i < (int)a.size() - 1; i++) {
ans.push_back(distSq(a[i], a[i - 1]));
ans.push_back(orientation1(a[i - 1], a[i], a[i + 1]));
}
return ans;
}
int kmp(vector<long long> &s, vector<long long> &pat) {
int n = pat.size();
long long lps[n];
lps[0] = 0;
int pos = 1;
int len = 0;
while (pos < n) {
if (pat[pos] == pat[len]) {
lps[pos] = len + 1;
pos++;
len++;
} else {
if (len == 0) {
lps[pos] = 0;
pos++;
} else {
len = lps[len - 1];
}
}
}
int m = s.size();
len = 0;
for (int i = 0; i < m; i++) {
if (s[i] == pat[len]) {
len++;
if (len == n) return i;
} else {
if (len != 0) {
i--;
len = lps[len - 1];
}
}
}
return -1;
}
bool sameConvexHull(vector<Point> a, vector<Point> b) {
if (a.size() != b.size()) return false;
vector<long long> first = reduce(a);
vector<long long> second = reduce(b);
int l = first.size();
for (int i = 0; i < l; i++) {
first.push_back(first[i]);
}
return kmp(first, second) == -1 ? false : true;
}
int main() {
std::ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n, m;
cin >> n >> m;
vector<Point> a, b;
for (int i = 0; i < n; i++) {
Point t;
cin >> t.first >> t.second;
a.push_back(t);
}
for (int i = 0; i < m; i++) {
Point t;
cin >> t.first >> t.second;
b.push_back(t);
}
a = convexHull(a);
b = convexHull(b);
if (sameConvexHull(a, b)) {
cout << "YES\n";
} else {
cout << "NO\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
struct point {
long long x, y;
point(long long x = 0, long long y = 0) : x(x), y(y) {}
point operator+(point p) { return point(x + p.x, y + p.y); }
point operator-(point p) { return point(x - p.x, y - p.y); }
point operator*(long long k) { return point(x * k, y * k); }
long long operator*(point p) { return x * p.y - y * p.x; }
long long operator^(point p) { return x * p.x + y * p.y; }
long long dist(point p) {
return (p.x - x) * (p.x - x) + (p.y - y) * (p.y - y);
}
point operator/(long long k) { return point(x / k, y / k); }
};
point c;
long long f(long long a) {
if (a < 0) {
return -1;
}
if (a > 0) {
return 1;
}
if (a == 0) {
return 0;
}
}
bool cmp(point a, point b) {
return ((a - c) * (b - c)) > 0 ||
((a - c) * (b - c) == 0 && c.dist(a) < c.dist(b));
}
void read(int n, vector<point> &st) {
vector<point> vec(n);
for (int i = 0; i < n; ++i) {
cin >> vec[i].x >> vec[i].y;
}
point ans = vec[0];
for (int i = 0; i < vec.size(); ++i) {
if (vec[i].x < ans.x || (vec[i].x == ans.x && vec[i].y < ans.y)) {
ans = vec[i];
swap(vec[i], vec[0]);
}
}
c = ans;
sort(vec.begin() + 1, vec.end(), cmp);
st.push_back(vec[0]);
st.push_back(vec[1]);
for (int i = 2; i < n; ++i) {
while (st.size() > 1 && (st[st.size() - 1] - st[st.size() - 2]) *
(vec[i] - st[st.size() - 1]) <=
0) {
st.pop_back();
}
st.push_back(vec[i]);
}
}
vector<pair<long long, pair<long long, long long> > > p1, p2;
bool pf() {
auto pp = p1;
pp.push_back({-1000000000000000000, {1000000000000000000, 1313241341252345}});
for (auto i : p2) {
pp.push_back(i);
}
for (auto i : p2) {
pp.push_back(i);
}
vector<int> p(pp.size());
for (int i = 1; i < p.size(); ++i) {
p[i] = p[i - 1];
while (p[i] > 0 && pp[i] != pp[p[i]]) {
p[i] = p[p[i] - 1];
}
if (pp[i] == pp[p[i]]) {
++p[i];
}
if (p[i] == p1.size()) {
return 1;
}
}
return 0;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int n, m;
cin >> n >> m;
vector<point> vec1, vec2;
read(n, vec1);
read(m, vec2);
if (vec1.size() != vec2.size()) {
cout << "NO" << endl;
return 0;
}
n = vec1.size();
p1.resize(n);
p2.resize(n);
for (int i = 0; i < vec1.size(); ++i) {
vec1[i] = vec1[i] * n;
vec2[i] = vec2[i] * n;
}
point c1(0, 0), c2(0, 0);
for (int i = 0; i < n; ++i) {
c1 = c1 + vec1[i];
c2 = c2 + vec2[i];
}
c1 = c1 / n;
c2 = c2 / n;
for (int i = 0; i < n; ++i) {
int j = (i + 1) % n;
p1[i] = {
vec1[i].dist(c1),
{(vec1[i] - c1) * (vec1[j] - c1), (vec1[i] - c1) ^ (vec1[j] - c1)}};
p2[i] = {
vec2[i].dist(c2),
{(vec2[i] - c2) * (vec2[j] - c2), (vec2[i] - c2) ^ (vec2[j] - c2)}};
}
if (pf()) {
cout << "YES" << endl;
} else {
cout << "NO" << endl;
}
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
struct point {
long long x, y;
point(long long x = 0, long long y = 0) : x(x), y(y) {}
point operator+(point p) { return point(x + p.x, y + p.y); }
point operator-(point p) { return point(x - p.x, y - p.y); }
point operator*(long long k) { return point(x * k, y * k); }
long long operator*(point p) { return x * p.y - y * p.x; }
long long operator^(point p) { return x * p.x + y * p.y; }
long long dist(point p) {
return (p.x - x) * (p.x - x) + (p.y - y) * (p.y - y);
}
point operator/(long long k) { return point(x / k, y / k); }
};
point c;
long long f(long long a) {
if (a < 0) {
return -1;
}
if (a > 0) {
return 1;
}
if (a == 0) {
return 0;
}
}
bool cmp(point a, point b) {
return ((a - c) * (b - c)) > 0 ||
((a - c) * (b - c) == 0 && c.dist(a) < c.dist(b));
}
void read(int n, vector<point> &st) {
vector<point> vec(n);
for (int i = 0; i < n; ++i) {
cin >> vec[i].x >> vec[i].y;
}
point ans = vec[0];
for (int i = 0; i < vec.size(); ++i) {
if (vec[i].x < ans.x || (vec[i].x == ans.x && vec[i].y < ans.y)) {
ans = vec[i];
swap(vec[i], vec[0]);
}
}
c = ans;
sort(vec.begin() + 1, vec.end(), cmp);
st.push_back(vec[0]);
st.push_back(vec[1]);
for (int i = 2; i < n; ++i) {
while (st.size() > 1 && (st[st.size() - 1] - st[st.size() - 2]) *
(vec[i] - st[st.size() - 1]) <=
0) {
st.pop_back();
}
st.push_back(vec[i]);
}
}
vector<pair<long long, pair<long long, long long> > > p1, p2;
bool pf() {
auto pp = p1;
pp.push_back({-1000000000000000000, {1000000000000000000, 1313241341252345}});
for (auto i : p2) {
pp.push_back(i);
}
for (auto i : p2) {
pp.push_back(i);
}
vector<int> p(pp.size());
for (int i = 1; i < p.size(); ++i) {
p[i] = p[i - 1];
while (p[i] > 0 && pp[i] != pp[p[i]]) {
p[i] = p[p[i] - 1];
}
if (pp[i] == pp[p[i]]) {
++p[i];
}
if (p[i] == p1.size()) {
return 1;
}
}
return 0;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int n, m;
cin >> n >> m;
vector<point> vec1, vec2;
read(n, vec1);
read(m, vec2);
if (vec1.size() != vec2.size()) {
cout << "NO" << endl;
return 0;
}
n = vec1.size();
p1.resize(n);
p2.resize(n);
for (int i = 0; i < vec1.size(); ++i) {
vec1[i] = vec1[i] * n;
vec2[i] = vec2[i] * n;
}
point c1(0, 0), c2(0, 0);
for (int i = 0; i < n; ++i) {
c1 = c1 + vec1[i];
c2 = c2 + vec2[i];
}
c1 = c1 / n;
c2 = c2 / n;
for (int i = 0; i < n; ++i) {
int j = (i + 1) % n;
p1[i] = {
vec1[i].dist(c1),
{(vec1[i] - c1) * (vec1[j] - c1), (vec1[i] - c1) ^ (vec1[j] - c1)}};
p2[i] = {
vec2[i].dist(c2),
{(vec2[i] - c2) * (vec2[j] - c2), (vec2[i] - c2) ^ (vec2[j] - c2)}};
}
if (pf()) {
cout << "YES" << endl;
} else {
cout << "NO" << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
const long double eps = 1e-9;
const long double pi = acos(-1.0);
struct vec {
long double x, y;
vec operator+(const vec &k) const { return {x + k.x, y + k.y}; }
vec operator-(const vec &k) const { return {x - k.x, y - k.y}; }
vec operator*(const long double &k) const { return {x * k, y * k}; }
long double dot(const vec &k) const { return x * k.x + y * k.y; }
long double cross(const vec &k) const { return x * k.y - y * k.x; }
long double len() const { return sqrt(x * x + y * y); }
long double angle() const { return atan2(y, x); }
bool operator<(const vec &k) const {
return x < k.x || (x == k.x && y < k.y);
}
void read() {
double u, v;
scanf("%lf%lf", &u, &v);
x = u;
y = v;
}
void print() { printf("%.8f %.8f\n", (double)x, (double)y); }
};
template <class C>
void convexHull(C p, C &q) {
std::sort(p.begin(), p.end());
q.clear();
for (int i = 0; i < p.size(); ++i) {
while (q.size() >= 2 &&
(q[q.size() - 1] - q[q.size() - 2]).cross(p[i] - q[q.size() - 1]) <=
eps) {
q.pop_back();
}
q.push_back(p[i]);
}
int protect = q.size() - 1;
for (int i = p.size() - 2; i >= 0; --i) {
while (q.size() - protect >= 2 &&
(q[q.size() - 1] - q[q.size() - 2]).cross(p[i] - q[q.size() - 1]) <=
eps) {
q.pop_back();
}
q.push_back(p[i]);
}
q.pop_back();
}
std::vector<int> lcs;
template <class C>
void kmpInit(const C &t) {
lcs.resize(t.size());
lcs.clear();
lcs[0] = 0;
for (int i = 1; i < t.size(); ++i) {
int k = lcs[i - 1];
while (k > 0 && fabs(t[k] - t[i]) > eps) k = lcs[k - 1];
lcs[i] = fabs(t[k] - t[i]) <= eps ? k + 1 : 0;
}
}
template <class C>
bool kmpSearch(const C &s, const C &t) {
kmpInit(t);
int l = 0;
for (int i = 0; i < s.size(); ++i) {
while (l > 0 && fabs(t[l] - s[i]) > eps) l = lcs[l - 1];
if (fabs(t[l] - s[i]) <= eps) ++l;
if (l >= t.size()) return true;
}
return false;
}
int n, m;
std::vector<vec> p, q;
std::vector<long double> s, t;
bool check() {
if (p.size() != q.size()) return false;
for (int i = 0; i < 2 * p.size(); ++i) {
s.push_back((p[(i + 1) % p.size()] - p[i % p.size()]).len());
double a = (p[(i + 2) % p.size()] - p[(i + 1) % p.size()]).angle() -
(p[(i + 1) % p.size()] - p[i % p.size()]).angle();
if (a < -eps) a += 2.0 * pi;
s.push_back(a);
}
for (int i = 0; i < q.size(); ++i) {
t.push_back((q[(i + 1) % q.size()] - q[i % q.size()]).len());
double a = (q[(i + 2) % q.size()] - q[(i + 1) % q.size()]).angle() -
(q[(i + 1) % q.size()] - q[i % q.size()]).angle();
if (a < -eps) a += 2.0 * pi;
t.push_back(a);
}
return kmpSearch(s, t);
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 0; i < n; ++i) {
vec v;
v.read();
p.emplace_back(v);
}
for (int i = 0; i < m; ++i) {
vec v;
v.read();
q.emplace_back(v);
}
convexHull(p, p);
convexHull(q, q);
puts(check() ? "YES" : "NO");
return 0;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
const long double eps = 1e-9;
const long double pi = acos(-1.0);
struct vec {
long double x, y;
vec operator+(const vec &k) const { return {x + k.x, y + k.y}; }
vec operator-(const vec &k) const { return {x - k.x, y - k.y}; }
vec operator*(const long double &k) const { return {x * k, y * k}; }
long double dot(const vec &k) const { return x * k.x + y * k.y; }
long double cross(const vec &k) const { return x * k.y - y * k.x; }
long double len() const { return sqrt(x * x + y * y); }
long double angle() const { return atan2(y, x); }
bool operator<(const vec &k) const {
return x < k.x || (x == k.x && y < k.y);
}
void read() {
double u, v;
scanf("%lf%lf", &u, &v);
x = u;
y = v;
}
void print() { printf("%.8f %.8f\n", (double)x, (double)y); }
};
template <class C>
void convexHull(C p, C &q) {
std::sort(p.begin(), p.end());
q.clear();
for (int i = 0; i < p.size(); ++i) {
while (q.size() >= 2 &&
(q[q.size() - 1] - q[q.size() - 2]).cross(p[i] - q[q.size() - 1]) <=
eps) {
q.pop_back();
}
q.push_back(p[i]);
}
int protect = q.size() - 1;
for (int i = p.size() - 2; i >= 0; --i) {
while (q.size() - protect >= 2 &&
(q[q.size() - 1] - q[q.size() - 2]).cross(p[i] - q[q.size() - 1]) <=
eps) {
q.pop_back();
}
q.push_back(p[i]);
}
q.pop_back();
}
std::vector<int> lcs;
template <class C>
void kmpInit(const C &t) {
lcs.resize(t.size());
lcs.clear();
lcs[0] = 0;
for (int i = 1; i < t.size(); ++i) {
int k = lcs[i - 1];
while (k > 0 && fabs(t[k] - t[i]) > eps) k = lcs[k - 1];
lcs[i] = fabs(t[k] - t[i]) <= eps ? k + 1 : 0;
}
}
template <class C>
bool kmpSearch(const C &s, const C &t) {
kmpInit(t);
int l = 0;
for (int i = 0; i < s.size(); ++i) {
while (l > 0 && fabs(t[l] - s[i]) > eps) l = lcs[l - 1];
if (fabs(t[l] - s[i]) <= eps) ++l;
if (l >= t.size()) return true;
}
return false;
}
int n, m;
std::vector<vec> p, q;
std::vector<long double> s, t;
bool check() {
if (p.size() != q.size()) return false;
for (int i = 0; i < 2 * p.size(); ++i) {
s.push_back((p[(i + 1) % p.size()] - p[i % p.size()]).len());
double a = (p[(i + 2) % p.size()] - p[(i + 1) % p.size()]).angle() -
(p[(i + 1) % p.size()] - p[i % p.size()]).angle();
if (a < -eps) a += 2.0 * pi;
s.push_back(a);
}
for (int i = 0; i < q.size(); ++i) {
t.push_back((q[(i + 1) % q.size()] - q[i % q.size()]).len());
double a = (q[(i + 2) % q.size()] - q[(i + 1) % q.size()]).angle() -
(q[(i + 1) % q.size()] - q[i % q.size()]).angle();
if (a < -eps) a += 2.0 * pi;
t.push_back(a);
}
return kmpSearch(s, t);
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 0; i < n; ++i) {
vec v;
v.read();
p.emplace_back(v);
}
for (int i = 0; i < m; ++i) {
vec v;
v.read();
q.emplace_back(v);
}
convexHull(p, p);
convexHull(q, q);
