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stringlengths 52
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#include <bits/stdc++.h>
using namespace std;
bool sBs(const pair<int, int> &a, const pair<int, int> &b) {
return (a.second < b.second);
}
long long SUMD(long long n);
long long BS(vector<pair<long long, long long> > &PS, long long s, long long e,
long long ser);
long long MI(long long a, long long m);
long long P[101LL + 1];
void Sieve(int n = 101LL);
long long Powb(long long b, long long n, long long m);
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
long long q = 1, t;
cin >> q;
t = q;
while (q--) {
long long i, j, n, m, k, l = 0, r = 0, a = 0, b = 0, c = 0, d = 0, e, g = 1,
p, u, v, w, x, y, flag = 1;
cin >> n >> x;
long long ans = 0;
vector<long long> V(n);
for (i = 0; i < n; i++) cin >> V[i];
sort(V.rbegin(), V.rend());
for (i = 0; i < n; i++) {
if (V[i] * (i - l + 1) >= x) {
ans += 1;
l = i + 1;
}
}
cout << ans << "\n";
}
return 0;
}
long long Powb(long long b, long long n, long long m) {
if (n == 0) return 1LL;
if (n == 1) return b;
long long temp = Powb(b, n / 2, m);
if (n % 2 == 0) {
return (temp * temp) % m;
} else {
return (b * ((temp * temp) % m)) % m;
}
}
long long SUMD(long long n) {
long long sum = 0;
while (n > 0) {
sum += n % 10;
n = n / 10;
}
return sum;
}
long long BS(vector<pair<long long, long long> > &PS, long long s, long long e,
long long ser) {
if (s > e) return s;
long long mid = (s + e) / 2;
if (PS[mid].first == ser) {
return mid;
} else if (PS[mid].first > ser) {
return BS(PS, s, mid - 1, ser);
} else
return BS(PS, mid + 1, e, ser);
}
long long MI(long long a, long long m) {
long long m0 = m;
long long y = 0, x = 1;
if (m == 1) return 0;
while (a > 1) {
long long q = a / m;
long long t = m;
m = a % m, a = t;
t = y;
y = x - q * y;
x = t;
}
if (x < 0) x += m0;
return x;
}
void Sieve(int n) {
for (int i = 1; i <= n; i++) P[i] = i;
for (long long i = 2; i * i <= n; i++) {
if (P[i] == i) {
for (long long j = i * i; j <= n; j += i) {
if (P[j] == j) P[j] = i;
}
}
}
}
| ### Prompt
Create a solution in CPP for the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
bool sBs(const pair<int, int> &a, const pair<int, int> &b) {
return (a.second < b.second);
}
long long SUMD(long long n);
long long BS(vector<pair<long long, long long> > &PS, long long s, long long e,
long long ser);
long long MI(long long a, long long m);
long long P[101LL + 1];
void Sieve(int n = 101LL);
long long Powb(long long b, long long n, long long m);
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
long long q = 1, t;
cin >> q;
t = q;
while (q--) {
long long i, j, n, m, k, l = 0, r = 0, a = 0, b = 0, c = 0, d = 0, e, g = 1,
p, u, v, w, x, y, flag = 1;
cin >> n >> x;
long long ans = 0;
vector<long long> V(n);
for (i = 0; i < n; i++) cin >> V[i];
sort(V.rbegin(), V.rend());
for (i = 0; i < n; i++) {
if (V[i] * (i - l + 1) >= x) {
ans += 1;
l = i + 1;
}
}
cout << ans << "\n";
}
return 0;
}
long long Powb(long long b, long long n, long long m) {
if (n == 0) return 1LL;
if (n == 1) return b;
long long temp = Powb(b, n / 2, m);
if (n % 2 == 0) {
return (temp * temp) % m;
} else {
return (b * ((temp * temp) % m)) % m;
}
}
long long SUMD(long long n) {
long long sum = 0;
while (n > 0) {
sum += n % 10;
n = n / 10;
}
return sum;
}
long long BS(vector<pair<long long, long long> > &PS, long long s, long long e,
long long ser) {
if (s > e) return s;
long long mid = (s + e) / 2;
if (PS[mid].first == ser) {
return mid;
} else if (PS[mid].first > ser) {
return BS(PS, s, mid - 1, ser);
} else
return BS(PS, mid + 1, e, ser);
}
long long MI(long long a, long long m) {
long long m0 = m;
long long y = 0, x = 1;
if (m == 1) return 0;
while (a > 1) {
long long q = a / m;
long long t = m;
m = a % m, a = t;
t = y;
y = x - q * y;
x = t;
}
if (x < 0) x += m0;
return x;
}
void Sieve(int n) {
for (int i = 1; i <= n; i++) P[i] = i;
for (long long i = 2; i * i <= n; i++) {
if (P[i] == i) {
for (long long j = i * i; j <= n; j += i) {
if (P[j] == j) P[j] = i;
}
}
}
}
``` |
#include <bits/stdc++.h>
const long long mod = 1000000007;
using namespace std;
long long MOD = 998244353;
void solve() {
long long n, x;
cin >> n >> x;
vector<long long> a(n);
for (auto &i : a) {
cin >> i;
}
sort(a.begin(), a.end(), greater<long long>());
long long curr = 1, ans = 0;
for (long long i = 0; i < n; i++) {
if (curr * a[i] >= x) {
ans++;
curr = 1;
} else {
curr++;
}
}
cout << ans << "\n";
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
| ### Prompt
Create a solution in cpp for the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
const long long mod = 1000000007;
using namespace std;
long long MOD = 998244353;
void solve() {
long long n, x;
cin >> n >> x;
vector<long long> a(n);
for (auto &i : a) {
cin >> i;
}
sort(a.begin(), a.end(), greater<long long>());
long long curr = 1, ans = 0;
for (long long i = 0; i < n; i++) {
if (curr * a[i] >= x) {
ans++;
curr = 1;
} else {
curr++;
}
}
cout << ans << "\n";
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long binomialCoeff(long long n, long long k) {
long long res = 1;
if (k > n - k) k = n - k;
for (long long i = 0; i < k; ++i) {
res *= (n - i);
res /= (i + 1);
}
return res;
}
struct MyComp {
bool operator()(const pair<int, int>& x, const pair<int, int>& y) const {
return x.second > y.second || (x.second == y.second && x.first > y.first);
}
};
long long M = 1e9 + 7;
long long mods(long long a, long long b) {
a = a % M;
b = b % M;
return (a + b) % M;
}
long long modp(long long a, long long b) {
a = a % M;
b = b % M;
return (a * b) % M;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while (t--) {
int n, x;
cin >> n >> x;
vector<long long> v(n);
for (int i = 0; i < n; i++) {
cin >> v[i];
}
sort(v.begin(), v.end());
int ans = 0;
long long count = 0;
for (int i = n - 1; i >= 0; i--) {
count += 1;
if (count * v[i] >= x) {
ans += 1;
count = 0;
}
}
cout << ans << endl;
}
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long binomialCoeff(long long n, long long k) {
long long res = 1;
if (k > n - k) k = n - k;
for (long long i = 0; i < k; ++i) {
res *= (n - i);
res /= (i + 1);
}
return res;
}
struct MyComp {
bool operator()(const pair<int, int>& x, const pair<int, int>& y) const {
return x.second > y.second || (x.second == y.second && x.first > y.first);
}
};
long long M = 1e9 + 7;
long long mods(long long a, long long b) {
a = a % M;
b = b % M;
return (a + b) % M;
}
long long modp(long long a, long long b) {
a = a % M;
b = b % M;
return (a * b) % M;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while (t--) {
int n, x;
cin >> n >> x;
vector<long long> v(n);
for (int i = 0; i < n; i++) {
cin >> v[i];
}
sort(v.begin(), v.end());
int ans = 0;
long long count = 0;
for (int i = n - 1; i >= 0; i--) {
count += 1;
if (count * v[i] >= x) {
ans += 1;
count = 0;
}
}
cout << ans << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long int t;
cin >> t;
while (t--) {
long long int n, x, ans = 0, j = 0;
cin >> n >> x;
vector<long long int> a(n);
for (long long int i = 0; i < n; i++) cin >> a[i];
sort(a.rbegin(), a.rend());
for (long long int i = 0; i < n; i++) {
if (a[i] > x) {
ans++;
j = i + 1;
} else {
if ((i - j + 1) >= ceil(double(x) / double(a[i]))) {
ans++;
j = i + 1;
}
}
}
cout << ans << endl;
}
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long int t;
cin >> t;
while (t--) {
long long int n, x, ans = 0, j = 0;
cin >> n >> x;
vector<long long int> a(n);
for (long long int i = 0; i < n; i++) cin >> a[i];
sort(a.rbegin(), a.rend());
for (long long int i = 0; i < n; i++) {
if (a[i] > x) {
ans++;
j = i + 1;
} else {
if ((i - j + 1) >= ceil(double(x) / double(a[i]))) {
ans++;
j = i + 1;
}
}
}
cout << ans << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
signed main() {
static const int _ = []() {
ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
return 0;
}();
long long t = 1;
cin >> t;
while (t--) {
long long n, x;
cin >> n >> x;
vector<long long> a(n);
for (auto &i : a) cin >> i;
sort(a.begin(), a.end());
long long curr_sz = 1, ans = 0;
long long mn_till_now = a[0];
for (long long i = n - 1; i >= 0; i--) {
if (a[i] >= x) {
ans++;
curr_sz = 1;
continue;
}
if (curr_sz * a[i] >= x) {
ans++;
curr_sz = 1;
} else {
curr_sz++;
}
}
cout << ans << "\n";
}
return 0;
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
signed main() {
static const int _ = []() {
ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
return 0;
}();
long long t = 1;
cin >> t;