puts(check() ? "YES" : "NO");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
inline int read() {
int X = 0, w = 1;
char ch = 0;
while (ch < '0' || ch > '9') {
if (ch == '-') w = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
X = (X << 3) + (X << 1) + ch - '0', ch = getchar();
return X * w;
}
struct poi {
double x, y;
};
typedef poi tub[100007];
int dcmp(double x) {
if (fabs(x) < 1e-8)
return 0;
else
return x < 0 ? -1 : 1;
}
poi operator+(poi A, poi B) { return (poi){A.x + B.x, A.y + B.y}; }
poi operator-(poi A, poi B) { return (poi){A.x - B.x, A.y - B.y}; }
poi operator*(poi A, double B) { return (poi){A.x * B, A.y * B}; }
poi operator/(poi A, double B) { return (poi){A.x / B, A.y / B}; }
bool operator==(const poi& a, const poi& b) {
return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}
double Dot(poi A, poi B) { return A.x * B.x + A.y * B.y; }
double Length(poi A) { return sqrt(A.x * A.x + A.y * A.y); }
double Angle(poi A, poi B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
double Cross(poi A, poi B) { return A.x * B.y - A.y * B.x; }
double GetS(poi A, poi B, poi C) { return Cross(B - A, C - A); }
bool operator<(poi a, poi b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
int Tub_maker(poi* p, int x, poi* ch) {
sort(p, p + x);
int sum = 0;
for (int i = 0; i < x; i++) {
while (sum > 1 &&
dcmp(Cross(ch[sum - 1] - ch[sum - 2], p[i] - ch[sum - 2])) >= 0) {
sum--;
}
ch[sum++] = p[i];
}
int k = sum;
for (int i = x - 2; i >= 0; i--) {
while (sum > k &&
dcmp(Cross(ch[sum - 1] - ch[sum - 2], p[i] - ch[sum - 2])) >= 0) {
sum--;
}
ch[sum++] = p[i];
}
if (x > 1) {
sum--;
}
return sum;
}
bool is0(double a, double b) { return fabs(a - b) <= 1e-8; }
int n, m, sum1, sum2, cnts, cntt, nxt[100007 << 1];
bool KmpSearch(double* s, double* p, int sLen, int pLen) {
int i = 0, j = 0;
while (i < sLen && j < pLen) {
if (!is0(s[i], p[j])) {
if (j == 0) {
i++;
}
j = nxt[j];
}
while (j < pLen && is0(s[i], p[j])) {
i++, j++;
}
if (j == pLen) {
return 1;
}
}
return 0;
}
void Getxext(double* p, int xxt[], int pLen) {
xxt[0] = -1;
xxt[1] = 0;
int j = 0, i = 1;
for (i = 1; i < pLen; i++) {
while (j > 0 && !is0(p[i], p[j])) {
j = xxt[j];
}
if (is0(p[i], p[j])) {
j++;
}
xxt[i + 1] = j;
}
}
tub p1, p2, c1, c2;
double S[100007 << 2], T[100007 << 1];
void print(double* s, int len) {
for (int i = 0; i < len; i++) {
printf("%.2lf%c", s[i], i == len - 1 ? '\n' : ' ');
}
}
int main() {
n = read(), m = read();
for (int i = 0; i < n; i++) {
scanf("%lf%lf", &p1[i].x, &p1[i].y);
}
for (int i = 0; i < m; i++) {
scanf("%lf%lf", &p2[i].x, &p2[i].y);
}
sum1 = Tub_maker(p1, n, c1);
sum2 = Tub_maker(p2, m, c2);
if (sum1 != sum2) {
puts("NO");
return 0;
}
c1[sum1] = c1[0];
c1[sum1 + 1] = c1[1];
c2[sum2] = c2[0];
c2[sum2 + 1] = c2[1];
for (int i = 0; i < sum1; i++) {
S[cnts++] = Length(c1[i + 1] - c1[i]);
S[cnts++] = Angle(c1[i + 2] - c1[i + 1], c1[i + 1] - c1[i]);
}
for (int i = 0; i < sum2; i++) {
T[cntt++] = Length(c2[i + 1] - c2[i]);
T[cntt++] = Angle(c2[i + 2] - c2[i + 1], c2[i + 1] - c2[i]);
}
for (int i = cnts; i < cnts * 2 - 1; i++) {
S[i] = S[i - cnts];
}
Getxext(T, nxt, cntt);
if (KmpSearch(S, T, cnts * 2 - 2, cntt - 1)) {
puts("YES");
} else {
puts("NO");
}
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
inline int read() {
int X = 0, w = 1;
char ch = 0;
while (ch < '0' || ch > '9') {
if (ch == '-') w = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
X = (X << 3) + (X << 1) + ch - '0', ch = getchar();
return X * w;
}
struct poi {
double x, y;
};
typedef poi tub[100007];
int dcmp(double x) {
if (fabs(x) < 1e-8)
return 0;
else
return x < 0 ? -1 : 1;
}
poi operator+(poi A, poi B) { return (poi){A.x + B.x, A.y + B.y}; }
poi operator-(poi A, poi B) { return (poi){A.x - B.x, A.y - B.y}; }
poi operator*(poi A, double B) { return (poi){A.x * B, A.y * B}; }
poi operator/(poi A, double B) { return (poi){A.x / B, A.y / B}; }
bool operator==(const poi& a, const poi& b) {
return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}
double Dot(poi A, poi B) { return A.x * B.x + A.y * B.y; }
double Length(poi A) { return sqrt(A.x * A.x + A.y * A.y); }
double Angle(poi A, poi B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
double Cross(poi A, poi B) { return A.x * B.y - A.y * B.x; }
double GetS(poi A, poi B, poi C) { return Cross(B - A, C - A); }
bool operator<(poi a, poi b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
int Tub_maker(poi* p, int x, poi* ch) {
sort(p, p + x);
int sum = 0;
for (int i = 0; i < x; i++) {
while (sum > 1 &&
dcmp(Cross(ch[sum - 1] - ch[sum - 2], p[i] - ch[sum - 2])) >= 0) {
sum--;
}
ch[sum++] = p[i];
}
int k = sum;
for (int i = x - 2; i >= 0; i--) {
while (sum > k &&
dcmp(Cross(ch[sum - 1] - ch[sum - 2], p[i] - ch[sum - 2])) >= 0) {
sum--;
}
ch[sum++] = p[i];
}
if (x > 1) {
sum--;
}
return sum;
}
bool is0(double a, double b) { return fabs(a - b) <= 1e-8; }
int n, m, sum1, sum2, cnts, cntt, nxt[100007 << 1];
bool KmpSearch(double* s, double* p, int sLen, int pLen) {
int i = 0, j = 0;
while (i < sLen && j < pLen) {
if (!is0(s[i], p[j])) {
if (j == 0) {
i++;
}
j = nxt[j];
}
while (j < pLen && is0(s[i], p[j])) {
i++, j++;
}
if (j == pLen) {
return 1;
}
}
return 0;
}
void Getxext(double* p, int xxt[], int pLen) {
xxt[0] = -1;
xxt[1] = 0;
int j = 0, i = 1;
for (i = 1; i < pLen; i++) {
while (j > 0 && !is0(p[i], p[j])) {
j = xxt[j];
}
if (is0(p[i], p[j])) {
j++;
}
xxt[i + 1] = j;
}
}
tub p1, p2, c1, c2;
double S[100007 << 2], T[100007 << 1];
void print(double* s, int len) {
for (int i = 0; i < len; i++) {
printf("%.2lf%c", s[i], i == len - 1 ? '\n' : ' ');
}
}
int main() {
n = read(), m = read();
for (int i = 0; i < n; i++) {
scanf("%lf%lf", &p1[i].x, &p1[i].y);
}
for (int i = 0; i < m; i++) {
scanf("%lf%lf", &p2[i].x, &p2[i].y);
}
sum1 = Tub_maker(p1, n, c1);
sum2 = Tub_maker(p2, m, c2);
if (sum1 != sum2) {
puts("NO");
return 0;
}
c1[sum1] = c1[0];
c1[sum1 + 1] = c1[1];
c2[sum2] = c2[0];
c2[sum2 + 1] = c2[1];
for (int i = 0; i < sum1; i++) {
S[cnts++] = Length(c1[i + 1] - c1[i]);
S[cnts++] = Angle(c1[i + 2] - c1[i + 1], c1[i + 1] - c1[i]);
}
for (int i = 0; i < sum2; i++) {
T[cntt++] = Length(c2[i + 1] - c2[i]);
T[cntt++] = Angle(c2[i + 2] - c2[i + 1], c2[i + 1] - c2[i]);
}
for (int i = cnts; i < cnts * 2 - 1; i++) {
S[i] = S[i - cnts];
}
Getxext(T, nxt, cntt);
if (KmpSearch(S, T, cnts * 2 - 2, cntt - 1)) {
puts("YES");
} else {
puts("NO");
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
vector<string> split(const string& s, char c) {
vector<string> v;
stringstream ss(s);
string x;
while (getline(ss, x, c)) v.emplace_back(x);
return move(v);
}
void err(vector<string>::iterator it) {}
template <typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
cerr << it->substr((*it)[0] == ' ', it->length()) << " = " << a << '\t';
err(++it, args...);
}
const int NN = 1e5 + 5, mod = 1e9 + 7;
struct point {
int x, y;
};
point source[2][NN], pivot;
long long int dist2(point a, point b) {
return ((long long int)a.x - b.x) * (a.x - b.x) +
((long long int)a.y - b.y) * ((long long int)a.y - b.y);
}
int ccw(point a, point b, point c) {
long long int dx1 = b.x - a.x, dx2 = c.x - a.x, dy1 = b.y - a.y,
dy2 = c.y - a.y;
if (dx1 * dy2 > dy1 * dx2) return 1;
if (dx1 * dy2 < dy1 * dx2) return -1;
return 0;
}
bool ccw_sort(point a, point b) {
int o = ccw(pivot, a, b);
if (o == 0) return dist2(pivot, a) > dist2(pivot, b);
return o > 0;
}
long long int dot(point a, point b) {
return (long long int)a.x * b.x + (long long int)a.y * b.y;
}
vector<point> convex_hull(point ar[], int n) {
int i, j;
for (i = 1; i < n; i++) {
if (ar[i].y < ar[0].y)
swap(ar[i], ar[0]);
else if (ar[i].y == ar[0].y) {
if (ar[i].x < ar[0].x) swap(ar[i], ar[0]);
}
}
pivot = ar[0];
sort(ar + 1, ar + n, ccw_sort);
vector<point> v;
for (i = 1; i < n; i++) {
for (j = i + 1; j < n; j++) {
if (ccw(ar[0], ar[i], ar[j])) break;
}
v.emplace_back(ar[i]);
i = j - 1;
}
int sz = v.size();
if (sz == 1) {
return {ar[0], v[0]};
}
vector<point> S;
S.emplace_back(ar[0]);
S.emplace_back(v[0]);
S.emplace_back(v[1]);
point nxt;
for (i = 2; i < sz; i++) {
nxt = S[S.size() - 2];
while (ccw(nxt, S[S.size() - 1], v[i]) <= 0) {
S.pop_back();
nxt = S[S.size() - 2];
}
S.emplace_back(v[i]);
}
return S;
}
string prefix_function(vector<long long int> Z) {
int n = Z.size(), i;
int F[n];
F[0] = 0;
for (i = 1; Z[i] < 1e18; ++i) {
int j = F[i - 1];
while (j > 0 && Z[i] != Z[j]) j = F[j - 1];
if (Z[i] == Z[j]) ++j;
F[i] = j;
}
F[i] = 0;
++i;
int j, m = i - 1;
if (Z[i] == Z[0]) {
F[i] = 1;
} else {
F[i] = 0;
}
for (++i; i < n; i++) {
j = F[i - 1];
while (j > 0 && Z[i] != Z[j]) j = F[j - 1];
if (Z[i] == Z[j]) ++j;
F[i] = j;
if (F[i] == m) return "YES";
}
return "NO";
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n, m, i, j, k;
cin >> n >> m;
for (i = 0; i < n; i++) {
cin >> j >> k;
source[0][i] = {j, k};
}
for (i = 0; i < m; i++) {
cin >> j >> k;
source[1][i] = {j, k};
}
vector<point> con1 = convex_hull(source[0], n);
vector<point> con2 = convex_hull(source[1], m);
con1.insert(con1.end(), con1.begin(), con1.end());
vector<long long int> sas1;
for (i = 0; i < con1.size() - 1; i++) {
if (i > 0 && i < con1.size() - 1) {
long long int d =
dot({con1[i].x - con1[i - 1].x, con1[i].y - con1[i - 1].y},
{con1[i + 1].x - con1[i].x, con1[i + 1].y - con1[i].y});
sas1.emplace_back(d);
}
sas1.emplace_back(dist2(con1[i], con1[i + 1]));
}
vector<long long int> sas2;
for (i = 0; i < con2.size() - 1; i++) {
if (i > 0 && i < con2.size() - 1) {
long long int d =
dot({con2[i].x - con2[i - 1].x, con2[i].y - con2[i - 1].y},
{con2[i + 1].x - con2[i].x, con2[i + 1].y - con2[i].y});
sas2.emplace_back(d);
}
sas2.emplace_back(dist2(con2[i], con2[i + 1]));
}
sas2.push_back(1e18);
sas2.insert(sas2.end(), sas1.begin(), sas1.end());
cout << prefix_function(sas2);
return 0;
}
| ### Prompt
Construct a CPP code solution to the problem outlined:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
vector<string> split(const string& s, char c) {
vector<string> v;
stringstream ss(s);
string x;
while (getline(ss, x, c)) v.emplace_back(x);
return move(v);
}
void err(vector<string>::iterator it) {}
template <typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
cerr << it->substr((*it)[0] == ' ', it->length()) << " = " << a << '\t';
err(++it, args...);
}
const int NN = 1e5 + 5, mod = 1e9 + 7;
struct point {
int x, y;
};
point source[2][NN], pivot;
long long int dist2(point a, point b) {
return ((long long int)a.x - b.x) * (a.x - b.x) +
((long long int)a.y - b.y) * ((long long int)a.y - b.y);
}
int ccw(point a, point b, point c) {
long long int dx1 = b.x - a.x, dx2 = c.x - a.x, dy1 = b.y - a.y,
dy2 = c.y - a.y;
if (dx1 * dy2 > dy1 * dx2) return 1;
if (dx1 * dy2 < dy1 * dx2) return -1;
return 0;
}
bool ccw_sort(point a, point b) {
int o = ccw(pivot, a, b);
if (o == 0) return dist2(pivot, a) > dist2(pivot, b);
return o > 0;
}
long long int dot(point a, point b) {
return (long long int)a.x * b.x + (long long int)a.y * b.y;
}
vector<point> convex_hull(point ar[], int n) {
int i, j;
for (i = 1; i < n; i++) {
if (ar[i].y < ar[0].y)
swap(ar[i], ar[0]);
else if (ar[i].y == ar[0].y) {
if (ar[i].x < ar[0].x) swap(ar[i], ar[0]);
}
}
pivot = ar[0];
sort(ar + 1, ar + n, ccw_sort);
vector<point> v;
for (i = 1; i < n; i++) {
for (j = i + 1; j < n; j++) {
if (ccw(ar[0], ar[i], ar[j])) break;
}
v.emplace_back(ar[i]);
i = j - 1;
}
int sz = v.size();
if (sz == 1) {
return {ar[0], v[0]};
}
vector<point> S;
S.emplace_back(ar[0]);
S.emplace_back(v[0]);
S.emplace_back(v[1]);
point nxt;
for (i = 2; i < sz; i++) {
nxt = S[S.size() - 2];
while (ccw(nxt, S[S.size() - 1], v[i]) <= 0) {
S.pop_back();
nxt = S[S.size() - 2];
}
S.emplace_back(v[i]);
}
return S;
}
string prefix_function(vector<long long int> Z) {
int n = Z.size(), i;
int F[n];
F[0] = 0;
for (i = 1; Z[i] < 1e18; ++i) {
int j = F[i - 1];
while (j > 0 && Z[i] != Z[j]) j = F[j - 1];
if (Z[i] == Z[j]) ++j;
F[i] = j;
}
F[i] = 0;
++i;
int j, m = i - 1;
if (Z[i] == Z[0]) {
F[i] = 1;
} else {
F[i] = 0;
}
for (++i; i < n; i++) {
j = F[i - 1];
while (j > 0 && Z[i] != Z[j]) j = F[j - 1];
if (Z[i] == Z[j]) ++j;
F[i] = j;