while (t--) {
long long n, x;
cin >> n >> x;
vector<long long> a(n);
for (auto &i : a) cin >> i;
sort(a.begin(), a.end());
long long curr_sz = 1, ans = 0;
long long mn_till_now = a[0];
for (long long i = n - 1; i >= 0; i--) {
if (a[i] >= x) {
ans++;
curr_sz = 1;
continue;
}
if (curr_sz * a[i] >= x) {
ans++;
curr_sz = 1;
} else {
curr_sz++;
}
}
cout << ans << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
int n, x;
while (t--) {
cin >> n >> x;
int a[n];
for (int i = 0; i < n; ++i) {
cin >> a[i];
}
sort(a, a + n);
int teamselect = 0;
int no_of_teams = 0;
for (int i = n - 1; i >= 0; --i) {
if ((n - teamselect - i) * a[i] >= x) {
teamselect = n - i;
no_of_teams++;
}
}
cout << no_of_teams << "\n";
}
return 0;
}
| ### Prompt
Please create a solution in CPP to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
int n, x;
while (t--) {
cin >> n >> x;
int a[n];
for (int i = 0; i < n; ++i) {
cin >> a[i];
}
sort(a, a + n);
int teamselect = 0;
int no_of_teams = 0;
for (int i = n - 1; i >= 0; --i) {
if ((n - teamselect - i) * a[i] >= x) {
teamselect = n - i;
no_of_teams++;
}
}
cout << no_of_teams << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using pi = pair<int, int>;
void setIO(string name = "1380C") {
ios_base::sync_with_stdio(0);
cin.tie(0);
freopen((name + ".in").c_str(), "r", stdin);
freopen((name + ".out").c_str(), "w", stdout);
}
int main() {
int t;
cin >> t;
while (t--) {
int n, x;
cin >> n >> x;
vector<int> v;
for (int i = 0; i < n; i++) {
int c;
cin >> c;
v.push_back(c);
}
sort(begin(v), end(v));
int peo = 0;
int minVal;
int count = 0;
for (int i = n - 1; i >= 0; i--) {
peo++;
minVal = v[i];
if (minVal * peo >= x) {
count++;
peo = 0;
}
}
cout << count << "\n";
}
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using pi = pair<int, int>;
void setIO(string name = "1380C") {
ios_base::sync_with_stdio(0);
cin.tie(0);
freopen((name + ".in").c_str(), "r", stdin);
freopen((name + ".out").c_str(), "w", stdout);
}
int main() {
int t;
cin >> t;
while (t--) {
int n, x;
cin >> n >> x;
vector<int> v;
for (int i = 0; i < n; i++) {
int c;
cin >> c;
v.push_back(c);
}
sort(begin(v), end(v));
int peo = 0;
int minVal;
int count = 0;
for (int i = n - 1; i >= 0; i--) {
peo++;
minVal = v[i];
if (minVal * peo >= x) {
count++;
peo = 0;
}
}
cout << count << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
void SOLVE() {
long long n, x;
cin >> n >> x;
long long a[n], last = n, ans = 0;
for (auto &i : a) cin >> i;
sort(a, a + n);
for (long long i = n - 1; i > -1; i--) {
if (a[i] * (last - i) >= x) {
ans++;
last = i;
}
}
cout << ans << '\n';
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int TC = 1;
cin >> TC;
while (TC--) SOLVE();
return 0;
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
void SOLVE() {
long long n, x;
cin >> n >> x;
long long a[n], last = n, ans = 0;
for (auto &i : a) cin >> i;
sort(a, a + n);
for (long long i = n - 1; i > -1; i--) {
if (a[i] * (last - i) >= x) {
ans++;
last = i;
}
}
cout << ans << '\n';
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int TC = 1;
cin >> TC;
while (TC--) SOLVE();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long int N = 200006;
long long int power2(long long int x, long long int n) {
long long int m;
if (n == 0) return 1;
if (n % 2 == 0) {
m = power2(x, n / 2);
return (m * m) % 998244353;
;
} else
return (x * (power2(x, n - 1))) % 998244353;
;
}
void solve() {
long long int i, j, k, q;
long long int n;
long long int mn = 1e13, r = 0, x, y;
cin >> n >> k;
long long int a[n];
long long int tm = 0;
for (i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
long long int l = 0;
for (i = n - 1; i >= 0; i--) {
l = l + 1;
if (l == 1)
mn = a[i];
else
mn = min(mn, a[i]);
if (mn * l >= k) tm++, l = 0;
}
cout << tm;
}
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long int tt = 1.0 * clock();
int t;
cin >> t;
for (int y = 0; y < t; y++) {
solve();
cout << endl;
}
}
| ### Prompt
Generate a Cpp solution to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long int N = 200006;
long long int power2(long long int x, long long int n) {
long long int m;
if (n == 0) return 1;
if (n % 2 == 0) {
m = power2(x, n / 2);
return (m * m) % 998244353;
;
} else
return (x * (power2(x, n - 1))) % 998244353;
;
}
void solve() {
long long int i, j, k, q;
long long int n;
long long int mn = 1e13, r = 0, x, y;
cin >> n >> k;
long long int a[n];
long long int tm = 0;
for (i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
long long int l = 0;
for (i = n - 1; i >= 0; i--) {
l = l + 1;
if (l == 1)
mn = a[i];
else
mn = min(mn, a[i]);
if (mn * l >= k) tm++, l = 0;
}
cout << tm;
}
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long int tt = 1.0 * clock();
int t;
cin >> t;
for (int y = 0; y < t; y++) {
solve();
cout << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
bool isprime(long long n) {
for (long long int i = 2; i * i <= n; i++) {
if (n % i == 0) return 0;
}
return 1;
}
long long gcd(long long int a, long long int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
long long lcm(int a, int b) { return (a / gcd(a, b)) * b; }
long long ncr(long long n, long long k) {
long long res = 1;
if (k > n - k) k = n - k;
for (long long i = 0; i < k; ++i) {
res *= (n - i);
res /= (i + 1);
}
return res;
}
vector<long long> primefactors(long long n) {
vector<long long> ans;
for (long long i = 2; i * i <= n; i++) {
long long c = 0;
while (n % i == 0) {
n = n / i;
c++;
ans.push_back(i);
}
}
if (n > 1) ans.push_back(n);
return ans;
}
bool compare(pair<int, int> a, pair<int, int> b) {
if (a.second > b.second) return 1;
return 0;
}
void solve() {
int n, x;
cin >> n >> x;
vector<int> v(n);
for (int i = 0; i < n; i++) {
cin >> v[i];
}
int teams = INT_MIN;
sort(v.begin(), v.end());
vector<int> dp(n + 1, 0);
for (int i = n - 1; i > -1; i--) {
int need = ceil((1.0 * x) / v[i]);
if (i + need > n)
dp[i] = 0;
else
dp[i] = 1 + dp[i + need];
teams = max(teams, dp[i]);
}
cout << teams << endl;
return;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
;
int t = 1;
cin >> t;
while (t--) {
solve();
}
return 0;
}
| ### Prompt
Please create a solution in cpp to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
bool isprime(long long n) {
for (long long int i = 2; i * i <= n; i++) {
if (n % i == 0) return 0;
}
return 1;
}
long long gcd(long long int a, long long int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
long long lcm(int a, int b) { return (a / gcd(a, b)) * b; }
long long ncr(long long n, long long k) {
long long res = 1;
if (k > n - k) k = n - k;
for (long long i = 0; i < k; ++i) {
res *= (n - i);
res /= (i + 1);
}
return res;
}
vector<long long> primefactors(long long n) {
vector<long long> ans;
for (long long i = 2; i * i <= n; i++) {
long long c = 0;
while (n % i == 0) {
n = n / i;
c++;
ans.push_back(i);
}
}
if (n > 1) ans.push_back(n);
return ans;
}
bool compare(pair<int, int> a, pair<int, int> b) {
if (a.second > b.second) return 1;
return 0;
}
void solve() {
int n, x;
cin >> n >> x;
vector<int> v(n);
for (int i = 0; i < n; i++) {
cin >> v[i];
}
int teams = INT_MIN;
sort(v.begin(), v.end());
vector<int> dp(n + 1, 0);
for (int i = n - 1; i > -1; i--) {
int need = ceil((1.0 * x) / v[i]);
if (i + need > n)
dp[i] = 0;
else
dp[i] = 1 + dp[i + need];
teams = max(teams, dp[i]);
}
cout << teams << endl;
return;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
;
int t = 1;
cin >> t;
while (t--) {
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <class A, class B>
ostream& operator<<(ostream& out, const pair<A, B>& a) {
return out << "(" << a.first << ", " << a.second << ")";
}
template <class A>
ostream& operator<<(ostream& out, const vector<A>& v) {
out << "[";
for (int i = 0; i < v.size(); i++) {
if (i) out << ", ";
out << v[i];
}
return out << "]";
}
void solve() {
int n, x;
cin >> n >> x;
vector<int> a(n);
for (auto& i : a) cin >> i;
sort(a.begin(), a.end());
int res = 0;
int len = n;
for (int i = n - 1; i >= 0; i--) {
if (1LL * (len - i) * a[i] >= x) {
res++;
len = i;
}
}
cout << res << '\n';
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int q = 1;
cin >> q;
for (int kase = 1; kase <= q; kase++) {
solve();
}
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <class A, class B>
ostream& operator<<(ostream& out, const pair<A, B>& a) {
return out << "(" << a.first << ", " << a.second << ")";
}
template <class A>
ostream& operator<<(ostream& out, const vector<A>& v) {
out << "[";
for (int i = 0; i < v.size(); i++) {
if (i) out << ", ";
out << v[i];