if (F[i] == m) return "YES";
}
return "NO";
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n, m, i, j, k;
cin >> n >> m;
for (i = 0; i < n; i++) {
cin >> j >> k;
source[0][i] = {j, k};
}
for (i = 0; i < m; i++) {
cin >> j >> k;
source[1][i] = {j, k};
}
vector<point> con1 = convex_hull(source[0], n);
vector<point> con2 = convex_hull(source[1], m);
con1.insert(con1.end(), con1.begin(), con1.end());
vector<long long int> sas1;
for (i = 0; i < con1.size() - 1; i++) {
if (i > 0 && i < con1.size() - 1) {
long long int d =
dot({con1[i].x - con1[i - 1].x, con1[i].y - con1[i - 1].y},
{con1[i + 1].x - con1[i].x, con1[i + 1].y - con1[i].y});
sas1.emplace_back(d);
}
sas1.emplace_back(dist2(con1[i], con1[i + 1]));
}
vector<long long int> sas2;
for (i = 0; i < con2.size() - 1; i++) {
if (i > 0 && i < con2.size() - 1) {
long long int d =
dot({con2[i].x - con2[i - 1].x, con2[i].y - con2[i - 1].y},
{con2[i + 1].x - con2[i].x, con2[i + 1].y - con2[i].y});
sas2.emplace_back(d);
}
sas2.emplace_back(dist2(con2[i], con2[i + 1]));
}
sas2.push_back(1e18);
sas2.insert(sas2.end(), sas1.begin(), sas1.end());
cout << prefix_function(sas2);
return 0;
}
``` |
#include <bits/stdc++.h>
int sig(long long p) { return (p > 0) - (p < 0); }
class P {
public:
long long x, y;
P(long long x = 0, long long y = 0) : x(x), y(y) {}
P operator-(const P &p) const { return P(x - p.x, y - p.y); }
long long operator^(const P &p) const { return x * p.y - y * p.x; }
long long operator%(const P &p) const {
return (x % (1000000009) * p.x % (1000000009) +
y % (1000000009) * p.y % (1000000009)) %
(1000000009);
}
bool operator<(const P &p) const {
if (sig(x - p.x)) return x < p.x;
return y < p.y;
}
void read() { scanf("%lld%lld", &x, &y); }
long long len() { return x * x + y * y; }
};
class L {
public:
P p, v;
L(P a, P b) : p(a), v(b - a) {}
};
long long onLeft(L l, P p) { return l.v ^ (p - l.p); }
std::vector<P> convexHull(std::vector<P> &ps) {
static P stack[(100010)];
std::sort(ps.begin(), ps.end());
int n = ps.size(), top = 0;
for (int i = 0; i < n; ++i) {
while (top > 1 &&
sig(onLeft(L(stack[top - 2], stack[top - 1]), ps[i])) <= 0) {
--top;
}
stack[top++] = ps[i];
}
int tmp = top;
for (int i = n - 2; i >= 0; --i) {
while (top > tmp &&
sig(onLeft(L(stack[top - 2], stack[top - 1]), ps[i])) <= 0) {
--top;
}
stack[top++] = ps[i];
}
if (n > 1) --top;
std::vector<P> ret;
for (int i = 0; i < top; ++i) {
ret.push_back(stack[i]);
}
return ret;
}
struct KMP {
int fail[(100010)];
bool searchForm(std::vector<std::pair<long long, int> > &s,
std::vector<std::pair<long long, int> > &t) {
int n = t.size(), m = s.size();
fail[0] = -1;
for (int i = 1, j = -1; i <= m; ++i) {
while (j >= 0 && s[j] != s[i - 1]) j = fail[j];
fail[i] = ++j;
}
for (int i = 0, j = 0; i < n + n; ++i) {
auto ti = i < n ? t[i] : t[i - n];
while (j >= 0 && s[j] != ti) j = fail[j];
if (++j == m) return true;
}
return false;
}
} kmp;
int powmod(int a, int b) {
int ret = 1;
for (; b; b >>= 1) {
if (b & 1) ret = 1ll * ret * a % (1000000009);
a = 1ll * a * a % (1000000009);
}
return ret;
}
int angle(P a, P b) {
return (a % b) % powmod(a.len(), (1000000009) - 2) % (1000000009) *
powmod(b.len(), (1000000009) - 2) % (1000000009);
}
int n, m;
std::vector<P> blue, red;
int main() {
scanf("%d%d", &n, &m);
blue.resize(n);
red.resize(m);
for (int i = 0; i < n; ++i) blue[i].read();
for (int i = 0; i < m; ++i) red[i].read();
blue = convexHull(blue);
red = convexHull(red);
std::vector<std::pair<long long, int> > s, t;
n = blue.size();
m = red.size();
for (int i = 0; i < n; ++i) {
int j = (i + 1) % n, k = (j + 1) % n;
s.push_back({(blue[j] - blue[i]).len(),
angle(blue[j] - blue[i], blue[k] - blue[j])});
}
for (int i = 0; i < m; ++i) {
int j = (i + 1) % m, k = (j + 1) % m;
t.push_back(
{(red[j] - red[i]).len(), angle(red[j] - red[i], red[k] - red[j])});
}
printf(kmp.searchForm(s, t) ? "YES\n" : "NO\n");
return 0;
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
int sig(long long p) { return (p > 0) - (p < 0); }
class P {
public:
long long x, y;
P(long long x = 0, long long y = 0) : x(x), y(y) {}
P operator-(const P &p) const { return P(x - p.x, y - p.y); }
long long operator^(const P &p) const { return x * p.y - y * p.x; }
long long operator%(const P &p) const {
return (x % (1000000009) * p.x % (1000000009) +
y % (1000000009) * p.y % (1000000009)) %
(1000000009);
}
bool operator<(const P &p) const {
if (sig(x - p.x)) return x < p.x;
return y < p.y;
}
void read() { scanf("%lld%lld", &x, &y); }
long long len() { return x * x + y * y; }
};
class L {
public:
P p, v;
L(P a, P b) : p(a), v(b - a) {}
};
long long onLeft(L l, P p) { return l.v ^ (p - l.p); }
std::vector<P> convexHull(std::vector<P> &ps) {
static P stack[(100010)];
std::sort(ps.begin(), ps.end());
int n = ps.size(), top = 0;
for (int i = 0; i < n; ++i) {
while (top > 1 &&
sig(onLeft(L(stack[top - 2], stack[top - 1]), ps[i])) <= 0) {
--top;
}
stack[top++] = ps[i];
}
int tmp = top;
for (int i = n - 2; i >= 0; --i) {
while (top > tmp &&
sig(onLeft(L(stack[top - 2], stack[top - 1]), ps[i])) <= 0) {
--top;
}
stack[top++] = ps[i];
}
if (n > 1) --top;
std::vector<P> ret;
for (int i = 0; i < top; ++i) {
ret.push_back(stack[i]);
}
return ret;
}
struct KMP {
int fail[(100010)];
bool searchForm(std::vector<std::pair<long long, int> > &s,
std::vector<std::pair<long long, int> > &t) {
int n = t.size(), m = s.size();
fail[0] = -1;
for (int i = 1, j = -1; i <= m; ++i) {
while (j >= 0 && s[j] != s[i - 1]) j = fail[j];
fail[i] = ++j;
}
for (int i = 0, j = 0; i < n + n; ++i) {
auto ti = i < n ? t[i] : t[i - n];
while (j >= 0 && s[j] != ti) j = fail[j];
if (++j == m) return true;
}
return false;
}
} kmp;
int powmod(int a, int b) {
int ret = 1;
for (; b; b >>= 1) {
if (b & 1) ret = 1ll * ret * a % (1000000009);
a = 1ll * a * a % (1000000009);
}
return ret;
}
int angle(P a, P b) {
return (a % b) % powmod(a.len(), (1000000009) - 2) % (1000000009) *
powmod(b.len(), (1000000009) - 2) % (1000000009);
}
int n, m;
std::vector<P> blue, red;
int main() {
scanf("%d%d", &n, &m);
blue.resize(n);
red.resize(m);
for (int i = 0; i < n; ++i) blue[i].read();
for (int i = 0; i < m; ++i) red[i].read();
blue = convexHull(blue);
red = convexHull(red);
std::vector<std::pair<long long, int> > s, t;
n = blue.size();
m = red.size();
for (int i = 0; i < n; ++i) {
int j = (i + 1) % n, k = (j + 1) % n;
s.push_back({(blue[j] - blue[i]).len(),
angle(blue[j] - blue[i], blue[k] - blue[j])});
}
for (int i = 0; i < m; ++i) {
int j = (i + 1) % m, k = (j + 1) % m;
t.push_back(
{(red[j] - red[i]).len(), angle(red[j] - red[i], red[k] - red[j])});
}
printf(kmp.searchForm(s, t) ? "YES\n" : "NO\n");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-10;
bool dcmp(double x, double y = 0) { return fabs(x - y) <= EPS; }
double add(double a, double b) {
if (abs(a + b) < EPS * (abs(a) + abs(b))) return 0;
return a + b;
}
struct P {
double x, y;
P() {}
P(double x, double y) : x(x), y(y) {}
P operator+(P p) { return P(add(x, p.x), add(y, p.y)); }
P operator-(P p) { return P(add(x, -p.x), add(y, -p.y)); }
P operator*(double d) { return P(x * d, y * d); }
double dot(P p) { return add(x * p.x, y * p.y); }
double det(P p) { return add(x * p.y, -y * p.x); }
};
vector<P> convex_hull(vector<P> &ps) {
int n = ps.size();
sort(ps.begin(), ps.end(), [](const P &a, const P &b) {
return a.x != b.x ? a.x < b.x : a.y < b.y;
});
int k = 0;
vector<P> qs(n << 1);
for (int i = 0; i < n; i++) {
while (k > 1 && (qs[k - 1] - qs[k - 2]).det(ps[i] - qs[k - 1]) <= 0) {
k--;
}
qs[k++] = ps[i];
}
for (int i = n - 2, t = k; i >= 0; i--) {
while (k > t && (qs[k - 1] - qs[k - 2]).det(ps[i] - qs[k - 1]) <= 0) {
k--;
}
qs[k++] = ps[i];
}
qs.resize(k - 1);
return qs;
}
double dis(P a, P b) {
double x = a.x - b.x, y = a.y - b.y;
return x * x + y * y;
}
struct V {
long long x, y, z;
bool operator<(const V &v) const {
if (x != v.x) return x < v.x;
if (y != v.y) return y < v.y;
return z < v.z;
}
bool operator==(const V &v) const { return x == v.x && y == v.y && z == v.z; }
bool operator!=(const V &v) const { return x != v.x || y != v.y || z != v.z; }
};
int kmp(vector<V> &p, vector<V> &t) {
if (p.size() > t.size()) {
return 0;
}
int lenp = p.size();
vector<int> next(lenp + 1);
for (int i = 1; i < lenp; i++) {
int j = next[i];
while (j && p[i] != p[j]) {
j = next[j];
}
next[i + 1] = (p[i] == p[j] ? j + 1 : 0);
}
int res = 0, lent = t.size();
for (int i = 0, j = 0; i < lent; i++) {
while (j && t[i] != p[j]) {
j = next[j];
}
if (t[i] == p[j]) {
j++;
}
if (j == lenp) {
res++;
}
}
return res;
}
int main() {
int n, m;
scanf("%d %d", &n, &m);
vector<P> a(n), b(m);
for (int i = 0; i < n; i++) {
scanf("%lf %lf", &a[i].x, &a[i].y);
}
for (int i = 0; i < m; i++) {
scanf("%lf %lf", &b[i].x, &b[i].y);
}
vector<P> ac = convex_hull(a), bc = convex_hull(b);
int an = ac.size(), bn = bc.size();
if (an != bn) {
printf("NO\n");
} else {
vector<V> av(an << 1), bv(bn);
for (int i = 0; i < an; i++) {
long long x = dis(ac[i], ac[(i + 1) % an]);
long long y = dis(ac[(i + 1) % an], ac[(i + 2) % an]);
long long z = dis(ac[i], ac[(i + 2) % an]);
av[i] = {x, y, z};
av[an + i] = av[i];
}
for (int i = 0; i < bn; i++) {
long long x = dis(bc[i], bc[(i + 1) % bn]);
long long y = dis(bc[(i + 1) % bn], bc[(i + 2) % bn]);
long long z = dis(bc[i], bc[(i + 2) % bn]);
bv[i] = {x, y, z};
}
int ans = kmp(bv, av);
printf("%s\n", ans ? "YES" : "NO");
}
return 0;
}
| ### Prompt
Please create a solution in Cpp to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-10;
bool dcmp(double x, double y = 0) { return fabs(x - y) <= EPS; }
double add(double a, double b) {
if (abs(a + b) < EPS * (abs(a) + abs(b))) return 0;
return a + b;
}
struct P {
double x, y;
P() {}
P(double x, double y) : x(x), y(y) {}
P operator+(P p) { return P(add(x, p.x), add(y, p.y)); }
P operator-(P p) { return P(add(x, -p.x), add(y, -p.y)); }
P operator*(double d) { return P(x * d, y * d); }
double dot(P p) { return add(x * p.x, y * p.y); }
double det(P p) { return add(x * p.y, -y * p.x); }
};
vector<P> convex_hull(vector<P> &ps) {
int n = ps.size();
sort(ps.begin(), ps.end(), [](const P &a, const P &b) {
return a.x != b.x ? a.x < b.x : a.y < b.y;
});
int k = 0;
vector<P> qs(n << 1);
for (int i = 0; i < n; i++) {
while (k > 1 && (qs[k - 1] - qs[k - 2]).det(ps[i] - qs[k - 1]) <= 0) {
k--;
}
qs[k++] = ps[i];
}
for (int i = n - 2, t = k; i >= 0; i--) {
while (k > t && (qs[k - 1] - qs[k - 2]).det(ps[i] - qs[k - 1]) <= 0) {
k--;
}
qs[k++] = ps[i];
}
qs.resize(k - 1);
return qs;
}
double dis(P a, P b) {
double x = a.x - b.x, y = a.y - b.y;
return x * x + y * y;
}
struct V {
long long x, y, z;
bool operator<(const V &v) const {
if (x != v.x) return x < v.x;
if (y != v.y) return y < v.y;
return z < v.z;
}
bool operator==(const V &v) const { return x == v.x && y == v.y && z == v.z; }
bool operator!=(const V &v) const { return x != v.x || y != v.y || z != v.z; }
};
int kmp(vector<V> &p, vector<V> &t) {
if (p.size() > t.size()) {
return 0;
}
int lenp = p.size();
vector<int> next(lenp + 1);
for (int i = 1; i < lenp; i++) {
int j = next[i];
while (j && p[i] != p[j]) {
j = next[j];
}
next[i + 1] = (p[i] == p[j] ? j + 1 : 0);
}
int res = 0, lent = t.size();
for (int i = 0, j = 0; i < lent; i++) {
while (j && t[i] != p[j]) {
j = next[j];
}
if (t[i] == p[j]) {
j++;
}
if (j == lenp) {
res++;
}
}
return res;
}
int main() {
int n, m;
scanf("%d %d", &n, &m);
vector<P> a(n), b(m);
for (int i = 0; i < n; i++) {
scanf("%lf %lf", &a[i].x, &a[i].y);
}
for (int i = 0; i < m; i++) {
scanf("%lf %lf", &b[i].x, &b[i].y);
}
vector<P> ac = convex_hull(a), bc = convex_hull(b);
int an = ac.size(), bn = bc.size();
if (an != bn) {
printf("NO\n");
} else {
vector<V> av(an << 1), bv(bn);
for (int i = 0; i < an; i++) {
long long x = dis(ac[i], ac[(i + 1) % an]);
long long y = dis(ac[(i + 1) % an], ac[(i + 2) % an]);
long long z = dis(ac[i], ac[(i + 2) % an]);
av[i] = {x, y, z};
av[an + i] = av[i];
}
for (int i = 0; i < bn; i++) {
long long x = dis(bc[i], bc[(i + 1) % bn]);
long long y = dis(bc[(i + 1) % bn], bc[(i + 2) % bn]);
long long z = dis(bc[i], bc[(i + 2) % bn]);
bv[i] = {x, y, z};
}