}
return out << "]";
}
void solve() {
int n, x;
cin >> n >> x;
vector<int> a(n);
for (auto& i : a) cin >> i;
sort(a.begin(), a.end());
int res = 0;
int len = n;
for (int i = n - 1; i >= 0; i--) {
if (1LL * (len - i) * a[i] >= x) {
res++;
len = i;
}
}
cout << res << '\n';
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int q = 1;
cin >> q;
for (int kase = 1; kase <= q; kase++) {
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
scanf("%d", &t);
while (t--) {
long long int n, x, count = 0, y, z;
scanf("%lld %lld", &n, &x);
long long int ara[n];
for (long long int i = 0; i < n; i++) scanf("%lld", &ara[i]);
sort(ara, ara + n);
for (long long int i = n - 1; i >= 0; i--) {
if (ara[i] >= x) {
count++;
} else {
y = 1;
for (long long int j = i - 1; j >= 0; j--) {
z = x / ara[j];
if (x % ara[j] != 0) {
z++;
}
if (z - y == 1) {
count++;
i = j;
break;
} else if (z - y - 1 > j) {
z = -1;
break;
} else {
y++;
}
}
if (z == -1) {
break;
}
}
}
printf("%lld\n", count);
;
}
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
scanf("%d", &t);
while (t--) {
long long int n, x, count = 0, y, z;
scanf("%lld %lld", &n, &x);
long long int ara[n];
for (long long int i = 0; i < n; i++) scanf("%lld", &ara[i]);
sort(ara, ara + n);
for (long long int i = n - 1; i >= 0; i--) {
if (ara[i] >= x) {
count++;
} else {
y = 1;
for (long long int j = i - 1; j >= 0; j--) {
z = x / ara[j];
if (x % ara[j] != 0) {
z++;
}
if (z - y == 1) {
count++;
i = j;
break;
} else if (z - y - 1 > j) {
z = -1;
break;
} else {
y++;
}
}
if (z == -1) {
break;
}
}
}
printf("%lld\n", count);
;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long t, n, x;
cin >> t;
while (t--) {
cin >> n >> x;
vector<long long> a(n);
for (long long i = 0; i < n; ++i) {
cin >> a[i];
}
sort(a.begin(), a.end());
long long r = n;
long long sum = 0;
for (long long i = n - 1; i >= 0; --i) {
if (a[i] * (r - i) >= x) {
++sum;
r = i;
}
}
cout << sum << endl;
}
}
| ### Prompt
Develop a solution in cpp to the problem described below:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long t, n, x;
cin >> t;
while (t--) {
cin >> n >> x;
vector<long long> a(n);
for (long long i = 0; i < n; ++i) {
cin >> a[i];
}
sort(a.begin(), a.end());
long long r = n;
long long sum = 0;
for (long long i = n - 1; i >= 0; --i) {
if (a[i] * (r - i) >= x) {
++sum;
r = i;
}
}
cout << sum << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long int max(long long int a, long long int b) {
if (a > b) {
return a;
}
return b;
}
long long int gcd(long long int a, long long int b) {
if (a == 0) {
return b;
}
return gcd(b % a, a);
}
long long int lcm(long long int a, long long int b) {
return (a * b) / gcd(a, b);
}
long long int min(long long int a, long long int b) {
if (a < b) {
return a;
}
return b;
}
long long int power(long long int x, long long int n) {
long long int res = 1;
while (n > 0) {
if (n % 2 == 1) {
res = (res * x);
}
x = (x * x);
n = n / 2;
}
return res;
}
long long int factorialNumInverse[1000001];
long long int naturalNumInverse[1000001];
long long int fact[1000001];
void InverseofNumber(long long int n, long long int p) {
naturalNumInverse[0] = naturalNumInverse[1] = 1;
for (long long int i = 2; i <= n; i++)
naturalNumInverse[i] = naturalNumInverse[p % i] * (p - p / i) % p;
}
void InverseofFactorial(long long int n, long long int p) {
factorialNumInverse[0] = factorialNumInverse[1] = 1;
for (long long int i = 2; i <= n; i++)
factorialNumInverse[i] =
(naturalNumInverse[i] * factorialNumInverse[i - 1]) % p;
}
void factorial(long long int n, long long int p) {
fact[0] = 1;
for (long long int i = 1; i <= n; i++) {
fact[i] = (fact[i - 1] * i) % p;
}
}
long long int nCrModp(long long int n, long long int r, long long int p) {
long long int ans =
((fact[n] * factorialNumInverse[r]) % p * factorialNumInverse[n - r]) % p;
return ans;
}
struct pos {
long long int val, paint;
};
bool compare(pos p1, pos p2) {
if (p1.paint == p2.paint) {
return p1.val > p2.val;
}
return p1.paint < p2.paint;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long int tc;
cin >> tc;
factorial(200005, 1000000007);
InverseofNumber(200005, 1000000007);
InverseofFactorial(200005, 1000000007);
while (tc--) {
long long int n, x;
cin >> n >> x;
long long int arr[n];
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
long long int ans = 0;
sort(arr, arr + n, greater<long long int>());
for (int i = 0; i < n; i++) {
if (arr[i] >= x) {
ans++;
} else {
long long int len = 1;
while (len * arr[i] < x && i < n) {
i++;
len++;
}
if (len * arr[i] >= x && i <= n - 1) {
ans++;
}
}
}
cout << ans << "\n";
}
return 0;
}
| ### Prompt
Create a solution in cpp for the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long int max(long long int a, long long int b) {
if (a > b) {
return a;
}
return b;
}
long long int gcd(long long int a, long long int b) {
if (a == 0) {
return b;
}
return gcd(b % a, a);
}
long long int lcm(long long int a, long long int b) {
return (a * b) / gcd(a, b);
}
long long int min(long long int a, long long int b) {
if (a < b) {
return a;
}
return b;
}
long long int power(long long int x, long long int n) {
long long int res = 1;
while (n > 0) {
if (n % 2 == 1) {
res = (res * x);
}
x = (x * x);
n = n / 2;
}
return res;
}
long long int factorialNumInverse[1000001];
long long int naturalNumInverse[1000001];
long long int fact[1000001];
void InverseofNumber(long long int n, long long int p) {
naturalNumInverse[0] = naturalNumInverse[1] = 1;
for (long long int i = 2; i <= n; i++)
naturalNumInverse[i] = naturalNumInverse[p % i] * (p - p / i) % p;
}
void InverseofFactorial(long long int n, long long int p) {
factorialNumInverse[0] = factorialNumInverse[1] = 1;
for (long long int i = 2; i <= n; i++)
factorialNumInverse[i] =
(naturalNumInverse[i] * factorialNumInverse[i - 1]) % p;
}
void factorial(long long int n, long long int p) {
fact[0] = 1;
for (long long int i = 1; i <= n; i++) {
fact[i] = (fact[i - 1] * i) % p;
}
}
long long int nCrModp(long long int n, long long int r, long long int p) {
long long int ans =
((fact[n] * factorialNumInverse[r]) % p * factorialNumInverse[n - r]) % p;
return ans;
}
struct pos {
long long int val, paint;
};
bool compare(pos p1, pos p2) {
if (p1.paint == p2.paint) {
return p1.val > p2.val;
}
return p1.paint < p2.paint;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long int tc;
cin >> tc;
factorial(200005, 1000000007);
InverseofNumber(200005, 1000000007);
InverseofFactorial(200005, 1000000007);
while (tc--) {
long long int n, x;
cin >> n >> x;
long long int arr[n];
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
long long int ans = 0;
sort(arr, arr + n, greater<long long int>());
for (int i = 0; i < n; i++) {
if (arr[i] >= x) {
ans++;
} else {
long long int len = 1;
while (len * arr[i] < x && i < n) {
i++;
len++;
}
if (len * arr[i] >= x && i <= n - 1) {
ans++;
}
}
}
cout << ans << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <class T>
void _R(T &x) {
cin >> x;
}
void _R(int &x) { scanf("%d", &x); }
void _R(int64_t &x) { scanf("%lld", &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template <class T, class... U>
void R(T &head, U &...tail) {
_R(head);
R(tail...);
}
template <class T>
void _W(const T &x) {
cout << x;
}
void _W(const int &x) { printf("%d", x); }
void _W(const int64_t &x) { printf("%lld", x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template <class T, class U>
void _W(const pair<T, U> &x) {
_W(x.first);
putchar(' ');
_W(x.second);
}
template <class T>
void _W(const vector<T> &x) {
for (auto i = x.begin(); i != x.end(); _W(*i++))
if (i != x.cbegin()) putchar(' ');
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
_W(head);
putchar(sizeof...(tail) ? ' ' : '\n');
W(tail...);
}
int MOD = 1e9 + 7;
void ADD(long long &x, long long v) {
x = (x + v) % MOD;
if (x < 0) x += MOD;
}
const int SIZE = 1 << 20;
void solve() {
int n, x;
R(n, x);
vector<long long> a(n);
for (int i = 0; i < (n); ++i) {
R(a[i]);
}
int an = 0;
sort((a).begin(), (a).end(), greater<int>());
for (int i = 0, j; i < n; i = j) {
for (j = i; j < n; j++) {
if (a[j] * (j - i + 1) >= x) {
an++;
j++;
break;
}
}
}
W(an);
}
int main() {
int ___T;
scanf("%d", &___T);
for (int cs = 1; cs <= ___T; cs++) {
solve();
}
return 0;
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
void _R(T &x) {
cin >> x;
}
void _R(int &x) { scanf("%d", &x); }
void _R(int64_t &x) { scanf("%lld", &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template <class T, class... U>
void R(T &head, U &...tail) {
_R(head);
R(tail...);
}
template <class T>
void _W(const T &x) {
cout << x;
}
void _W(const int &x) { printf("%d", x); }
void _W(const int64_t &x) { printf("%lld", x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template <class T, class U>
void _W(const pair<T, U> &x) {
_W(x.first);
putchar(' ');
_W(x.second);
}
template <class T>
void _W(const vector<T> &x) {
for (auto i = x.begin(); i != x.end(); _W(*i++))
if (i != x.cbegin()) putchar(' ');
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
_W(head);
putchar(sizeof...(tail) ? ' ' : '\n');
W(tail...);
}
int MOD = 1e9 + 7;
void ADD(long long &x, long long v) {
x = (x + v) % MOD;
if (x < 0) x += MOD;
}
const int SIZE = 1 << 20;
void solve() {
int n, x;
R(n, x);
vector<long long> a(n);
for (int i = 0; i < (n); ++i) {
R(a[i]);
}
int an = 0;
sort((a).begin(), (a).end(), greater<int>());
for (int i = 0, j; i < n; i = j) {
for (j = i; j < n; j++) {
if (a[j] * (j - i + 1) >= x) {
an++;
j++;
break;
}
}
}
W(an);
}
int main() {
int ___T;
scanf("%d", &___T);
for (int cs = 1; cs <= ___T; cs++) {
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long n, t, k;
int main() {
cin >> t;
while (t--) {
cin >> n;
cin >> k;
vector<long long> a(n);
for (long long i = 0; i < n; i++) cin >> a[i];
sort((a).begin(), (a).end(), greater<long long>());
int sol = 0;
int cmin;
long long len = 0;
for (long long i = 0; i < n; i++) {
cmin = a[i];
len++;
if (cmin * len >= k) {
sol++;
len = 0;
}
}
cout << sol << endl;
}
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long n, t, k;
int main() {
cin >> t;
while (t--) {
cin >> n;
cin >> k;
vector<long long> a(n);
for (long long i = 0; i < n; i++) cin >> a[i];
sort((a).begin(), (a).end(), greater<long long>());
int sol = 0;
int cmin;
long long len = 0;
for (long long i = 0; i < n; i++) {
cmin = a[i];
len++;
if (cmin * len >= k) {
sol++;
len = 0;
}
}
cout << sol << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n, m;
cin >> n >> m;
int a[n + 1];
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
reverse(a, a + n);
int ans = 0, lst = -1;
for (int i = 0; i < n; i++)
if (a[i] * (i - lst) >= m) {
ans++;
lst = i;
}
cout << ans << endl;
}
}
| ### Prompt
Develop a solution in CPP to the problem described below:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n, m;
cin >> n >> m;
int a[n + 1];
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
reverse(a, a + n);
int ans = 0, lst = -1;
for (int i = 0; i < n; i++)
if (a[i] * (i - lst) >= m) {
ans++;
lst = i;
}
cout << ans << endl;
}
}
``` |
#include <bits/stdc++.h>
#pragma GCC optimize "03"
using namespace std;
const long long int inf = 1e15;
const long long int MOD = 1e9 + 7;
const long long int N = 2e5 + 5;
signed main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
;
long long int t;
cin >> t;
while (t--) {
long long int n, x;
cin >> n >> x;
long long int a[n];
for (long long int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n, greater<long long int>());
long long int c = 0, s = 0, cnt = 0, mn = inf;
for (long long int i = 0; i < n; i++) {
mn = min(mn, a[i]);
cnt++;
if (mn * cnt >= x) {
mn = inf;
cnt = 0;
c++;
}
}
cout << c << '\n';
}
return 0;
}
| ### Prompt
Construct a CPP code solution to the problem outlined:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC optimize "03"
using namespace std;
const long long int inf = 1e15;
const long long int MOD = 1e9 + 7;
const long long int N = 2e5 + 5;
signed main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
;
long long int t;
cin >> t;
while (t--) {
long long int n, x;
cin >> n >> x;
long long int a[n];
for (long long int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n, greater<long long int>());
long long int c = 0, s = 0, cnt = 0, mn = inf;
for (long long int i = 0; i < n; i++) {
mn = min(mn, a[i]);
cnt++;
if (mn * cnt >= x) {
mn = inf;
cnt = 0;
c++;
}
}
cout << c << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n, x;
cin >> n >> x;
vector<int> a(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
int k = n, ans = 0;
sort(a.begin(), a.end());
for (int i = n - 1; i >= 0; i--) {
if (a[i] * (k - i) >= x) {
ans++;
k = i;
}
}
cout << ans << endl;
}
return 0;
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n, x;
cin >> n >> x;
vector<int> a(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
int k = n, ans = 0;
sort(a.begin(), a.end());
for (int i = n - 1; i >= 0; i--) {
if (a[i] * (k - i) >= x) {
ans++;
k = i;
}
}
cout << ans << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long t;
cin >> t;
while (t--) {
long long n, x;
cin >> n >> x;
vector<long long> v(n);
for (long long i = 0; i < n; i++) {
cin >> v[i];
}
sort(v.begin(), v.end());
long long ans = 0, i = n - 1, curr = 0;
while (i >= 0) {
if ((curr + 1) * v[i] >= x) {
ans++;
curr = 0;
} else {
curr++;
}
i--;
}
cout << ans << endl;
}
return 0;
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long t;
cin >> t;
while (t--) {
long long n, x;
cin >> n >> x;
vector<long long> v(n);
for (long long i = 0; i < n; i++) {
cin >> v[i];
}
sort(v.begin(), v.end());
long long ans = 0, i = n - 1, curr = 0;
while (i >= 0) {
if ((curr + 1) * v[i] >= x) {
ans++;
curr = 0;
} else {
curr++;
}
i--;
}
cout << ans << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long t;
cin >> t;
while (t--) {
long long n, x;
cin >> n >> x;
vector<long long> a(n);
for (__typeof(n) i = 0; i < n; i++) {
cin >> a[i];
}
sort(a.rbegin(), a.rend());
long long val, no = 0, mul = 1;
for (__typeof(n) i = 0; i < n; i++) {
val = a[i];
if (val * mul >= x) {
mul = 1;
no++;
} else {
mul++;
}
}
cout << no << "\n";
}
return 0;
}
| ### Prompt
Please create a solution in CPP to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long t;
cin >> t;
while (t--) {
long long n, x;
cin >> n >> x;
vector<long long> a(n);
for (__typeof(n) i = 0; i < n; i++) {
cin >> a[i];
}
sort(a.rbegin(), a.rend());
long long val, no = 0, mul = 1;
for (__typeof(n) i = 0; i < n; i++) {
val = a[i];
if (val * mul >= x) {
mul = 1;
no++;
} else {
mul++;
}
}
cout << no << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int a, b, c, d, i, j, k, l, p, q, r = 0, s = 1;
scanf("%lld", &d);
while (d--) {
scanf("%lld %lld", &a, &b);
long long int arr1[a + 5], arr2[a + 5], arr3[a + 5];
for (i = 1; i <= a; i++) {
scanf("%lld", &arr3[i]);
}
sort(arr3 + 1, arr3 + a + 1);
r = 0;
s = 0;
for (i = a; i >= 1; i--) {
s++;
if (b <= s * arr3[i]) {
r++;
s = 0;
}
}
printf("%lld\n", r);
}
return 0;
}
| ### Prompt
Develop a solution in CPP to the problem described below:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int a, b, c, d, i, j, k, l, p, q, r = 0, s = 1;
scanf("%lld", &d);
while (d--) {
scanf("%lld %lld", &a, &b);
long long int arr1[a + 5], arr2[a + 5], arr3[a + 5];
for (i = 1; i <= a; i++) {
scanf("%lld", &arr3[i]);
}
sort(arr3 + 1, arr3 + a + 1);
r = 0;
s = 0;
for (i = a; i >= 1; i--) {
s++;
if (b <= s * arr3[i]) {
r++;
s = 0;
}
}
printf("%lld\n", r);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
long long n, x;
cin >> n >> x;
long long i;
long long ar[n];
for (i = 0; i < n; i++) {
cin >> ar[i];
}
sort(ar, ar + n, greater<int>());
long long count = 0;
long long p = 1;
i = 0;
while (i < n) {
if (ar[i] >= x) {
count++;
i++;
} else {
if (i == n - 1) {
break;
}
i++;
p++;
if (ar[i] * p >= x) {
count++;
i++;
p = 1;
}
}
}
cout << count << endl;
}
return 0;
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
long long n, x;
cin >> n >> x;
long long i;
long long ar[n];
for (i = 0; i < n; i++) {
cin >> ar[i];
}
sort(ar, ar + n, greater<int>());
long long count = 0;
long long p = 1;
i = 0;
while (i < n) {
if (ar[i] >= x) {
count++;
i++;
} else {
if (i == n - 1) {
break;
}
i++;
p++;
if (ar[i] * p >= x) {
count++;
i++;
p = 1;
}
}
}
cout << count << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void solve() {
int n;
long long int x;
cin >> n >> x;
long long int a[n];
for (int i = 0; i < n; i++) cin >> a[i];
;
sort(a, a + n, greater<int>());
int ans = 0;
int curr_len = 0;
for (int i = 0; i < n; i++) {
if (a[i] >= x) {
ans++;
continue;
}
curr_len++;
if (curr_len * a[i] >= x) ans++, curr_len = 0;
}
cout << ans << "\n";
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while (t--) solve();
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
int n;
long long int x;
cin >> n >> x;
long long int a[n];
for (int i = 0; i < n; i++) cin >> a[i];
;
sort(a, a + n, greater<int>());
int ans = 0;
int curr_len = 0;
for (int i = 0; i < n; i++) {
if (a[i] >= x) {
ans++;
continue;
}
curr_len++;