int ans = kmp(bv, av);
printf("%s\n", ans ? "YES" : "NO");
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <typename T>
using maxHeap = priority_queue<T, vector<T>, less<T>>;
template <typename T>
using minHeap = priority_queue<T, vector<T>, greater<T>>;
template <typename Iter>
ostream& _out(ostream& s, Iter b, Iter e) {
s << "[";
for (auto it = b; it != e; it++) s << (it == b ? "" : " ") << *it;
s << "]";
return s;
}
template <typename A, typename B>
ostream& operator<<(ostream& s, const pair<A, B>& p) {
return s << "(" << p.first << "," << p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& s, const vector<T>& c) {
return _out(s, begin(c), end(c));
}
bool debug = 0;
template <typename T>
void DEBUG(const T& x) {
if (debug) cerr << x;
}
template <typename T, typename... Args>
void DEBUG(const T& head, const Args&... tail) {
if (debug) {
cerr << head;
DEBUG(tail...);
}
}
template <typename T>
struct Point {
static constexpr long double EPS = 1e-8;
T x, y;
Point(T _ = 0, T __ = 0) : x(_), y(__) {}
template <typename T2>
Point(const Point<T2>& a) : x(a.x), y(a.y) {}
inline long double theta() const {
return atan2((long double)y, (long double)x);
}
inline long double dis() const {
return hypot((long double)x, (long double)y);
}
inline long double dis(const Point& o) const {
return hypot((long double)(x - o.x), (long double)(y - o.y));
}
Point operator-(const Point& o) const { return Point(x - o.x, y - o.y); }
Point operator-=(const Point& o) {
x -= o.x, y -= o.y;
return *this;
}
Point operator+(const Point& o) const { return Point(x + o.x, y + o.y); }
Point operator+=(const Point& o) {
x += o.x, y += o.y;
return *this;
}
Point operator*(const T& k) const { return Point(x * k, y * k); }
Point operator*=(const T& k) {
x *= k, y *= k;
return *this;
}
Point operator/(const T& k) const { return Point(x / k, y / k); }
Point operator/=(const T& k) {
x /= k, y /= k;
return *this;
}
Point operator-() const { return Point(-x, -y); }
Point rot90() const { return Point(-y, x); }
bool equal(const Point& o, true_type) const {
return fabs(x - o.x) < EPS and fabs(y - o.y) < EPS;
}
bool equal(const Point& o, false_type) const {
return tie(x, y) == tie(o.x, o.y);
}
bool operator==(const Point& o) const {
return equal(o, is_floating_point<T>());
}
bool operator!=(const Point& o) const { return !(*this == o); }
bool operator<(const Point& o) const { return theta() < o.theta(); }
friend inline T cross(const Point& a, const Point& b) {
return a.x * b.y - b.x * a.y;
}
friend inline T dot(const Point& a, const Point& b) {
return a.x * b.x + a.y * b.y;
}
friend ostream& operator<<(ostream& ss, const Point& o) {
ss << "(" << o.x << ", " << o.y << ")";
return ss;
}
};
template <typename T>
class ConvexHull_2D {
private:
vector<Point<T>> dots;
struct myhash {
uint64_t operator()(const Point<T>& a) const {
uint64_t xx = 0, yy = 0;
memcpy(&xx, &a.x, sizeof(a.x));
memcpy(&yy, &a.y, sizeof(a.y));
uint64_t ret = xx * 17 + yy * 31;
ret = (ret ^ (ret >> 16)) * 0x9E3779B1;
ret = (ret ^ (ret >> 13)) * 0xC2B2AE35;
ret = ret ^ xx;
return (ret ^ (ret << 3)) * yy;
}
};
unordered_set<Point<T>, myhash> in_hull;
public:
inline void init() {
in_hull.clear();
dots.clear();
}
void insert(const Point<T>& x) { dots.push_back(x); }
void solve() {
sort(begin(dots), end(dots), [](const Point<T>& a, const Point<T>& b) {
return tie(a.x, a.y) < tie(b.x, b.y);
});
vector<Point<T>> stk((int)((dots).size()) << 1);
int top = 0;
for (auto p : dots) {
while (top >= 2 and
cross(p - stk[top - 2], stk[top - 1] - stk[top - 2]) <= 0)
top--;
stk[top++] = p;
}
for (int i = (int)((dots).size()) - 2, t = top + 1; i >= 0; i--) {
while (top >= t and
cross(dots[i] - stk[top - 2], stk[top - 1] - stk[top - 2]) <= 0)
top--;
stk[top++] = dots[i];
}
stk.resize(top - 1);
swap(stk, dots);
for (auto i : stk) in_hull.insert(i);
}
vector<Point<T>> get() { return dots; }
inline bool in_it(const Point<T>& x) {
return in_hull.find(x) != in_hull.end();
}
};
using PTL = Point<int64_t>;
ConvexHull_2D<int64_t> cv;
template <typename T>
bool check(const vector<T>&, const vector<T>&);
vector<int64_t> mcp(vector<int64_t> s) {
int n = (int)((s).size());
for (int i = 0; i < n; i++) s.push_back(s[i]);
int i = 0, j = 1;
while (i < n && j < n) {
int k = 0;
while (k < n && s[i + k] == s[j + k]) k++;
if (s[i + k] <= s[j + k])
j += k + 1;
else
i += k + 1;
if (i == j) j++;
}
int ans = i < n ? i : j;
return vector<int64_t>(next(s.begin(), ans), next(s.begin(), ans + n));
}
int main(int argc, char* argv[]) {
if (argc > 1 and string(argv[1]) == "-D") debug = 1;
if (!debug) {
ios_base::sync_with_stdio(0);
cin.tie(0);
}
auto kill = []() {
cout << "NO" << '\n';
exit(0);
};
int n, m;
cin >> n >> m;
cv.init();
for (int i = 0; i < n; i++) {
int64_t x, y;
cin >> x >> y;
cv.insert(PTL(x, y));
}
cv.solve();
vector<PTL> one = cv.get();
cv.init();
for (int i = 0; i < m; i++) {
int64_t x, y;
cin >> x >> y;
cv.insert(PTL(x, y));
}
cv.solve();
vector<PTL> two = cv.get();
if (debug)
cerr << __PRETTY_FUNCTION__ << ":" << 221 << " - " << ("one") << "="
<< (one) << '\n';
if (debug)
cerr << __PRETTY_FUNCTION__ << ":" << 222 << " - " << ("two") << "="
<< (two) << '\n';
if ((int)((one).size()) != (int)((two).size())) kill();
vector<int64_t> d1, d2;
for (int i = 1; i < (int)((one).size()); i++)
d1.push_back(dot(one[i] - one[i - 1], one[i] - one[i - 1]));
d1.push_back(dot(one.back() - one[0], one.back() - one[0]));
for (int i = 1; i < (int)((two).size()); i++)
d2.push_back(dot(two[i] - two[i - 1], two[i] - two[i - 1]));
d2.push_back(dot(two.back() - two[0], two.back() - two[0]));
if (debug)
cerr << __PRETTY_FUNCTION__ << ":" << 231 << " - " << ("d1") << "=" << (d1)
<< '\n';
if (debug)
cerr << __PRETTY_FUNCTION__ << ":" << 232 << " - " << ("d2") << "=" << (d2)
<< '\n';
if (!check(d1, d2)) kill();
vector<int64_t> t1, t2;
t1.push_back(
dot(one[0] - one.back(), one[(int)((one).size()) - 2] - one.back()));
t1.push_back(dot(one[1] - one[0], one.back() - one[0]));
for (int i = 2; i < (int)((one).size()); i++)
t1.push_back(dot(one[i] - one[i - 1], one[i - 2] - one[i - 1]));
t2.push_back(
dot(two[0] - two.back(), two[(int)((two).size()) - 2] - two.back()));
t2.push_back(dot(two[1] - two[0], two.back() - two[0]));
for (int i = 2; i < (int)((two).size()); i++)
t2.push_back(dot(two[i] - two[i - 1], two[i - 2] - two[i - 1]));
if (debug)
cerr << __PRETTY_FUNCTION__ << ":" << 243 << " - " << ("t1") << "=" << (t1)
<< '\n';
if (debug)
cerr << __PRETTY_FUNCTION__ << ":" << 244 << " - " << ("t2") << "=" << (t2)
<< '\n';
if (!check(t1, t2)) kill();
cout << "YES" << '\n';
return 0;
}
template <typename T>
bool check(const vector<T>& a, const vector<T>& b) {
return mcp(a) == mcp(b);
}
| ### Prompt
Please create a solution in CPP to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
using maxHeap = priority_queue<T, vector<T>, less<T>>;
template <typename T>
using minHeap = priority_queue<T, vector<T>, greater<T>>;
template <typename Iter>
ostream& _out(ostream& s, Iter b, Iter e) {
s << "[";
for (auto it = b; it != e; it++) s << (it == b ? "" : " ") << *it;
s << "]";
return s;
}
template <typename A, typename B>
ostream& operator<<(ostream& s, const pair<A, B>& p) {
return s << "(" << p.first << "," << p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& s, const vector<T>& c) {
return _out(s, begin(c), end(c));
}
bool debug = 0;
template <typename T>
void DEBUG(const T& x) {
if (debug) cerr << x;
}
template <typename T, typename... Args>
void DEBUG(const T& head, const Args&... tail) {
if (debug) {
cerr << head;
DEBUG(tail...);
}
}
template <typename T>
struct Point {
static constexpr long double EPS = 1e-8;
T x, y;
Point(T _ = 0, T __ = 0) : x(_), y(__) {}
template <typename T2>
Point(const Point<T2>& a) : x(a.x), y(a.y) {}
inline long double theta() const {
return atan2((long double)y, (long double)x);
}
inline long double dis() const {
return hypot((long double)x, (long double)y);
}
inline long double dis(const Point& o) const {
return hypot((long double)(x - o.x), (long double)(y - o.y));
}
Point operator-(const Point& o) const { return Point(x - o.x, y - o.y); }
Point operator-=(const Point& o) {
x -= o.x, y -= o.y;
return *this;
}
Point operator+(const Point& o) const { return Point(x + o.x, y + o.y); }
Point operator+=(const Point& o) {
x += o.x, y += o.y;
return *this;
}
Point operator*(const T& k) const { return Point(x * k, y * k); }
Point operator*=(const T& k) {
x *= k, y *= k;
return *this;
}
Point operator/(const T& k) const { return Point(x / k, y / k); }
Point operator/=(const T& k) {
x /= k, y /= k;
return *this;
}
Point operator-() const { return Point(-x, -y); }
Point rot90() const { return Point(-y, x); }
bool equal(const Point& o, true_type) const {
return fabs(x - o.x) < EPS and fabs(y - o.y) < EPS;
}
bool equal(const Point& o, false_type) const {
return tie(x, y) == tie(o.x, o.y);
}
bool operator==(const Point& o) const {
return equal(o, is_floating_point<T>());
}
bool operator!=(const Point& o) const { return !(*this == o); }
bool operator<(const Point& o) const { return theta() < o.theta(); }
friend inline T cross(const Point& a, const Point& b) {
return a.x * b.y - b.x * a.y;
}
friend inline T dot(const Point& a, const Point& b) {
return a.x * b.x + a.y * b.y;
}
friend ostream& operator<<(ostream& ss, const Point& o) {
ss << "(" << o.x << ", " << o.y << ")";
return ss;
}
};
template <typename T>
class ConvexHull_2D {
private:
vector<Point<T>> dots;
struct myhash {
uint64_t operator()(const Point<T>& a) const {
uint64_t xx = 0, yy = 0;
memcpy(&xx, &a.x, sizeof(a.x));
memcpy(&yy, &a.y, sizeof(a.y));
uint64_t ret = xx * 17 + yy * 31;
ret = (ret ^ (ret >> 16)) * 0x9E3779B1;
ret = (ret ^ (ret >> 13)) * 0xC2B2AE35;
ret = ret ^ xx;
return (ret ^ (ret << 3)) * yy;
}
};
unordered_set<Point<T>, myhash> in_hull;
public:
inline void init() {
in_hull.clear();
dots.clear();
}
void insert(const Point<T>& x) { dots.push_back(x); }
void solve() {
sort(begin(dots), end(dots), [](const Point<T>& a, const Point<T>& b) {
return tie(a.x, a.y) < tie(b.x, b.y);
});
vector<Point<T>> stk((int)((dots).size()) << 1);
int top = 0;
for (auto p : dots) {
while (top >= 2 and
cross(p - stk[top - 2], stk[top - 1] - stk[top - 2]) <= 0)
top--;
stk[top++] = p;
}
for (int i = (int)((dots).size()) - 2, t = top + 1; i >= 0; i--) {
while (top >= t and
cross(dots[i] - stk[top - 2], stk[top - 1] - stk[top - 2]) <= 0)
top--;
stk[top++] = dots[i];
}
stk.resize(top - 1);
swap(stk, dots);
for (auto i : stk) in_hull.insert(i);
}
vector<Point<T>> get() { return dots; }
inline bool in_it(const Point<T>& x) {
return in_hull.find(x) != in_hull.end();
}
};
using PTL = Point<int64_t>;
ConvexHull_2D<int64_t> cv;
template <typename T>
bool check(const vector<T>&, const vector<T>&);
vector<int64_t> mcp(vector<int64_t> s) {
int n = (int)((s).size());
for (int i = 0; i < n; i++) s.push_back(s[i]);
int i = 0, j = 1;
while (i < n && j < n) {
int k = 0;
while (k < n && s[i + k] == s[j + k]) k++;
if (s[i + k] <= s[j + k])
j += k + 1;
else
i += k + 1;
if (i == j) j++;
}
int ans = i < n ? i : j;
return vector<int64_t>(next(s.begin(), ans), next(s.begin(), ans + n));
}
int main(int argc, char* argv[]) {
if (argc > 1 and string(argv[1]) == "-D") debug = 1;
if (!debug) {
ios_base::sync_with_stdio(0);
cin.tie(0);
}
auto kill = []() {
cout << "NO" << '\n';
exit(0);
};
int n, m;
cin >> n >> m;
cv.init();
for (int i = 0; i < n; i++) {
int64_t x, y;
cin >> x >> y;
cv.insert(PTL(x, y));
}
cv.solve();
vector<PTL> one = cv.get();
cv.init();
for (int i = 0; i < m; i++) {
int64_t x, y;
cin >> x >> y;
cv.insert(PTL(x, y));
}
cv.solve();
vector<PTL> two = cv.get();
if (debug)
cerr << __PRETTY_FUNCTION__ << ":" << 221 << " - " << ("one") << "="
<< (one) << '\n';
if (debug)
cerr << __PRETTY_FUNCTION__ << ":" << 222 << " - " << ("two") << "="
<< (two) << '\n';
if ((int)((one).size()) != (int)((two).size())) kill();
vector<int64_t> d1, d2;
for (int i = 1; i < (int)((one).size()); i++)
d1.push_back(dot(one[i] - one[i - 1], one[i] - one[i - 1]));
d1.push_back(dot(one.back() - one[0], one.back() - one[0]));