if (curr_len * a[i] >= x) ans++, curr_len = 0;
}
cout << ans << "\n";
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while (t--) solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <class c>
struct rge {
c b, e;
};
template <class c>
rge<c> range(c i, c j) {
return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
template <class c>
debug& operator<<(const c&) {
return *this;
}
};
template <class T>
void ckmin(T& a, T b) {
a = min(a, b);
}
template <class T>
void ckmax(T& a, T b) {
a = max(a, b);
}
void solve() {
int n, x;
cin >> n >> x;
vector<int> a(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
sort(a.rbegin(), a.rend());
int ans = 0;
int currmin = -1;
int pre = 0;
for (int i = 0; i < n; i++) {
currmin = a[i];
if (currmin * (i - pre + 1) >= x) {
currmin = -1;
pre = i + 1;
ans++;
}
}
cout << ans << endl;
}
int main() {
int t;
cin >> t;
while (t--) solve();
return 0;
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <class c>
struct rge {
c b, e;
};
template <class c>
rge<c> range(c i, c j) {
return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
template <class c>
debug& operator<<(const c&) {
return *this;
}
};
template <class T>
void ckmin(T& a, T b) {
a = min(a, b);
}
template <class T>
void ckmax(T& a, T b) {
a = max(a, b);
}
void solve() {
int n, x;
cin >> n >> x;
vector<int> a(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
sort(a.rbegin(), a.rend());
int ans = 0;
int currmin = -1;
int pre = 0;
for (int i = 0; i < n; i++) {
currmin = a[i];
if (currmin * (i - pre + 1) >= x) {
currmin = -1;
pre = i + 1;
ans++;
}
}
cout << ans << endl;
}
int main() {
int t;
cin >> t;
while (t--) solve();
return 0;
}
``` |
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
using namespace std;
double pi = 1.57079632679489661923;
long long modu = 1e9 + 7;
long long t = 1, n = 1, x = 1;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> t;
for (int ti = 1; ti <= t; ti++) {
cin >> n >> x;
vector<long long> arr(n);
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
sort(arr.begin(), arr.end());
long long ans = 0;
long long j = n;
for (long long i = n - 1; i >= 0; i--) {
if (x <= arr[i] * (j - i)) {
ans++;
j = i;
}
}
cout << ans << endl;
}
}
| ### Prompt
In CPP, your task is to solve the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
using namespace std;
double pi = 1.57079632679489661923;
long long modu = 1e9 + 7;
long long t = 1, n = 1, x = 1;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> t;
for (int ti = 1; ti <= t; ti++) {
cin >> n >> x;
vector<long long> arr(n);
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
sort(arr.begin(), arr.end());
long long ans = 0;
long long j = n;
for (long long i = n - 1; i >= 0; i--) {
if (x <= arr[i] * (j - i)) {
ans++;
j = i;
}
}
cout << ans << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long int tt;
vector<long long int> a;
long long int check(long long int x) {
long long int n = a.size();
long long int s = 0;
long long int ans = 0;
for (long long int i = n - 1; i >= 0; i--) {
s++;
long long int x1;
x1 = ceil(x / double(a[i]));
if (s >= x1) {
ans++;
s = s - x1;
}
}
return ans;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long int t;
cin >> t;
while (t--) {
long long int n, x;
cin >> n >> x;
a.resize(n);
for (long long int i = 0; i < n; i++) {
cin >> a[i];
}
sort(a.begin(), a.end());
long long int ans = check(x);
cout << ans << endl;
}
return 0;
}
| ### Prompt
Please create a solution in Cpp to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long int tt;
vector<long long int> a;
long long int check(long long int x) {
long long int n = a.size();
long long int s = 0;
long long int ans = 0;
for (long long int i = n - 1; i >= 0; i--) {
s++;
long long int x1;
x1 = ceil(x / double(a[i]));
if (s >= x1) {
ans++;
s = s - x1;
}
}
return ans;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long int t;
cin >> t;
while (t--) {
long long int n, x;
cin >> n >> x;
a.resize(n);
for (long long int i = 0; i < n; i++) {
cin >> a[i];
}
sort(a.begin(), a.end());
long long int ans = check(x);
cout << ans << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long int mod = 998244353;
long long int ceil1(long long int n, long long int k) {
if (n % k) return n / k + 1;
return n / k;
}
long long int pow1(long long int n, long long int k) {
long long int ans = 1;
while (k != 0) {
if (k % 2) ans = (ans * n) % mod;
n = (n * n) % mod;
k = k / 2;
}
return ans;
}
bool ispal(string s) {
long long int i = 0;
long long int j = s.size() - 1;
while (i < j) {
if (s[i] != s[j]) return false;
i++;
j--;
}
return true;
}
long long int gcdd(long long int a, long long int b) {
if (b == 0) return a;
return gcdd(b, a % b);
}
long long int fact(long long int n) {
if (n == 1 || n == 0)
return 1;
else
return (n * fact(n - 1)) % mod;
}
bool isprime(long long int n) {
if (n <= 1) return false;
if (n <= 3) return true;
if (n % 2 == 0 || n % 3 == 0) return false;
for (long long int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0) return false;
return true;
}
void solve() {
long long int n, k;
cin >> n >> k;
vector<long long int> a(n);
for (long long int i = 0; i < n; i++) cin >> a[i];
sort(a.begin(), a.end(), greater<long long int>());
long long int ans = 0, c = 1;
for (long long int i = 0; i < n; i++) {
if (a[i] * c >= k) {
ans++;
c = 1;
} else
c++;
}
cout << ans << endl;
return;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long int q = 1;
cin >> q;
while (q--) {
solve();
}
return 0;
}
| ### Prompt
In cpp, your task is to solve the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long int mod = 998244353;
long long int ceil1(long long int n, long long int k) {
if (n % k) return n / k + 1;
return n / k;
}
long long int pow1(long long int n, long long int k) {
long long int ans = 1;
while (k != 0) {
if (k % 2) ans = (ans * n) % mod;
n = (n * n) % mod;
k = k / 2;
}
return ans;
}
bool ispal(string s) {
long long int i = 0;
long long int j = s.size() - 1;
while (i < j) {
if (s[i] != s[j]) return false;
i++;
j--;
}
return true;
}
long long int gcdd(long long int a, long long int b) {
if (b == 0) return a;
return gcdd(b, a % b);
}
long long int fact(long long int n) {
if (n == 1 || n == 0)
return 1;
else
return (n * fact(n - 1)) % mod;
}
bool isprime(long long int n) {
if (n <= 1) return false;
if (n <= 3) return true;
if (n % 2 == 0 || n % 3 == 0) return false;
for (long long int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0) return false;
return true;
}
void solve() {
long long int n, k;
cin >> n >> k;
vector<long long int> a(n);
for (long long int i = 0; i < n; i++) cin >> a[i];
sort(a.begin(), a.end(), greater<long long int>());
long long int ans = 0, c = 1;
for (long long int i = 0; i < n; i++) {
if (a[i] * c >= k) {
ans++;
c = 1;
} else
c++;
}
cout << ans << endl;
return;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long int q = 1;
cin >> q;
while (q--) {
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void input(long long int n, long long int arr[]) {
for (__typeof(n) i = (0) - ((0) > (n)); i != (n) - ((0) > (n));
i += 1 - 2 * ((0) > (n)))
cin >> arr[i];
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long int t;
cin >> t;
while (t--) {
long long int n, x;
cin >> n >> x;
long long int a[n];
input(n, a);
sort(a, a + n, greater<long long int>());
long long int ans = 0;
long long int i = 0;
while (i < n) {
long long int j = i;
while (j < n) {
if (a[j] * (j - i + 1) >= x) {
ans++;
break;
}
j++;
}
i = j;
i++;
}
cout << ans << endl;
}
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void input(long long int n, long long int arr[]) {
for (__typeof(n) i = (0) - ((0) > (n)); i != (n) - ((0) > (n));
i += 1 - 2 * ((0) > (n)))
cin >> arr[i];
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long int t;
cin >> t;
while (t--) {
long long int n, x;
cin >> n >> x;
long long int a[n];
input(n, a);
sort(a, a + n, greater<long long int>());
long long int ans = 0;
long long int i = 0;
while (i < n) {
long long int j = i;
while (j < n) {
if (a[j] * (j - i + 1) >= x) {
ans++;
break;
}
j++;
}
i = j;
i++;
}
cout << ans << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long seive[1000001];
void buildseive() {
for (int i = 0; i < 1000001; i++) seive[i] = 1;
for (int i = 2; i * i < 1000001; i++) {
if (seive[i]) {
for (int j = i; j * i < 1000001; j++) {
seive[i * j] = 0;
}
}
}
}
int main() {
int t;
cin >> t;
while (t--) {
long long n, x;
cin >> n >> x;
long long arr[n];
for (int i = 0; i < n; i++) cin >> arr[i];
int count = 0;