for (int i = 1; i < (int)((two).size()); i++)
d2.push_back(dot(two[i] - two[i - 1], two[i] - two[i - 1]));
d2.push_back(dot(two.back() - two[0], two.back() - two[0]));
if (debug)
cerr << __PRETTY_FUNCTION__ << ":" << 231 << " - " << ("d1") << "=" << (d1)
<< '\n';
if (debug)
cerr << __PRETTY_FUNCTION__ << ":" << 232 << " - " << ("d2") << "=" << (d2)
<< '\n';
if (!check(d1, d2)) kill();
vector<int64_t> t1, t2;
t1.push_back(
dot(one[0] - one.back(), one[(int)((one).size()) - 2] - one.back()));
t1.push_back(dot(one[1] - one[0], one.back() - one[0]));
for (int i = 2; i < (int)((one).size()); i++)
t1.push_back(dot(one[i] - one[i - 1], one[i - 2] - one[i - 1]));
t2.push_back(
dot(two[0] - two.back(), two[(int)((two).size()) - 2] - two.back()));
t2.push_back(dot(two[1] - two[0], two.back() - two[0]));
for (int i = 2; i < (int)((two).size()); i++)
t2.push_back(dot(two[i] - two[i - 1], two[i - 2] - two[i - 1]));
if (debug)
cerr << __PRETTY_FUNCTION__ << ":" << 243 << " - " << ("t1") << "=" << (t1)
<< '\n';
if (debug)
cerr << __PRETTY_FUNCTION__ << ":" << 244 << " - " << ("t2") << "=" << (t2)
<< '\n';
if (!check(t1, t2)) kill();
cout << "YES" << '\n';
return 0;
}
template <typename T>
bool check(const vector<T>& a, const vector<T>& b) {
return mcp(a) == mcp(b);
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <typename T>
using maxHeap = priority_queue<T, vector<T>, less<T>>;
template <typename T>
using minHeap = priority_queue<T, vector<T>, greater<T>>;
template <typename Iter>
ostream& _out(ostream& s, Iter b, Iter e) {
s << "[";
for (auto it = b; it != e; it++) s << (it == b ? "" : " ") << *it;
s << "]";
return s;
}
template <typename A, typename B>
ostream& operator<<(ostream& s, const pair<A, B>& p) {
return s << "(" << p.first << "," << p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& s, const vector<T>& c) {
return _out(s, begin(c), end(c));
}
bool debug = 0;
template <typename T>
void DEBUG(const T& x) {
if (debug) cerr << x;
}
template <typename T, typename... Args>
void DEBUG(const T& head, const Args&... tail) {
if (debug) {
cerr << head;
DEBUG(tail...);
}
}
template <typename T>
struct Point {
static constexpr long double EPS = 1e-8;
T x, y;
Point(T _ = 0, T __ = 0) : x(_), y(__) {}
template <typename T2>
Point(const Point<T2>& a) : x(a.x), y(a.y) {}
inline long double theta() const {
return atan2((long double)y, (long double)x);
}
inline long double dis() const {
return hypot((long double)x, (long double)y);
}
inline long double dis(const Point& o) const {
return hypot((long double)(x - o.x), (long double)(y - o.y));
}
Point operator-(const Point& o) const { return Point(x - o.x, y - o.y); }
Point operator-=(const Point& o) {
x -= o.x, y -= o.y;
return *this;
}
Point operator+(const Point& o) const { return Point(x + o.x, y + o.y); }
Point operator+=(const Point& o) {
x += o.x, y += o.y;
return *this;
}
Point operator*(const T& k) const { return Point(x * k, y * k); }
Point operator*=(const T& k) {
x *= k, y *= k;
return *this;
}
Point operator/(const T& k) const { return Point(x / k, y / k); }
Point operator/=(const T& k) {
x /= k, y /= k;
return *this;
}
Point operator-() const { return Point(-x, -y); }
Point rot90() const { return Point(-y, x); }
bool equal(const Point& o, true_type) const {
return fabs(x - o.x) < EPS and fabs(y - o.y) < EPS;
}
bool equal(const Point& o, false_type) const {
return tie(x, y) == tie(o.x, o.y);
}
bool operator==(const Point& o) const {
return equal(o, is_floating_point<T>());
}
bool operator!=(const Point& o) const { return !(*this == o); }
bool operator<(const Point& o) const { return theta() < o.theta(); }
friend inline T cross(const Point& a, const Point& b) {
return a.x * b.y - b.x * a.y;
}
friend inline T dot(const Point& a, const Point& b) {
return a.x * b.x + a.y * b.y;
}
friend ostream& operator<<(ostream& ss, const Point& o) {
ss << "(" << o.x << ", " << o.y << ")";
return ss;
}
};
template <typename T>
class ConvexHull_2D {
private:
vector<Point<T>> dots;
struct myhash {
uint64_t operator()(const Point<T>& a) const {
uint64_t xx = 0, yy = 0;
memcpy(&xx, &a.x, sizeof(a.x));
memcpy(&yy, &a.y, sizeof(a.y));
uint64_t ret = xx * 17 + yy * 31;
ret = (ret ^ (ret >> 16)) * 0x9E3779B1;
ret = (ret ^ (ret >> 13)) * 0xC2B2AE35;
ret = ret ^ xx;
return (ret ^ (ret << 3)) * yy;
}
};
unordered_set<Point<T>, myhash> in_hull;
public:
inline void init() {
in_hull.clear();
dots.clear();
}
void insert(const Point<T>& x) { dots.push_back(x); }
void solve() {
sort(begin(dots), end(dots), [](const Point<T>& a, const Point<T>& b) {
return tie(a.x, a.y) < tie(b.x, b.y);
});
vector<Point<T>> stk((int)((dots).size()) << 1);
int top = 0;
for (auto p : dots) {
while (top >= 2 and
cross(p - stk[top - 2], stk[top - 1] - stk[top - 2]) <= 0)
top--;
stk[top++] = p;
}
for (int i = (int)((dots).size()) - 2, t = top + 1; i >= 0; i--) {
while (top >= t and
cross(dots[i] - stk[top - 2], stk[top - 1] - stk[top - 2]) <= 0)
top--;
stk[top++] = dots[i];
}
stk.resize(top - 1);
swap(stk, dots);
for (auto i : stk) in_hull.insert(i);
}
vector<Point<T>> get() { return dots; }
inline bool in_it(const Point<T>& x) {
return in_hull.find(x) != in_hull.end();
}
};
using PTL = Point<int64_t>;
ConvexHull_2D<int64_t> cv;
template <typename T>
bool check(const vector<T>&, const vector<T>&);
vector<int64_t> mcp(vector<int64_t> s) {
int n = (int)((s).size());
for (int i = 0; i < n; i++) s.push_back(s[i]);
int i = 0, j = 1;
while (i < n && j < n) {
int k = 0;
while (k < n && s[i + k] == s[j + k]) k++;
if (s[i + k] <= s[j + k])
j += k + 1;
else
i += k + 1;
if (i == j) j++;
}
int ans = i < n ? i : j;
return vector<int64_t>(next(s.begin(), ans), next(s.begin(), ans + n));
}
vector<int64_t> get_it(const vector<PTL>& a) {
vector<int64_t> ret;
ret.push_back(dot(a.back() - a[0], a.back() - a[0]));
ret.push_back(dot(a.back() - a[0], a[1] - a[0]));
for (int i = 1; i < (int)((a).size()) - 1; i++) {
ret.push_back(dot(a[i - 1] - a[i], a[i - 1] - a[i]));
ret.push_back(dot(a[i - 1] - a[i], a[i + 1] - a[i]));
}
ret.push_back(dot(a[(int)((a).size()) - 2] - a[(int)((a).size()) - 1],
a[(int)((a).size()) - 2] - a[(int)((a).size()) - 1]));
ret.push_back(dot(a[(int)((a).size()) - 2] - a[(int)((a).size()) - 1],
a[0] - a[(int)((a).size()) - 1]));
return ret;
}
int main(int argc, char* argv[]) {
if (argc > 1 and string(argv[1]) == "-D") debug = 1;
if (!debug) {
ios_base::sync_with_stdio(0);
cin.tie(0);
}
auto kill = []() {
cout << "NO" << '\n';
exit(0);
};
int n, m;
cin >> n >> m;
cv.init();
for (int i = 0; i < n; i++) {
int64_t x, y;
cin >> x >> y;
cv.insert(PTL(x, y));
}
cv.solve();
vector<PTL> one = cv.get();
cv.init();
for (int i = 0; i < m; i++) {
int64_t x, y;
cin >> x >> y;
cv.insert(PTL(x, y));
}
cv.solve();
vector<PTL> two = cv.get();
if (debug)
cerr << __PRETTY_FUNCTION__ << ":" << 233 << " - " << ("one") << "="
<< (one) << '\n';
if (debug)
cerr << __PRETTY_FUNCTION__ << ":" << 234 << " - " << ("two") << "="
<< (two) << '\n';
if ((int)((one).size()) != (int)((two).size())) kill();
if (!check(get_it(one), get_it(two))) kill();
cout << "YES" << '\n';
return 0;
}
template <typename T>
bool check(const vector<T>& a, const vector<T>& b) {
return mcp(a) == mcp(b);
}
| ### Prompt
Generate a Cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
using maxHeap = priority_queue<T, vector<T>, less<T>>;
template <typename T>
using minHeap = priority_queue<T, vector<T>, greater<T>>;
template <typename Iter>
ostream& _out(ostream& s, Iter b, Iter e) {
s << "[";
for (auto it = b; it != e; it++) s << (it == b ? "" : " ") << *it;
s << "]";
return s;
}
template <typename A, typename B>
ostream& operator<<(ostream& s, const pair<A, B>& p) {
return s << "(" << p.first << "," << p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& s, const vector<T>& c) {
return _out(s, begin(c), end(c));
}
bool debug = 0;
template <typename T>
void DEBUG(const T& x) {
if (debug) cerr << x;
}
template <typename T, typename... Args>
void DEBUG(const T& head, const Args&... tail) {
if (debug) {
cerr << head;
DEBUG(tail...);
}
}
template <typename T>
struct Point {
static constexpr long double EPS = 1e-8;
T x, y;
Point(T _ = 0, T __ = 0) : x(_), y(__) {}
template <typename T2>
Point(const Point<T2>& a) : x(a.x), y(a.y) {}
inline long double theta() const {
return atan2((long double)y, (long double)x);
}
inline long double dis() const {
return hypot((long double)x, (long double)y);
}
inline long double dis(const Point& o) const {
return hypot((long double)(x - o.x), (long double)(y - o.y));
}
Point operator-(const Point& o) const { return Point(x - o.x, y - o.y); }
Point operator-=(const Point& o) {
x -= o.x, y -= o.y;
return *this;
}
Point operator+(const Point& o) const { return Point(x + o.x, y + o.y); }
Point operator+=(const Point& o) {
x += o.x, y += o.y;
return *this;
}
Point operator*(const T& k) const { return Point(x * k, y * k); }
Point operator*=(const T& k) {
x *= k, y *= k;
return *this;
}
Point operator/(const T& k) const { return Point(x / k, y / k); }
Point operator/=(const T& k) {
x /= k, y /= k;
return *this;
}
Point operator-() const { return Point(-x, -y); }
Point rot90() const { return Point(-y, x); }
bool equal(const Point& o, true_type) const {
return fabs(x - o.x) < EPS and fabs(y - o.y) < EPS;
}
bool equal(const Point& o, false_type) const {
return tie(x, y) == tie(o.x, o.y);
}
bool operator==(const Point& o) const {
return equal(o, is_floating_point<T>());
}
bool operator!=(const Point& o) const { return !(*this == o); }
bool operator<(const Point& o) const { return theta() < o.theta(); }
friend inline T cross(const Point& a, const Point& b) {
return a.x * b.y - b.x * a.y;
}
friend inline T dot(const Point& a, const Point& b) {
return a.x * b.x + a.y * b.y;
}
friend ostream& operator<<(ostream& ss, const Point& o) {
ss << "(" << o.x << ", " << o.y << ")";
return ss;
}
};
template <typename T>
class ConvexHull_2D {
private:
vector<Point<T>> dots;
struct myhash {
uint64_t operator()(const Point<T>& a) const {
uint64_t xx = 0, yy = 0;
memcpy(&xx, &a.x, sizeof(a.x));
memcpy(&yy, &a.y, sizeof(a.y));
uint64_t ret = xx * 17 + yy * 31;
ret = (ret ^ (ret >> 16)) * 0x9E3779B1;
ret = (ret ^ (ret >> 13)) * 0xC2B2AE35;
ret = ret ^ xx;
return (ret ^ (ret << 3)) * yy;
}
};
unordered_set<Point<T>, myhash> in_hull;
public:
inline void init() {
in_hull.clear();
dots.clear();
}
void insert(const Point<T>& x) { dots.push_back(x); }
void solve() {
sort(begin(dots), end(dots), [](const Point<T>& a, const Point<T>& b) {
return tie(a.x, a.y) < tie(b.x, b.y);
});
vector<Point<T>> stk((int)((dots).size()) << 1);
int top = 0;
for (auto p : dots) {
while (top >= 2 and
cross(p - stk[top - 2], stk[top - 1] - stk[top - 2]) <= 0)
top--;
stk[top++] = p;
}
for (int i = (int)((dots).size()) - 2, t = top + 1; i >= 0; i--) {
while (top >= t and
cross(dots[i] - stk[top - 2], stk[top - 1] - stk[top - 2]) <= 0)
top--;
stk[top++] = dots[i];
}
stk.resize(top - 1);
swap(stk, dots);
for (auto i : stk) in_hull.insert(i);
}
vector<Point<T>> get() { return dots; }
inline bool in_it(const Point<T>& x) {
return in_hull.find(x) != in_hull.end();
}
};
using PTL = Point<int64_t>;
ConvexHull_2D<int64_t> cv;
template <typename T>
bool check(const vector<T>&, const vector<T>&);
vector<int64_t> mcp(vector<int64_t> s) {
int n = (int)((s).size());
for (int i = 0; i < n; i++) s.push_back(s[i]);
int i = 0, j = 1;
while (i < n && j < n) {
int k = 0;
while (k < n && s[i + k] == s[j + k]) k++;
if (s[i + k] <= s[j + k])
j += k + 1;
else
i += k + 1;
if (i == j) j++;
}
int ans = i < n ? i : j;
return vector<int64_t>(next(s.begin(), ans), next(s.begin(), ans + n));
}
vector<int64_t> get_it(const vector<PTL>& a) {
vector<int64_t> ret;
ret.push_back(dot(a.back() - a[0], a.back() - a[0]));
ret.push_back(dot(a.back() - a[0], a[1] - a[0]));
for (int i = 1; i < (int)((a).size()) - 1; i++) {
ret.push_back(dot(a[i - 1] - a[i], a[i - 1] - a[i]));
ret.push_back(dot(a[i - 1] - a[i], a[i + 1] - a[i]));
}
ret.push_back(dot(a[(int)((a).size()) - 2] - a[(int)((a).size()) - 1],
a[(int)((a).size()) - 2] - a[(int)((a).size()) - 1]));