sort(arr, arr + n);
int grp = 0;
for (int i = n - 1; i >= 0; i--) {
long long z;
z = (x / arr[i]) + (x % arr[i] != 0 ? 1 : 0);
if (count + 1 >= z) {
grp++;
count = 0;
} else
count++;
}
cout << grp << endl;
}
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long seive[1000001];
void buildseive() {
for (int i = 0; i < 1000001; i++) seive[i] = 1;
for (int i = 2; i * i < 1000001; i++) {
if (seive[i]) {
for (int j = i; j * i < 1000001; j++) {
seive[i * j] = 0;
}
}
}
}
int main() {
int t;
cin >> t;
while (t--) {
long long n, x;
cin >> n >> x;
long long arr[n];
for (int i = 0; i < n; i++) cin >> arr[i];
int count = 0;
sort(arr, arr + n);
int grp = 0;
for (int i = n - 1; i >= 0; i--) {
long long z;
z = (x / arr[i]) + (x % arr[i] != 0 ? 1 : 0);
if (count + 1 >= z) {
grp++;
count = 0;
} else
count++;
}
cout << grp << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
vector<int> v;
int dp[100005];
int main() {
int t;
cin >> t;
while (t--) {
int n, x;
cin >> n >> x;
v.clear();
for (int i = 1; i <= n; i++) {
int a;
cin >> a;
int k = x / a;
if (x % a > 0) k++;
v.push_back(k);
}
sort(v.begin(), v.end());
for (int i = 0; i < v.size(); i++) {
dp[i] = dp[i - 1];
if (i + 1 >= v[i]) dp[i] = max(dp[i], dp[i - v[i]] + 1);
}
int ans = 0;
cout << dp[n - 1] << endl;
for (int i = 0; i < n; i++) dp[i] = 0;
}
}
| ### Prompt
Generate a CPP solution to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
vector<int> v;
int dp[100005];
int main() {
int t;
cin >> t;
while (t--) {
int n, x;
cin >> n >> x;
v.clear();
for (int i = 1; i <= n; i++) {
int a;
cin >> a;
int k = x / a;
if (x % a > 0) k++;
v.push_back(k);
}
sort(v.begin(), v.end());
for (int i = 0; i < v.size(); i++) {
dp[i] = dp[i - 1];
if (i + 1 >= v[i]) dp[i] = max(dp[i], dp[i - v[i]] + 1);
}
int ans = 0;
cout << dp[n - 1] << endl;
for (int i = 0; i < n; i++) dp[i] = 0;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
void __print(int x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
cerr << '{';
__print(x.first);
cerr << ',';
__print(x.second);
cerr << '}';
}
template <typename T>
void __print(const T &x) {
int f = 0;
cerr << '{';
for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
__print(t);
if (sizeof...(v)) cerr << ", ";
_print(v...);
}
void testcase() {
int n, x;
cin >> n >> x;
vector<int> v(n);
for (int i = 0; i < n; i++) {
cin >> v[i];
}
sort(v.begin(), v.end(), greater<int>());
int team = 0;
for (int i = 0; i < n; i++) {
int mem = 1;
for (int j = i; j < n; j++) {
if (mem * v[j] >= x) {
team++;
i = j;
break;
} else {
mem++;
i = j;
}
}
}
cout << team << "\n";
;
}
int main() {
ios_base::sync_with_stdio(false), cin.tie(NULL);
int t;
cin >> t;
while (t--) {
testcase();
}
return 0;
}
| ### Prompt
Please formulate a CPP solution to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void __print(int x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
cerr << '{';
__print(x.first);
cerr << ',';
__print(x.second);
cerr << '}';
}
template <typename T>
void __print(const T &x) {
int f = 0;
cerr << '{';
for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
__print(t);
if (sizeof...(v)) cerr << ", ";
_print(v...);
}
void testcase() {
int n, x;
cin >> n >> x;
vector<int> v(n);
for (int i = 0; i < n; i++) {
cin >> v[i];
}
sort(v.begin(), v.end(), greater<int>());
int team = 0;
for (int i = 0; i < n; i++) {
int mem = 1;
for (int j = i; j < n; j++) {
if (mem * v[j] >= x) {
team++;
i = j;
break;
} else {
mem++;
i = j;
}
}
}
cout << team << "\n";
;
}
int main() {
ios_base::sync_with_stdio(false), cin.tie(NULL);
int t;
cin >> t;
while (t--) {
testcase();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while (t--) {
int n, x;
cin >> n >> x;
vector<int> arr;
for (int i = 0; i < n; i++) {
int temp;
cin >> temp;
arr.push_back(temp);
}
sort(arr.begin(), arr.end(), greater<int>());
int nteams = 0;
int count = 1;
for (int i = 0; i < n; i++) {
if (count * arr[i] >= x) {
nteams++;
count = 1;
continue;
}
count++;
}
cout << nteams << "\n";
}
return 0;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while (t--) {
int n, x;
cin >> n >> x;
vector<int> arr;
for (int i = 0; i < n; i++) {
int temp;
cin >> temp;
arr.push_back(temp);
}
sort(arr.begin(), arr.end(), greater<int>());
int nteams = 0;
int count = 1;
for (int i = 0; i < n; i++) {
if (count * arr[i] >= x) {
nteams++;
count = 1;
continue;
}
count++;
}
cout << nteams << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 1;
void solve() {
int n, x;
cin >> n >> x;
vector<int> a(n);
for (int i = 0; i < n; i++) cin >> a[i];
sort(a.begin(), a.end(), greater<int>());
int ans = 0;
int j = 1;
for (int i = 0; i < n; i++) {
if ((x + a[i] - 1) / a[i] == j) {
ans++;
j = 1;
} else
j++;
}
cout << ans << endl;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
;
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 1;
void solve() {
int n, x;
cin >> n >> x;
vector<int> a(n);
for (int i = 0; i < n; i++) cin >> a[i];
sort(a.begin(), a.end(), greater<int>());
int ans = 0;
int j = 1;
for (int i = 0; i < n; i++) {
if ((x + a[i] - 1) / a[i] == j) {
ans++;
j = 1;
} else
j++;
}
cout << ans << endl;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
;
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long int N = 2e5 + 5;
void solve() {
long long int n, x;
cin >> n >> x;
vector<long long int> v(n);
for (auto &it : v) cin >> it;
;
sort(v.begin(), v.end());
long long int ans = 0;
vector<long long int> a;
for (long long int i = 0; i < n; i++) {
if (v[i] >= x) {
ans++;
} else {
a.push_back(v[i]);
}
}
reverse(a.begin(), a.end());
for (long long int i = 0; i < a.size(); i++) {
long long int mini = a[i];
long long int count = 1;
while (mini * count < x) {
i++;
if (i >= a.size()) {
break;
}
mini = min(mini, a[i]);
count++;
}
if (mini * count >= x) {
ans++;
}
}
cout << ans << "\n";
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long int t = 1;
cin >> t;
while (t--) solve();
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long int N = 2e5 + 5;
void solve() {
long long int n, x;
cin >> n >> x;
vector<long long int> v(n);
for (auto &it : v) cin >> it;
;
sort(v.begin(), v.end());
long long int ans = 0;
vector<long long int> a;
for (long long int i = 0; i < n; i++) {
if (v[i] >= x) {
ans++;
} else {
a.push_back(v[i]);
}
}
reverse(a.begin(), a.end());
for (long long int i = 0; i < a.size(); i++) {
long long int mini = a[i];
long long int count = 1;
while (mini * count < x) {
i++;
if (i >= a.size()) {
break;
}
mini = min(mini, a[i]);
count++;
}
if (mini * count >= x) {
ans++;
}
}
cout << ans << "\n";
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long int t = 1;
cin >> t;
while (t--) solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void tmkc() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(0);
}
int main() {
tmkc();
int t;
cin >> t;
while (t--) {
long long n, x, i, j, k, r = 0;
cin >> n >> x;
vector<long long> a;
for (i = 0; i < n; i++) {
cin >> j;
a.push_back(j);
}
sort(a.begin(), a.end(), greater<int>());
j = -1;
for (i = 0; i < n; i++) {
if ((i - j) * a[i] >= x) {
r++;
j = i;
}
}
cout << r << endl;
}
}
| ### Prompt
Please create a solution in CPP to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void tmkc() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(0);
}
int main() {
tmkc();
int t;
cin >> t;
while (t--) {
long long n, x, i, j, k, r = 0;
cin >> n >> x;
vector<long long> a;
for (i = 0; i < n; i++) {
cin >> j;
a.push_back(j);
}
sort(a.begin(), a.end(), greater<int>());
j = -1;
for (i = 0; i < n; i++) {
if ((i - j) * a[i] >= x) {
r++;
j = i;
}
}
cout << r << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
void Open() {
freopen("Input.txt", "r", stdin);
freopen("Output.txt", "w", stdout);
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
void madesetprobblem() {}
long long n, *a, x;
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int t;
cin >> t;
while (t--) {
cin >> n >> x;
a = new long long[n];
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
int z = n - 1, c = 0, v = n;
for (int i = n - 1; i >= 0; i--) {
if (a[i] * (v - i) >= x) {
c++;
v = i;
}
}
cout << c << "\n";
}
}
| ### Prompt
Generate a cpp solution to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void Open() {
freopen("Input.txt", "r", stdin);
freopen("Output.txt", "w", stdout);
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
void madesetprobblem() {}
long long n, *a, x;
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int t;
cin >> t;
while (t--) {
cin >> n >> x;
a = new long long[n];