ret.push_back(dot(a[(int)((a).size()) - 2] - a[(int)((a).size()) - 1],
a[0] - a[(int)((a).size()) - 1]));
return ret;
}
int main(int argc, char* argv[]) {
if (argc > 1 and string(argv[1]) == "-D") debug = 1;
if (!debug) {
ios_base::sync_with_stdio(0);
cin.tie(0);
}
auto kill = []() {
cout << "NO" << '\n';
exit(0);
};
int n, m;
cin >> n >> m;
cv.init();
for (int i = 0; i < n; i++) {
int64_t x, y;
cin >> x >> y;
cv.insert(PTL(x, y));
}
cv.solve();
vector<PTL> one = cv.get();
cv.init();
for (int i = 0; i < m; i++) {
int64_t x, y;
cin >> x >> y;
cv.insert(PTL(x, y));
}
cv.solve();
vector<PTL> two = cv.get();
if (debug)
cerr << __PRETTY_FUNCTION__ << ":" << 233 << " - " << ("one") << "="
<< (one) << '\n';
if (debug)
cerr << __PRETTY_FUNCTION__ << ":" << 234 << " - " << ("two") << "="
<< (two) << '\n';
if ((int)((one).size()) != (int)((two).size())) kill();
if (!check(get_it(one), get_it(two))) kill();
cout << "YES" << '\n';
return 0;
}
template <typename T>
bool check(const vector<T>& a, const vector<T>& b) {
return mcp(a) == mcp(b);
}
``` |
#include <bits/stdc++.h>
using namespace std;
struct vec {
long long first, second;
vec() {}
vec(long long first, long long second) : first(first), second(second) {}
vec operator+(const vec& T) const {
return {first + T.first, second + T.second};
}
vec operator-(const vec& T) const {
return {first - T.first, second - T.second};
}
vec operator*(const long long& T) const { return {first * T, second * T}; }
long long dot(const vec& T) const {
return first * T.first + second * T.second;
}
long long cross(const vec& T) const {
return first * T.second - second * T.first;
}
long long len() { return first * first + second * second; }
bool operator==(const vec& T) const {
return first == T.first && second == T.second;
}
};
bool CHcmp(vec u, vec v) {
if (u.first != v.first) return u.first < v.first;
return u.second < v.second;
}
vector<vec> convex_hull(vector<vec> p) {
sort(p.begin(), p.end(), CHcmp);
vector<vec> hull;
for (int i = 0; i < p.size(); i++) {
while (hull.size() > 1 && (hull.back() - hull[hull.size() - 2])
.cross(p[i] - hull[hull.size() - 2]) <= 0)
hull.pop_back();
hull.push_back(p[i]);
}
int tmp = (int)hull.size();
for (int i = p.size() - 2; i >= 0; i--) {
while (hull.size() > tmp &&
(hull.back() - hull[hull.size() - 2])
.cross(p[i] - hull[hull.size() - 2]) <= 0)
hull.pop_back();
hull.push_back(p[i]);
}
if (p.size() > 1) hull.pop_back();
return hull;
}
int n, m, sz;
vector<vec> ina, inb, ha, hb;
vector<pair<long long, long long> > sa, sb;
long long val[200055], ps[200055];
bool solve(long long p1, long long p2, long long md) {
long long tar = 0;
for (int i = 0; i < sz; i++) {
long long num = sb[i].first * p1 + sb[i].second * p2;
num %= md;
tar = (tar + num * (i + 1)) % md;
}
long long cur = 0;
for (int i = 0; i < sz + sz; i++) {
val[i] = sa[i].first * p1 + sa[i].second * p2;
val[i] %= md;
if (i < sz) cur = (cur + val[i] * (i + 1)) % md;
if (i > 0)
ps[i] = (ps[i - 1] + val[i]) % md;
else
ps[i] = val[i];
}
if (cur == tar) return true;
for (int i = sz; i < sz + sz; i++) {
cur -= ps[i - 1] - (i - 1 - sz >= 0 ? ps[i - 1 - sz] : 0);
cur %= md;
cur = (cur + md) % md;
cur = (cur + val[i] * sz) % md;
if (cur == tar) return true;
}
return false;
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++) {
long long first, second;
scanf("%lld%lld", &first, &second);
ina.push_back(vec(first, second));
}
for (int i = 0; i < m; i++) {
long long first, second;
scanf("%lld%lld", &first, &second);
inb.push_back(vec(first, second));
}
ha = convex_hull(ina);
hb = convex_hull(inb);
if (ha.size() != hb.size()) {
printf("no\n");
return 0;
}
sz = ha.size();
ha.push_back(ha[0]);
ha.push_back(ha[1]);
hb.push_back(hb[0]);
hb.push_back(hb[1]);
for (int i = 0; i < sz; i++) {
long long len = (ha[i] - ha[i + 1]).len();
long long ang = (ha[i] - ha[i + 1]).dot(ha[i] - ha[i + 2]);
sa.push_back({len, ang});
}
for (int i = 0; i < sz; i++) {
long long len = (hb[i] - hb[i + 1]).len();
long long ang = (hb[i] - hb[i + 1]).dot(hb[i] - hb[i + 2]);
sb.push_back({len, ang});
}
for (int i = 0; i < sz; i++) sa.push_back(sa[i]);
if (solve(29, 31, 12582917) && solve(57, 41, 12582917) &&
solve(67, 07, 12582917) && solve(37, 97, 12582917) &&
solve(29, 31, 402653189) && solve(57, 41, 402653189) &&
solve(67, 07, 402653189) && solve(37, 97, 402653189) &&
solve(29, 31, 1610612741) && solve(57, 41, 1610612741) &&
solve(67, 07, 1610612741) && solve(37, 97, 1610612741)) {
printf("YES\n");
} else {
printf("NO\n");
}
return 0;
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
struct vec {
long long first, second;
vec() {}
vec(long long first, long long second) : first(first), second(second) {}
vec operator+(const vec& T) const {
return {first + T.first, second + T.second};
}
vec operator-(const vec& T) const {
return {first - T.first, second - T.second};
}
vec operator*(const long long& T) const { return {first * T, second * T}; }
long long dot(const vec& T) const {
return first * T.first + second * T.second;
}
long long cross(const vec& T) const {
return first * T.second - second * T.first;
}
long long len() { return first * first + second * second; }
bool operator==(const vec& T) const {
return first == T.first && second == T.second;
}
};
bool CHcmp(vec u, vec v) {
if (u.first != v.first) return u.first < v.first;
return u.second < v.second;
}
vector<vec> convex_hull(vector<vec> p) {
sort(p.begin(), p.end(), CHcmp);
vector<vec> hull;
for (int i = 0; i < p.size(); i++) {
while (hull.size() > 1 && (hull.back() - hull[hull.size() - 2])
.cross(p[i] - hull[hull.size() - 2]) <= 0)
hull.pop_back();
hull.push_back(p[i]);
}
int tmp = (int)hull.size();
for (int i = p.size() - 2; i >= 0; i--) {
while (hull.size() > tmp &&
(hull.back() - hull[hull.size() - 2])
.cross(p[i] - hull[hull.size() - 2]) <= 0)
hull.pop_back();
hull.push_back(p[i]);
}
if (p.size() > 1) hull.pop_back();
return hull;
}
int n, m, sz;
vector<vec> ina, inb, ha, hb;
vector<pair<long long, long long> > sa, sb;
long long val[200055], ps[200055];
bool solve(long long p1, long long p2, long long md) {
long long tar = 0;
for (int i = 0; i < sz; i++) {
long long num = sb[i].first * p1 + sb[i].second * p2;
num %= md;
tar = (tar + num * (i + 1)) % md;
}
long long cur = 0;
for (int i = 0; i < sz + sz; i++) {
val[i] = sa[i].first * p1 + sa[i].second * p2;
val[i] %= md;
if (i < sz) cur = (cur + val[i] * (i + 1)) % md;
if (i > 0)
ps[i] = (ps[i - 1] + val[i]) % md;
else
ps[i] = val[i];
}
if (cur == tar) return true;
for (int i = sz; i < sz + sz; i++) {
cur -= ps[i - 1] - (i - 1 - sz >= 0 ? ps[i - 1 - sz] : 0);
cur %= md;
cur = (cur + md) % md;
cur = (cur + val[i] * sz) % md;
if (cur == tar) return true;
}
return false;
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++) {
long long first, second;
scanf("%lld%lld", &first, &second);
ina.push_back(vec(first, second));
}
for (int i = 0; i < m; i++) {
long long first, second;
scanf("%lld%lld", &first, &second);
inb.push_back(vec(first, second));
}
ha = convex_hull(ina);
hb = convex_hull(inb);
if (ha.size() != hb.size()) {
printf("no\n");
return 0;
}
sz = ha.size();
ha.push_back(ha[0]);
ha.push_back(ha[1]);
hb.push_back(hb[0]);
hb.push_back(hb[1]);
for (int i = 0; i < sz; i++) {
long long len = (ha[i] - ha[i + 1]).len();
long long ang = (ha[i] - ha[i + 1]).dot(ha[i] - ha[i + 2]);
sa.push_back({len, ang});
}
for (int i = 0; i < sz; i++) {
long long len = (hb[i] - hb[i + 1]).len();
long long ang = (hb[i] - hb[i + 1]).dot(hb[i] - hb[i + 2]);
sb.push_back({len, ang});
}
for (int i = 0; i < sz; i++) sa.push_back(sa[i]);
if (solve(29, 31, 12582917) && solve(57, 41, 12582917) &&
solve(67, 07, 12582917) && solve(37, 97, 12582917) &&
solve(29, 31, 402653189) && solve(57, 41, 402653189) &&
solve(67, 07, 402653189) && solve(37, 97, 402653189) &&
solve(29, 31, 1610612741) && solve(57, 41, 1610612741) &&
solve(67, 07, 1610612741) && solve(37, 97, 1610612741)) {
printf("YES\n");
} else {
printf("NO\n");
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-9;
inline int cmp(double x, double y = 0, double tol = EPS) {
return ((x <= y + tol) ? (x + tol < y) ? -1 : 0 : 1);
}
struct point {
long long x, y;
point(long long x = 0, long long y = 0) : x(x), y(y) {}
point operator+(point q) { return point(x + q.x, y + q.y); }
point operator-(point q) { return point(x - q.x, y - q.y); }
int cmp(point q) const {
if (int t = ::cmp(x, q.x)) return t;
return ::cmp(y, q.y);
}
bool operator==(point q) const { return cmp(q) == 0; };
bool operator!=(point q) const { return cmp(q) != 0; };
bool operator<(point q) const { return cmp(q) < 0; };
};
ostream &operator<<(ostream &os, const point &p) {
os << "(" << p.x << "," << p.y << ")";
}
long long dot(point a, point b) { return a.x * b.x + a.y * b.y; }
long long cross(point a, point b) { return a.x * b.y - a.y * b.x; }
long long abs2(point a) { return dot(a, a); }
bool collinear(point a, point b, point c) {
return cmp(cross((b - a), (c - a))) == 0;
}
int ccw(point a, point b, point p) { return cmp(cross((b - a), (p - a))); }
point pivot;
bool radial_lt(point a, point b) {
int R = ccw(pivot, a, b);
if (R == 0)
return dot(pivot - a, pivot - a) < dot(pivot - b, pivot - b);
else
return (R == 1);
}
vector<point> convexhull(vector<point> T) {
sort(T.begin(), T.end());
T.resize(unique(T.begin(), T.end()) - T.begin());
int tam = 0, n = T.size();
vector<point> U;
int idx = min_element(T.begin(), T.end()) - T.begin();
pivot = T[idx];
swap(T[0], T[idx]);
sort(++T.begin(), T.end(), radial_lt);
for (int i = 0; i < T.size(); i++) {
while (tam > 1 && ccw(U[tam - 2], U[tam - 1], T[i]) <= 0)
U.pop_back(), tam--;
U.push_back(T[i]);
tam++;
}
return U;
}
void show(pair<pair<long long, long long>, pair<long long, long long> > s) {
cout << s.first.first << " " << s.first.second << " " << s.second.first << " "
<< s.second.second << endl;
}
int least_rot(
vector<pair<pair<long long, long long>, pair<long long, long long> > >
second) {
int n = second.size();
for (int i = (0); i < (n); i++) second.push_back(second[i]);
vector<int> f(second.size(), -1);
int k = 0;
for (int j = 1; j < second.size(); j++) {
pair<pair<long long, long long>, pair<long long, long long> > sj =
second[j];
int i = f[j - k - 1];
while (i != -1 && sj != second[k + i + 1]) {
if (sj < second[k + i + 1]) k = j - i - 1;
i = f[i];
}
if (sj != second[k + i + 1]) {
if (sj < second[k]) k = j;
f[j - k] = -1;
} else
f[j - k] = i + 1;
}
return k;
}
int main() {
int n, m;
scanf("%d %d", &n, &m);
vector<point> A(n), B(m);
point p;
for (int i = (0); i < (n); i++) {
scanf("%lld %lld", &p.x, &p.y);
A[i] = p;
}
for (int i = (0); i < (m); i++) {
scanf("%lld %lld", &p.x, &p.y);
B[i] = p;
}
A = convexhull(A);
B = convexhull(B);
n = A.size();
m = B.size();
if (n != m) {
cout << "NO\n";
return 0;
}
vector<pair<pair<long long, long long>, pair<long long, long long> > > AA, BB;
point esq, dir;
int ee, dd;
for (int i = (0); i < (n); i++) {
ee = i - 1;
if (ee < 0) ee += n;
dd = i + 1;
if (dd >= n) dd %= n;
esq = A[ee] - A[i], dir = A[dd] - A[i];
AA.push_back(make_pair(make_pair(cross(esq, dir), dot(esq, dir)),
make_pair(abs2(esq), abs2(dir))));
}
for (int i = (0); i < (n); i++) {
ee = i - 1;
if (ee < 0) ee += n;
dd = i + 1;
if (dd >= n) dd %= n;
esq = B[ee] - B[i], dir = B[dd] - B[i];
BB.push_back(make_pair(make_pair(cross(esq, dir), dot(esq, dir)),
make_pair(abs2(esq), abs2(dir))));
}
int u = least_rot(AA);
int v = least_rot(BB);
bool ok = true;
for (int i = (0); i < (n); i++) {
if (AA[u] != BB[v]) {
ok = false;
break;
}
u = (u + 1) % n;
v = (v + 1) % n;
}
if (ok)
cout << "YES\n";
else
cout << "NO\n";
return 0;
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-9;
inline int cmp(double x, double y = 0, double tol = EPS) {
return ((x <= y + tol) ? (x + tol < y) ? -1 : 0 : 1);
}
struct point {
long long x, y;
point(long long x = 0, long long y = 0) : x(x), y(y) {}
point operator+(point q) { return point(x + q.x, y + q.y); }
point operator-(point q) { return point(x - q.x, y - q.y); }
int cmp(point q) const {
if (int t = ::cmp(x, q.x)) return t;
return ::cmp(y, q.y);
}
bool operator==(point q) const { return cmp(q) == 0; };
bool operator!=(point q) const { return cmp(q) != 0; };
bool operator<(point q) const { return cmp(q) < 0; };
};
ostream &operator<<(ostream &os, const point &p) {
os << "(" << p.x << "," << p.y << ")";
}
long long dot(point a, point b) { return a.x * b.x + a.y * b.y; }
long long cross(point a, point b) { return a.x * b.y - a.y * b.x; }
long long abs2(point a) { return dot(a, a); }
bool collinear(point a, point b, point c) {
return cmp(cross((b - a), (c - a))) == 0;
}
int ccw(point a, point b, point p) { return cmp(cross((b - a), (p - a))); }
point pivot;
bool radial_lt(point a, point b) {
int R = ccw(pivot, a, b);
if (R == 0)
return dot(pivot - a, pivot - a) < dot(pivot - b, pivot - b);
else
return (R == 1);
}
vector<point> convexhull(vector<point> T) {
sort(T.begin(), T.end());
T.resize(unique(T.begin(), T.end()) - T.begin());
int tam = 0, n = T.size();
vector<point> U;
int idx = min_element(T.begin(), T.end()) - T.begin();
pivot = T[idx];
swap(T[0], T[idx]);
sort(++T.begin(), T.end(), radial_lt);
for (int i = 0; i < T.size(); i++) {
while (tam > 1 && ccw(U[tam - 2], U[tam - 1], T[i]) <= 0)
U.pop_back(), tam--;
U.push_back(T[i]);
tam++;
}
return U;
}
void show(pair<pair<long long, long long>, pair<long long, long long> > s) {
cout << s.first.first << " " << s.first.second << " " << s.second.first << " "
<< s.second.second << endl;
}
int least_rot(
vector<pair<pair<long long, long long>, pair<long long, long long> > >
second) {
int n = second.size();
for (int i = (0); i < (n); i++) second.push_back(second[i]);
vector<int> f(second.size(), -1);
int k = 0;
for (int j = 1; j < second.size(); j++) {
pair<pair<long long, long long>, pair<long long, long long> > sj =
second[j];
int i = f[j - k - 1];
while (i != -1 && sj != second[k + i + 1]) {
if (sj < second[k + i + 1]) k = j - i - 1;
i = f[i];
}
if (sj != second[k + i + 1]) {
if (sj < second[k]) k = j;
f[j - k] = -1;
} else
f[j - k] = i + 1;
}
return k;
}
int main() {
int n, m;
scanf("%d %d", &n, &m);
vector<point> A(n), B(m);
point p;
for (int i = (0); i < (n); i++) {
scanf("%lld %lld", &p.x, &p.y);
A[i] = p;
}
for (int i = (0); i < (m); i++) {
scanf("%lld %lld", &p.x, &p.y);
B[i] = p;
}
A = convexhull(A);
B = convexhull(B);
n = A.size();
m = B.size();
if (n != m) {
cout << "NO\n";
return 0;
}
vector<pair<pair<long long, long long>, pair<long long, long long> > > AA, BB;
point esq, dir;
int ee, dd;
for (int i = (0); i < (n); i++) {
ee = i - 1;
if (ee < 0) ee += n;
dd = i + 1;
if (dd >= n) dd %= n;
esq = A[ee] - A[i], dir = A[dd] - A[i];
AA.push_back(make_pair(make_pair(cross(esq, dir), dot(esq, dir)),
make_pair(abs2(esq), abs2(dir))));
}
for (int i = (0); i < (n); i++) {
ee = i - 1;
if (ee < 0) ee += n;
dd = i + 1;
if (dd >= n) dd %= n;
esq = B[ee] - B[i], dir = B[dd] - B[i];
BB.push_back(make_pair(make_pair(cross(esq, dir), dot(esq, dir)),
make_pair(abs2(esq), abs2(dir))));
}
int u = least_rot(AA);
int v = least_rot(BB);
bool ok = true;
for (int i = (0); i < (n); i++) {
if (AA[u] != BB[v]) {
ok = false;
break;
}
u = (u + 1) % n;
v = (v + 1) % n;
}
if (ok)
cout << "YES\n";
else
cout << "NO\n";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
struct Point {
long long x, y;
Point(long long x = 0, long long y = 0) : x(x), y(y) {}
void input() { scanf("%I64d%I64d", &x, &y); }
bool operator<(const Point& R) const {
if (x == R.x)
return y < R.y;
else
return x < R.x;
}
Point operator-(const Point& R) const { return Point(x - R.x, y - R.y); }
long long operator^(const Point& R) const { return x * R.y - y * R.x; }
long long operator%(const Point& R) const { return x * R.x + y * R.y; }
long long len() { return (*this) % (*this); }
};
inline long long OnLeft(Point P, Point A, Point B) { return (B - A) ^ (P - A); }
int Convex(Point* p, Point* s, int n) {
sort(p, p + n);
int top = 0;
for (int i = 0; i < n; i++) {
while (top > 1 && OnLeft(p[i], s[top - 2], s[top - 1]) <= 0) top--;
s[top++] = p[i];
}
int tmp = top;
for (int i = n - 2; i >= 0; i--) {
while (top > tmp && OnLeft(p[i], s[top - 2], s[top - 1]) <= 0) top--;
s[top++] = p[i];
}
if (n > 1) top--;
return top;
}
Point p1[100005], p2[100005], s1[100005], s2[100005];
int sz1, sz2;
struct edge {
long long len1;
long long len2;
long long angle;
edge(long long len1 = 0, long long len2 = 0, long long angle = 0)
: len1(len1), len2(len2), angle(angle) {}
bool operator==(const edge& R) const {
return len1 == R.len1 && len2 == R.len2 && angle == R.angle;
}
bool operator!=(const edge& R) const { return !((*this) == R); }
};
edge ss1[100005], ss2[200005];
int nxt[100005];
void kmp_pre(edge* x, int m, int* nxt) {
int i, j;
j = nxt[0] = -1;
i = 0;
while (i < m) {
while (j != -1 && x[i] != x[j]) j = nxt[j];
nxt[++i] = ++j;
}
}
bool kmp(edge* x, int m, edge* y, int n) {
int i, j;
kmp_pre(x, m, nxt);
i = j = 0;
while (i < n) {
while (j != -1 && y[i] != x[j]) j = nxt[j];
i++;
j++;
if (j >= m) return true;
}
return false;
}
int main() {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++) p1[i].input();
for (int i = 0; i < m; i++) p2[i].input();
sz1 = Convex(p1, s1, n);
sz2 = Convex(p2, s2, m);
if (sz1 != sz2)
printf("NO\n");
else {
for (int i = 0; i < sz1; i++) {
ss1[i].len1 = (s1[i] - s1[(i + sz1 - 1) % sz1]).len();
ss1[i].len2 = (s1[(i + 1) % sz1] - s1[i]).len();
ss1[i].angle =
(s1[i] - s1[(i + sz1 - 1) % sz1]) % (s1[(i + 1) % sz1] - s1[i]);
}
for (int i = 0; i < sz2; i++) {
ss2[i].len1 = (s2[i] - s2[(i + sz2 - 1) % sz2]).len();
ss2[i].len2 = (s2[(i + 1) % sz2] - s2[i]).len();
ss2[i].angle =
(s2[i] - s2[(i + sz2 - 1) % sz2]) % (s2[(i + 1) % sz2] - s2[i]);
}
for (int i = 0; i < sz2; i++) ss2[i + sz2] = ss2[i];
if (kmp(ss1, sz1, ss2, sz2 + sz2))
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
| ### Prompt
Your challenge is to write a cpp solution to the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
struct Point {
long long x, y;
Point(long long x = 0, long long y = 0) : x(x), y(y) {}
void input() { scanf("%I64d%I64d", &x, &y); }
bool operator<(const Point& R) const {
if (x == R.x)
return y < R.y;
else
return x < R.x;
}
Point operator-(const Point& R) const { return Point(x - R.x, y - R.y); }
long long operator^(const Point& R) const { return x * R.y - y * R.x; }
long long operator%(const Point& R) const { return x * R.x + y * R.y; }
long long len() { return (*this) % (*this); }
};
inline long long OnLeft(Point P, Point A, Point B) { return (B - A) ^ (P - A); }
int Convex(Point* p, Point* s, int n) {
sort(p, p + n);
int top = 0;
for (int i = 0; i < n; i++) {
while (top > 1 && OnLeft(p[i], s[top - 2], s[top - 1]) <= 0) top--;
s[top++] = p[i];
}
int tmp = top;
for (int i = n - 2; i >= 0; i--) {
while (top > tmp && OnLeft(p[i], s[top - 2], s[top - 1]) <= 0) top--;
s[top++] = p[i];
}
if (n > 1) top--;
return top;
}
Point p1[100005], p2[100005], s1[100005], s2[100005];
int sz1, sz2;
struct edge {
long long len1;
long long len2;
long long angle;
edge(long long len1 = 0, long long len2 = 0, long long angle = 0)
: len1(len1), len2(len2), angle(angle) {}
bool operator==(const edge& R) const {
return len1 == R.len1 && len2 == R.len2 && angle == R.angle;
}
bool operator!=(const edge& R) const { return !((*this) == R); }
};
edge ss1[100005], ss2[200005];
int nxt[100005];
void kmp_pre(edge* x, int m, int* nxt) {
int i, j;
j = nxt[0] = -1;
i = 0;
while (i < m) {
while (j != -1 && x[i] != x[j]) j = nxt[j];
nxt[++i] = ++j;
}
}
bool kmp(edge* x, int m, edge* y, int n) {
int i, j;
kmp_pre(x, m, nxt);
i = j = 0;
while (i < n) {
while (j != -1 && y[i] != x[j]) j = nxt[j];
i++;
j++;
if (j >= m) return true;
}
return false;
}
int main() {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++) p1[i].input();
for (int i = 0; i < m; i++) p2[i].input();
sz1 = Convex(p1, s1, n);
sz2 = Convex(p2, s2, m);
if (sz1 != sz2)
printf("NO\n");
else {
for (int i = 0; i < sz1; i++) {
ss1[i].len1 = (s1[i] - s1[(i + sz1 - 1) % sz1]).len();
ss1[i].len2 = (s1[(i + 1) % sz1] - s1[i]).len();
ss1[i].angle =
(s1[i] - s1[(i + sz1 - 1) % sz1]) % (s1[(i + 1) % sz1] - s1[i]);
}
for (int i = 0; i < sz2; i++) {
ss2[i].len1 = (s2[i] - s2[(i + sz2 - 1) % sz2]).len();
ss2[i].len2 = (s2[(i + 1) % sz2] - s2[i]).len();
ss2[i].angle =
(s2[i] - s2[(i + sz2 - 1) % sz2]) % (s2[(i + 1) % sz2] - s2[i]);
}
for (int i = 0; i < sz2; i++) ss2[i + sz2] = ss2[i];
if (kmp(ss1, sz1, ss2, sz2 + sz2))
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
``` |
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:512000000")
using namespace std;
using li = long long;
using ld = long double;
void solve(bool);
void precalc();
clock_t start;
int main() {
start = clock();
int t = 1;
cout.sync_with_stdio(0);
cin.tie(0);
cout.precision(20);
cout << fixed;
precalc();
while (t--) {
solve(true);
}
cout.flush();
return 0;
}
template <typename T>
T binpow(T q, T w, T mod) {
if (!w) return 1 % mod;
if (w & 1) return q * 1LL * binpow(q, w - 1, mod) % mod;
return binpow(q * 1LL * q % mod, w / 2, mod);
}
template <typename T>
T gcd(T q, T w) {
while (w) {
q %= w;
swap(q, w);
}
return q;
}
template <typename T>
T lcm(T q, T w) {
return q / gcd(q, w) * w;
}
template <typename T>
void make_unique(vector<T>& vec) {
sort(vec.begin(), vec.end());
vec.erase(unique(vec.begin(), vec.end()), vec.end());
}
template <typename T>
void relax_min(T& cur, T val) {
cur = min(cur, val);
}
template <typename T>
void relax_max(T& cur, T val) {
cur = max(cur, val);
}
void precalc() {}
struct Point {
li x, y;
Point() {}
Point(li x, li y) : x(x), y(y) {}
Point operator-(const Point& ot) const { return Point(x - ot.x, y - ot.y); }
li operator*(const Point& ot) const { return x * ot.y - y * ot.x; }
void scan() { cin >> x >> y; }
bool operator<(const Point& ot) const {
return make_pair(x, y) < make_pair(ot.x, ot.y);
}
li sqr_len() const { return x * x + y * y; }
long double get_len() const { return sqrtl(sqr_len()); }
li operator%(const Point& ot) const { return x * ot.x + y * ot.y; }
};
struct Polygon {
vector<Point> hull;
Polygon(li n) {
vector<Point> points(n);
for (li i = 0; i < n; ++i) {
points[i].scan();
}
sort(points.begin(), points.end());
vector<Point> up, down;
for (auto& pt : points) {
while (up.size() > 1 &&
(up[up.size() - 2] - up.back()) * (pt - up.back()) >= 0) {
up.pop_back();
}
up.push_back(pt);
while (down.size() > 1 &&
(down[down.size() - 2] - down.back()) * (pt - down.back()) <= 0) {
down.pop_back();
}
down.push_back(pt);
}
hull = up;
for (li i = (li)down.size() - 2; i > 0; --i) {
hull.push_back(down[i]);
}
}
li sqr_len(li pos) const {
return (hull[(pos + 1) % hull.size()] - hull[pos]).sqr_len();
}
li size() const { return (li)hull.size(); }
long double get_angle(li pos) const {
auto a = hull[(pos + 1) % hull.size()] - hull[pos],
b = hull[(pos + hull.size() - 1) % hull.size()] - hull[pos];
long double co = (a % b) / a.get_len() / b.get_len();
return co;
}
};
void solve(bool read) {
vector<Polygon> polys;
li n[2];
cin >> n[0] >> n[1];
for (li i = 0; i < 2; ++i) {
polys.emplace_back(n[i]);
}
if (polys[0].size() != polys[1].size()) {
cout << "NO\n";
return;
}
vector<long double> all_lens;
li m = polys[0].size();
for (li i = 0; i < m; ++i) {
all_lens.push_back(polys[0].sqr_len(i));
all_lens.push_back(polys[0].get_angle(i));
}
all_lens.push_back(-1);
for (li j = 0; j < 2; ++j) {
for (li i = 0; i < m; ++i) {
all_lens.push_back(polys[1].sqr_len(i));
all_lens.push_back(polys[1].get_angle(i));
}
}
vector<li> p(all_lens.size());
for (li i = 1; i < p.size(); ++i) {
li j = p[i - 1];
while (j > 0 && all_lens[i] != all_lens[j]) {
j = p[j - 1];
}
if (all_lens[i] == all_lens[j]) {
++j;
}
p[i] = j;
if (p[i] == 2 * m) {
cout << "YES\n";
return;
}
}
cout << "NO\n";
}
| ### Prompt
Create a solution in CPP for the following problem:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:512000000")
using namespace std;
using li = long long;
using ld = long double;
void solve(bool);
void precalc();
clock_t start;
int main() {
start = clock();
int t = 1;
cout.sync_with_stdio(0);
cin.tie(0);
cout.precision(20);
cout << fixed;
precalc();
while (t--) {
solve(true);
}
cout.flush();
return 0;
}
template <typename T>
T binpow(T q, T w, T mod) {
if (!w) return 1 % mod;