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
int z = n - 1, c = 0, v = n;
for (int i = n - 1; i >= 0; i--) {
if (a[i] * (v - i) >= x) {
c++;
v = i;
}
}
cout << c << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int L;
cin >> L;
for (int LL = 1; LL <= L; LL++) {
int n, x;
cin >> n >> x;
int s[n + 1];
for (int i = 1; i <= n; i++) cin >> s[i];
sort(s + 1, s + n + 1);
int ans = 0, f = 0, xx;
for (int i = n; i >= 1; i--) {
f++;
xx = s[i];
if (xx * f >= x) {
ans++;
f = 0;
}
}
cout << ans << endl;
}
return 0;
}
| ### Prompt
Generate a Cpp solution to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int L;
cin >> L;
for (int LL = 1; LL <= L; LL++) {
int n, x;
cin >> n >> x;
int s[n + 1];
for (int i = 1; i <= n; i++) cin >> s[i];
sort(s + 1, s + n + 1);
int ans = 0, f = 0, xx;
for (int i = n; i >= 1; i--) {
f++;
xx = s[i];
if (xx * f >= x) {
ans++;
f = 0;
}
}
cout << ans << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void solve() {
int n, k;
cin >> n >> k;
int a[n];
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
int m = 1;
int team = 0;
for (int i = n - 1; i >= 0; i--) {
if (a[i] * m >= k) {
team++;
m = 1;
} else {
m++;
}
}
cout << team << "\n";
}
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) solve();
return 0;
}
| ### Prompt
Develop a solution in CPP to the problem described below:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
int n, k;
cin >> n >> k;
int a[n];
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
int m = 1;
int team = 0;
for (int i = n - 1; i >= 0; i--) {
if (a[i] * m >= k) {
team++;
m = 1;
} else {
m++;
}
}
cout << team << "\n";
}
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
bool compare(long long int a, long long int b) { return a > b; }
int32_t main() {
int t;
cin >> t;
while (t--) {
long long int n, x;
cin >> n >> x;
long long int a[n + 1];
a[0] = 0;
for (int i = 1; i < n + 1; i++) {
cin >> a[i];
a[i] = (x / a[i]) + ((x % a[i]) != 0 ? 1 : 0);
}
sort(a, a + n + 1);
long long int dp[n + 1];
dp[0] = 0;
for (int i = 1; i < n + 1; i++) {
long long int p = i - a[i];
long long int z = 1;
if (p < 0) {
z = 0;
p = 0;
}
dp[i] = max(dp[i - 1], z + dp[p]);
}
cout << dp[n] << endl;
}
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
bool compare(long long int a, long long int b) { return a > b; }
int32_t main() {
int t;
cin >> t;
while (t--) {
long long int n, x;
cin >> n >> x;
long long int a[n + 1];
a[0] = 0;
for (int i = 1; i < n + 1; i++) {
cin >> a[i];
a[i] = (x / a[i]) + ((x % a[i]) != 0 ? 1 : 0);
}
sort(a, a + n + 1);
long long int dp[n + 1];
dp[0] = 0;
for (int i = 1; i < n + 1; i++) {
long long int p = i - a[i];
long long int z = 1;
if (p < 0) {
z = 0;
p = 0;
}
dp[i] = max(dp[i - 1], z + dp[p]);
}
cout << dp[n] << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
cin.tie(0);
ios_base::sync_with_stdio(false);
long long t;
cin >> t;
while (t--) {
long long n, x, c = 0, i, j, k = 1;
cin >> n >> x;
vector<long long> v(n);
for (auto &it : v) cin >> it;
sort((v).rbegin(), (v).rend());
for (i = 0; i < n; i++) {
if (v[i] * k >= x) {
k = 0;
c++;
}
k++;
}
cout << c << "\n";
}
}
| ### Prompt
Please create a solution in Cpp to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
cin.tie(0);
ios_base::sync_with_stdio(false);
long long t;
cin >> t;
while (t--) {
long long n, x, c = 0, i, j, k = 1;
cin >> n >> x;
vector<long long> v(n);
for (auto &it : v) cin >> it;
sort((v).rbegin(), (v).rend());
for (i = 0; i < n; i++) {
if (v[i] * k >= x) {
k = 0;
c++;
}
k++;
}
cout << c << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int a[100005];
int main() {
int T;
scanf("%d", &T);
while (T--) {
int n, x;
scanf("%d%d", &n, &x);
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
sort(a + 1, a + 1 + n);
int l = 1, ans = 0;
for (int i = n; i >= 1; i--) {
if (a[i] * l >= x) {
ans++;
l = 0;
}
l++;
}
printf("%d\n", ans);
}
return 0;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a[100005];
int main() {
int T;
scanf("%d", &T);
while (T--) {
int n, x;
scanf("%d%d", &n, &x);
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
sort(a + 1, a + 1 + n);
int l = 1, ans = 0;
for (int i = n; i >= 1; i--) {
if (a[i] * l >= x) {
ans++;
l = 0;
}
l++;
}
printf("%d\n", ans);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b) {
long long r = 1;
while (b != 0) {
if (b % 2 != 0) {
r = r * a;
r %= 998244353ll;
}
a = a * a;
a = a % 998244353ll;
b = b / 2;
}
return r;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t = 1;
cin >> t;
for (long long lp = (1); lp <= (t); lp++) {
long long n, x, ans = 0, cr = 0;
cin >> n >> x;
long long ast[n + 1];
for (long long i = (1); i <= (n); i++) {
cin >> ast[i];
}
sort(ast + 1, ast + n + 1);
for (long long i = (1); i <= (n); i++) {
long long tgt;
if (x % i == 0) {
tgt = x / i;
} else {
tgt = (x / i);
tgt += 1ll;
}
long long p = lower_bound(ast + 1, ast + n + 1, tgt) - (ast);
p = (n - p + 1ll) - cr;
long long make_pair = (p / i);
ans = ans + make_pair;
cr = cr + (i * make_pair);
}
cout << ans << endl;
}
}
| ### Prompt
Generate a CPP solution to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b) {
long long r = 1;
while (b != 0) {
if (b % 2 != 0) {
r = r * a;
r %= 998244353ll;
}
a = a * a;
a = a % 998244353ll;
b = b / 2;
}
return r;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t = 1;
cin >> t;
for (long long lp = (1); lp <= (t); lp++) {
long long n, x, ans = 0, cr = 0;
cin >> n >> x;
long long ast[n + 1];
for (long long i = (1); i <= (n); i++) {
cin >> ast[i];
}
sort(ast + 1, ast + n + 1);
for (long long i = (1); i <= (n); i++) {
long long tgt;
if (x % i == 0) {
tgt = x / i;
} else {
tgt = (x / i);
tgt += 1ll;
}
long long p = lower_bound(ast + 1, ast + n + 1, tgt) - (ast);
p = (n - p + 1ll) - cr;
long long make_pair = (p / i);
ans = ans + make_pair;
cr = cr + (i * make_pair);
}
cout << ans << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long tc;
cin >> tc;
long long n, x;
while (tc--) {
cin >> n >> x;
long long a[n];
for (long long i = 0; i < n; i++) cin >> a[i];
sort(a, a + n, greater<long long>());
long long cnt = 0, pn = 1;
for (long long i = 0; i < n; i++) {
if (pn * a[i] >= x) {
cnt++;
pn = 1;
} else
pn++;
}
cout << cnt << endl;
}
}
| ### Prompt
In cpp, your task is to solve the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long tc;
cin >> tc;
long long n, x;
while (tc--) {
cin >> n >> x;
long long a[n];
for (long long i = 0; i < n; i++) cin >> a[i];
sort(a, a + n, greater<long long>());
long long cnt = 0, pn = 1;
for (long long i = 0; i < n; i++) {
if (pn * a[i] >= x) {
cnt++;
pn = 1;
} else
pn++;
}
cout << cnt << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxN = 1e5 + 10;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(nullptr);
int t;
cin >> t;
while (t--) {
int n, x, a[maxN], cur = 1, res = 0;
cin >> n >> x;
for (int i = 1; i <= n; ++i) cin >> a[i];
sort(a + 1, a + 1 + n, greater<int>());
for (int i = 1; i <= n; ++i) {
if (cur * a[i] >= x) {
res++;
cur = 0;
}
cur += 1;
}
cout << res << '\n';
}
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxN = 1e5 + 10;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(nullptr);
int t;
cin >> t;
while (t--) {
int n, x, a[maxN], cur = 1, res = 0;
cin >> n >> x;
for (int i = 1; i <= n; ++i) cin >> a[i];
sort(a + 1, a + 1 + n, greater<int>());
for (int i = 1; i <= n; ++i) {
if (cur * a[i] >= x) {
res++;
cur = 0;
}
cur += 1;
}
cout << res << '\n';
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T gcd(T a, T b) {
T t;
if (a < b) {
while (a) {
t = a;
a = b % a;
b = t;
}
return b;
} else {
while (b) {
t = b;
b = a % b;
a = t;
}
return a;
}
}
template <typename T>
inline T sqr(T x) {
return x * x;
}
long long a[100005];
int cmp(int a, int b) { return a > b; }
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
int n, x;
int tot = 0, pos = 0;
int lastpos = -1;
long long m;
long long v;
for (int o = 0, oend = t; o < oend; o++) {
cin >> n >> x;
tot = 0;
pos = 0;
v = 0;
lastpos = -1;
for (int i = 0, iend = n; i < iend; i++) cin >> a[i];
sort(a, a + n, cmp);
m = a[0];
while (true) {
m = a[pos];
v = m * (pos - lastpos);
if (v >= x) {
tot++;
lastpos = pos;
}
pos++;
if (pos >= n) break;
}
cout << tot << endl;
}
return 0;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T gcd(T a, T b) {
T t;
if (a < b) {
while (a) {
t = a;
a = b % a;
b = t;
}
return b;
} else {
while (b) {
t = b;
b = a % b;
a = t;
}
return a;
}
}
template <typename T>
inline T sqr(T x) {
return x * x;
}
long long a[100005];
int cmp(int a, int b) { return a > b; }