if (w & 1) return q * 1LL * binpow(q, w - 1, mod) % mod;
return binpow(q * 1LL * q % mod, w / 2, mod);
}
template <typename T>
T gcd(T q, T w) {
while (w) {
q %= w;
swap(q, w);
}
return q;
}
template <typename T>
T lcm(T q, T w) {
return q / gcd(q, w) * w;
}
template <typename T>
void make_unique(vector<T>& vec) {
sort(vec.begin(), vec.end());
vec.erase(unique(vec.begin(), vec.end()), vec.end());
}
template <typename T>
void relax_min(T& cur, T val) {
cur = min(cur, val);
}
template <typename T>
void relax_max(T& cur, T val) {
cur = max(cur, val);
}
void precalc() {}
struct Point {
li x, y;
Point() {}
Point(li x, li y) : x(x), y(y) {}
Point operator-(const Point& ot) const { return Point(x - ot.x, y - ot.y); }
li operator*(const Point& ot) const { return x * ot.y - y * ot.x; }
void scan() { cin >> x >> y; }
bool operator<(const Point& ot) const {
return make_pair(x, y) < make_pair(ot.x, ot.y);
}
li sqr_len() const { return x * x + y * y; }
long double get_len() const { return sqrtl(sqr_len()); }
li operator%(const Point& ot) const { return x * ot.x + y * ot.y; }
};
struct Polygon {
vector<Point> hull;
Polygon(li n) {
vector<Point> points(n);
for (li i = 0; i < n; ++i) {
points[i].scan();
}
sort(points.begin(), points.end());
vector<Point> up, down;
for (auto& pt : points) {
while (up.size() > 1 &&
(up[up.size() - 2] - up.back()) * (pt - up.back()) >= 0) {
up.pop_back();
}
up.push_back(pt);
while (down.size() > 1 &&
(down[down.size() - 2] - down.back()) * (pt - down.back()) <= 0) {
down.pop_back();
}
down.push_back(pt);
}
hull = up;
for (li i = (li)down.size() - 2; i > 0; --i) {
hull.push_back(down[i]);
}
}
li sqr_len(li pos) const {
return (hull[(pos + 1) % hull.size()] - hull[pos]).sqr_len();
}
li size() const { return (li)hull.size(); }
long double get_angle(li pos) const {
auto a = hull[(pos + 1) % hull.size()] - hull[pos],
b = hull[(pos + hull.size() - 1) % hull.size()] - hull[pos];
long double co = (a % b) / a.get_len() / b.get_len();
return co;
}
};
void solve(bool read) {
vector<Polygon> polys;
li n[2];
cin >> n[0] >> n[1];
for (li i = 0; i < 2; ++i) {
polys.emplace_back(n[i]);
}
if (polys[0].size() != polys[1].size()) {
cout << "NO\n";
return;
}
vector<long double> all_lens;
li m = polys[0].size();
for (li i = 0; i < m; ++i) {
all_lens.push_back(polys[0].sqr_len(i));
all_lens.push_back(polys[0].get_angle(i));
}
all_lens.push_back(-1);
for (li j = 0; j < 2; ++j) {
for (li i = 0; i < m; ++i) {
all_lens.push_back(polys[1].sqr_len(i));
all_lens.push_back(polys[1].get_angle(i));
}
}
vector<li> p(all_lens.size());
for (li i = 1; i < p.size(); ++i) {
li j = p[i - 1];
while (j > 0 && all_lens[i] != all_lens[j]) {
j = p[j - 1];
}
if (all_lens[i] == all_lens[j]) {
++j;
}
p[i] = j;
if (p[i] == 2 * m) {
cout << "YES\n";
return;
}
}
cout << "NO\n";
}
``` |
#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-6;
struct point {
double x, y;
point() {}
point(double x, double y) : x(x), y(y) {}
};
point operator-(point a, point b) { return point(a.x - b.x, a.y - b.y); }
int dcmp(double x) { return fabs(x) < EPS ? 0 : (x > 0 ? 1 : -1); }
double Cross(point a, point b) { return a.x * b.y - b.x * a.y; }
double Distance(point a, point b) {
return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
point p[2][100009], ans[2][100009];
int n, m, top1, top2;
point Tmp;
bool cmp(point a, point b) {
double ans = Cross(a - Tmp, b - Tmp);
if (dcmp(ans) == 0) return dcmp(Distance(a, Tmp) - Distance(b, Tmp)) < 0;
return ans > 0;
}
void Graham(point p[], int n, point ans[], int &top) {
if (n < 3) return;
for (int i = 2; i <= n; i++) {
if (p[i].y < p[1].y || (dcmp(p[i].y - p[1].y) == 0 && p[i].x < p[1].x))
swap(p[1], p[i]);
}
Tmp = p[1];
sort(p + 2, p + 1 + n, cmp);
ans[1] = p[1], ans[2] = p[2], top = 2;
for (int i = 3; i <= n; i++) {
while (top > 2 &&
dcmp(Cross(ans[top] - ans[top - 1], p[i] - ans[top])) <= 0)
top--;
ans[++top] = p[i];
if (top > 2 &&
dcmp(Cross(ans[top] - ans[top - 1], ans[top - 1] - ans[top - 2])) == 0)
ans[top - 1] = ans[top], --top;
}
}
struct node {
double x, y, z;
node() {}
node(double x, double y, double z) : x(x), y(y), z(z) {}
void out() { printf("(%.1f %.1f %.1f)\n", x, y, z); }
bool operator==(const node &R) const {
return dcmp(x - R.x) == 0 && dcmp(y - R.y) == 0 && dcmp(z - R.z) == 0;
}
bool operator!=(const node &R) const { return !(*this == R); }
} e[100009], b[100009];
int fail[100009];
void KMP(int len) {
fail[1] = 0;
int i = 1, j = 0;
while (i <= len) {
while (j && e[i] != e[j]) j = fail[j];
fail[++i] = ++j;
if (e[i] == e[j]) fail[i] = fail[j];
}
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) {
scanf("%lf%lf", &p[0][i].x, &p[0][i].y);
}
for (int i = 1; i <= m; i++) {
scanf("%lf%lf", &p[1][i].x, &p[1][i].y);
}
Graham(p[0], n, ans[0], top1);
Graham(p[1], m, ans[1], top2);
if (top1 - top2) return 0 * printf("NO\n");
if (top1 == 1) return 0 * printf("YES\n");
if (top1 == 2) {
if (dcmp(Distance(ans[0][1], ans[0][2]) - Distance(ans[1][1], ans[1][2])) ==
0)
return 0 * printf("YES\n");
return 0 * printf("NO\n");
}
for (int i = 1; i <= top1; i++) {
int l = i - 1, r = i + 1;
if (l == 0) l = top1;
if (r > top1) r = 1;
e[i] = node(Distance(ans[0][i], ans[0][l]), Distance(ans[0][i], ans[0][r]),
Distance(ans[0][l], ans[0][r]));
}
for (int i = 1; i <= top2; i++) {
int l = i - 1, r = i + 1;
if (l == 0) l = top2;
if (r > top2) r = 1;
b[i] = node(Distance(ans[1][i], ans[1][l]), Distance(ans[1][i], ans[1][r]),
Distance(ans[1][l], ans[1][r]));
}
KMP(top1);
int j = 1;
for (int i = 1; i <= top2; i++) {
while (e[j] != b[i] && j) j = fail[j];
if (j == top1) {
return 0 * printf("YES\n");
}
j++;
}
for (int i = 1; i <= top2; i++) {
while (e[j] != b[i] && j) j = fail[j];
if (j == top1) {
return 0 * printf("YES\n");
}
j++;
}
printf("NO\n");
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
After the war, the supersonic rocket became the most common public transportation.
Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has n power sources, and the second one has m power sources. A power source can be described as a point (x_i, y_i) on a 2-D plane. All points in each engine are different.
You can manipulate each engine separately. There are two operations that you can do with each engine. You can do each operation as many times as you want.
1. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i+a, y_i+b), a and b can be any real numbers. In other words, all power sources will be shifted.
2. For every power source as a whole in that engine: (x_i, y_i) becomes (x_i cos θ - y_i sin θ, x_i sin θ + y_i cos θ), θ can be any real number. In other words, all power sources will be rotated.
The engines work as follows: after the two engines are powered, their power sources are being combined (here power sources of different engines may coincide). If two power sources A(x_a, y_a) and B(x_b, y_b) exist, then for all real number k that 0 < k < 1, a new power source will be created C_k(kx_a+(1-k)x_b,ky_a+(1-k)y_b). Then, this procedure will be repeated again with all new and old power sources. After that, the "power field" from all power sources will be generated (can be considered as an infinite set of all power sources occurred).
A supersonic rocket is "safe" if and only if after you manipulate the engines, destroying any power source and then power the engine, the power field generated won't be changed (comparing to the situation where no power source erased). Two power fields are considered the same if and only if any power source in one field belongs to the other one as well.
Given a supersonic rocket, check whether it is safe or not.
Input
The first line contains two integers n, m (3 ≤ n, m ≤ 10^5) — the number of power sources in each engine.
Each of the next n lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the first engine.
Each of the next m lines contains two integers x_i and y_i (0≤ x_i, y_i≤ 10^8) — the coordinates of the i-th power source in the second engine.
It is guaranteed that there are no two or more power sources that are located in the same point in each engine.
Output
Print "YES" if the supersonic rocket is safe, otherwise "NO".
You can print each letter in an arbitrary case (upper or lower).
Examples
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
1 1
Output
YES
Input
3 4
0 0
0 2
2 0
0 2
2 2
2 0
0 0
Output
NO
Note
The first sample:
<image> Those near pairs of blue and orange points actually coincide.
First, manipulate the first engine: use the second operation with θ = π (to rotate all power sources 180 degrees).
The power sources in the first engine become (0, 0), (0, -2), and (-2, 0).
<image>
Second, manipulate the second engine: use the first operation with a = b = -2.
The power sources in the second engine become (-2, 0), (0, 0), (0, -2), and (-1, -1).
<image>
You can examine that destroying any point, the power field formed by the two engines are always the solid triangle (0, 0), (-2, 0), (0, -2).
In the second sample, no matter how you manipulate the engines, there always exists a power source in the second engine that power field will shrink if you destroy it.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-6;
struct point {
double x, y;
point() {}
point(double x, double y) : x(x), y(y) {}
};
point operator-(point a, point b) { return point(a.x - b.x, a.y - b.y); }
int dcmp(double x) { return fabs(x) < EPS ? 0 : (x > 0 ? 1 : -1); }
double Cross(point a, point b) { return a.x * b.y - b.x * a.y; }
double Distance(point a, point b) {
return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
point p[2][100009], ans[2][100009];
int n, m, top1, top2;
point Tmp;
bool cmp(point a, point b) {
double ans = Cross(a - Tmp, b - Tmp);
if (dcmp(ans) == 0) return dcmp(Distance(a, Tmp) - Distance(b, Tmp)) < 0;
return ans > 0;
}
void Graham(point p[], int n, point ans[], int &top) {
if (n < 3) return;
for (int i = 2; i <= n; i++) {
if (p[i].y < p[1].y || (dcmp(p[i].y - p[1].y) == 0 && p[i].x < p[1].x))
swap(p[1], p[i]);
}
Tmp = p[1];
sort(p + 2, p + 1 + n, cmp);
ans[1] = p[1], ans[2] = p[2], top = 2;
for (int i = 3; i <= n; i++) {
while (top > 2 &&
dcmp(Cross(ans[top] - ans[top - 1], p[i] - ans[top])) <= 0)
top--;
ans[++top] = p[i];
if (top > 2 &&
dcmp(Cross(ans[top] - ans[top - 1], ans[top - 1] - ans[top - 2])) == 0)
ans[top - 1] = ans[top], --top;
}
}
struct node {
double x, y, z;
node() {}
node(double x, double y, double z) : x(x), y(y), z(z) {}
void out() { printf("(%.1f %.1f %.1f)\n", x, y, z); }
bool operator==(const node &R) const {
return dcmp(x - R.x) == 0 && dcmp(y - R.y) == 0 && dcmp(z - R.z) == 0;
}
bool operator!=(const node &R) const { return !(*this == R); }
} e[100009], b[100009];
int fail[100009];
void KMP(int len) {
fail[1] = 0;
int i = 1, j = 0;
while (i <= len) {
while (j && e[i] != e[j]) j = fail[j];
fail[++i] = ++j;
if (e[i] == e[j]) fail[i] = fail[j];
}
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) {
scanf("%lf%lf", &p[0][i].x, &p[0][i].y);
}
for (int i = 1; i <= m; i++) {
scanf("%lf%lf", &p[1][i].x, &p[1][i].y);
}
Graham(p[0], n, ans[0], top1);
Graham(p[1], m, ans[1], top2);
if (top1 - top2) return 0 * printf("NO\n");
if (top1 == 1) return 0 * printf("YES\n");
if (top1 == 2) {
if (dcmp(Distance(ans[0][1], ans[0][2]) - Distance(ans[1][1], ans[1][2])) ==
0)
return 0 * printf("YES\n");
return 0 * printf("NO\n");
}
for (int i = 1; i <= top1; i++) {
int l = i - 1, r = i + 1;
if (l == 0) l = top1;
if (r > top1) r = 1;
e[i] = node(Distance(ans[0][i], ans[0][l]), Distance(ans[0][i], ans[0][r]),
Distance(ans[0][l], ans[0][r]));
}
for (int i = 1; i <= top2; i++) {
int l = i - 1, r = i + 1;
if (l == 0) l = top2;
if (r > top2) r = 1;
b[i] = node(Distance(ans[1][i], ans[1][l]), Distance(ans[1][i], ans[1][r]),
Distance(ans[1][l], ans[1][r]));
}
KMP(top1);
int j = 1;
for (int i = 1; i <= top2; i++) {
while (e[j] != b[i] && j) j = fail[j];
if (j == top1) {
return 0 * printf("YES\n");
}
j++;
}
for (int i = 1; i <= top2; i++) {
while (e[j] != b[i] && j) j = fail[j];
if (j == top1) {
return 0 * printf("YES\n");
}
j++;
}
printf("NO\n");
}
``` |
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