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
int n, x;
int tot = 0, pos = 0;
int lastpos = -1;
long long m;
long long v;
for (int o = 0, oend = t; o < oend; o++) {
cin >> n >> x;
tot = 0;
pos = 0;
v = 0;
lastpos = -1;
for (int i = 0, iend = n; i < iend; i++) cin >> a[i];
sort(a, a + n, cmp);
m = a[0];
while (true) {
m = a[pos];
v = m * (pos - lastpos);
if (v >= x) {
tot++;
lastpos = pos;
}
pos++;
if (pos >= n) break;
}
cout << tot << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
const int MOD = 1e9 + 7;
const int v1 = 1e5 + 5;
const int N = 3e5 + 6;
using namespace std;
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
long long t = 0;
cin >> t;
while (t--) {
long long n = 0;
long long k = 0;
cin >> n >> k;
vector<long long> a(n);
for (long long i = 0; i < n; i++) cin >> a[i];
sort((a).begin(), (a).end(), greater<int>());
long long count = 0;
long long ans = 0;
for (long long i = 0; i < n; i++) {
if (a[i] >= k) {
a[i] = 0;
ans++;
} else {
long long m = k / a[i];
if (k % a[i]) m++;
a[i] = m;
if (count >= m) {
count = 0;
ans++;
} else {
count++;
if (count == m) {
count = 0;
ans++;
}
}
}
}
cout << ans << "\n";
}
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
const int MOD = 1e9 + 7;
const int v1 = 1e5 + 5;
const int N = 3e5 + 6;
using namespace std;
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
long long t = 0;
cin >> t;
while (t--) {
long long n = 0;
long long k = 0;
cin >> n >> k;
vector<long long> a(n);
for (long long i = 0; i < n; i++) cin >> a[i];
sort((a).begin(), (a).end(), greater<int>());
long long count = 0;
long long ans = 0;
for (long long i = 0; i < n; i++) {
if (a[i] >= k) {
a[i] = 0;
ans++;
} else {
long long m = k / a[i];
if (k % a[i]) m++;
a[i] = m;
if (count >= m) {
count = 0;
ans++;
} else {
count++;
if (count == m) {
count = 0;
ans++;
}
}
}
}
cout << ans << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long int mod = 1e9 + 7;
long double Pi = 3.14159265359;
auto clk = clock();
mt19937_64 rang(
chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int lim) {
uniform_int_distribution<int> uid(0, lim - 1);
return uid(rang);
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.precision(17);
int t;
t = 1;
cin >> t;
for (long long int test = 0; test < (t); test++) {
long long int n, x;
cin >> n >> x;
vector<long long int> a(n);
for (long long int i = 0; i < (n); i++) cin >> a[i];
sort((a).begin(), (a).end());
long long int ans = 0, ct = 0;
for (int i = n - 1; i >= 0; i--) {
ct++;
if (a[i] * ct >= x) {
ans++;
ct = 0;
}
}
cout << ans << endl;
}
}
| ### Prompt
Develop a solution in CPP to the problem described below:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long int mod = 1e9 + 7;
long double Pi = 3.14159265359;
auto clk = clock();
mt19937_64 rang(
chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int lim) {
uniform_int_distribution<int> uid(0, lim - 1);
return uid(rang);
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.precision(17);
int t;
t = 1;
cin >> t;
for (long long int test = 0; test < (t); test++) {
long long int n, x;
cin >> n >> x;
vector<long long int> a(n);
for (long long int i = 0; i < (n); i++) cin >> a[i];
sort((a).begin(), (a).end());
long long int ans = 0, ct = 0;
for (int i = n - 1; i >= 0; i--) {
ct++;
if (a[i] * ct >= x) {
ans++;
ct = 0;
}
}
cout << ans << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long t = 0;
cin >> t;
while (t--) {
long long n, x, ans = 0, end;
cin >> n >> x;
long long a[n];
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
end = n;
for (long long i = n - 1; i >= 0; i--) {
long long y = x / a[i];
if (x % a[i] != 0) y++;
if (i + y - 1 < end) {
ans += 1;
end = end - y;
}
}
cout << ans << endl;
}
return 0;
}
| ### Prompt
Create a solution in CPP for the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long t = 0;
cin >> t;
while (t--) {
long long n, x, ans = 0, end;
cin >> n >> x;
long long a[n];
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
end = n;
for (long long i = n - 1; i >= 0; i--) {
long long y = x / a[i];
if (x % a[i] != 0) y++;
if (i + y - 1 < end) {
ans += 1;
end = end - y;
}
}
cout << ans << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int a[100005], t, n, x;
int main() {
scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &x);
for (int i = 0; i < n; ++i) scanf("%d", a + i);
a[n] = 0;
sort(a, a + n, greater<int>());
int ans = 0, i = 0, now = 1;
while (i < n) {
while (i < n && 1ll * now * a[i] < x) ++i, ++now;
if (1ll * now * a[i] >= x) ++ans;
++i, now = 1;
}
printf("%d\n", ans);
}
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a[100005], t, n, x;
int main() {
scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &x);
for (int i = 0; i < n; ++i) scanf("%d", a + i);
a[n] = 0;
sort(a, a + n, greater<int>());
int ans = 0, i = 0, now = 1;
while (i < n) {
while (i < n && 1ll * now * a[i] < x) ++i, ++now;
if (1ll * now * a[i] >= x) ++ans;
++i, now = 1;
}
printf("%d\n", ans);
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int a[100010];
int main() {
int t;
cin >> t;
while (t--) {
int n, k, ans = 0;
cin >> n >> k;
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
for (int i = n - 1; i >= 0; i--) {
int mm = a[i], g = 0;
if (mm >= k) {
ans++;
} else {
while (mm * g < k && i >= 0) {
mm = a[i];
g++;
i--;
}
i++;
if (mm * g >= k) ans++;
}
}
cout << ans << endl;
}
return 0;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a[100010];
int main() {
int t;
cin >> t;
while (t--) {
int n, k, ans = 0;
cin >> n >> k;
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
for (int i = n - 1; i >= 0; i--) {
int mm = a[i], g = 0;
if (mm >= k) {
ans++;
} else {
while (mm * g < k && i >= 0) {
mm = a[i];
g++;
i--;
}
i++;
if (mm * g >= k) ans++;
}
}
cout << ans << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
for (int q = 1; q <= t; q++) {
long long int n;
long long int x;
cin >> n >> x;
vector<long long int> skls;
for (int i = 0; i < n; i++) {
long long int inp;
cin >> inp;
skls.push_back(inp);
}
sort(skls.begin(), skls.end());
long long int cnt = 0;
long long int sk = 0;
long long int ptr = 1;
for (int i = n - 1; i >= 0; i--) {
sk = ptr * skls[i];
if (sk >= x) {
cnt++;
ptr = 1;
sk = 0;
} else {
ptr++;
}
}
cout << cnt << endl;
}
}
| ### Prompt
Please formulate a CPP solution to the following problem:
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer — the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
for (int q = 1; q <= t; q++) {
long long int n;
long long int x;
cin >> n >> x;
vector<long long int> skls;
for (int i = 0; i < n; i++) {
long long int inp;
cin >> inp;
skls.push_back(inp);
}
sort(skls.begin(), skls.end());
long long int cnt = 0;
long long int sk = 0;
long long int ptr = 1;
for (int i = n - 1; i >= 0; i--) {
sk = ptr * skls[i];
if (sk >= x) {
cnt++;
ptr = 1;
sk = 0;
} else {
ptr++;
}
}
cout << cnt << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int T;
cin >> T;
while (T--) {
string s;
cin >> s;
int n = s.size();
int x;
cin >> x;
vector<char> ans(n, '1');
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i + x < n) ans[i + x] = '0';
if (i - x >= 0) ans[i - x] = '0';
}
}
int flag = 0;
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
if (i + x < n && i - x >= 0) {
if (ans[i + x] == '0' && ans[i - x] == '0') {
flag = 1;
break;
}
} else if (i + x < n) {
if (ans[i + x] == '0') {
flag = 1;
break;
}
} else if (i - x >= 0) {
if (ans[i - x] == '0') {
flag = 1;
break;
}
} else {
flag = 1;
break;
}
}
}
if (flag)
cout << "-1\n";
else {
for (int i = 0; i < n; i++) cout << ans[i];
cout << "\n";
}
}
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int T;
cin >> T;
while (T--) {
string s;
cin >> s;
int n = s.size();
int x;
cin >> x;
vector<char> ans(n, '1');
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i + x < n) ans[i + x] = '0';
if (i - x >= 0) ans[i - x] = '0';
}
}
int flag = 0;
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
if (i + x < n && i - x >= 0) {
if (ans[i + x] == '0' && ans[i - x] == '0') {
flag = 1;
break;
}
} else if (i + x < n) {
if (ans[i + x] == '0') {
flag = 1;
break;
}
} else if (i - x >= 0) {
if (ans[i - x] == '0') {
flag = 1;
break;
}
} else {
flag = 1;
break;
}
}
}
if (flag)
cout << "-1\n";
else {
for (int i = 0; i < n; i++) cout << ans[i];
cout << "\n";
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long p, f, s, a, sw, aw, ps = 0, pa = 0, fs = 0, fa = 0;
int main() {
int t;
cin >> t;
while (t--) {
string a;
cin >> a;
int n = a.length(), x;
cin >> x;
vector<int> s(n + 1, -1);
for (int i = 1; i <= n; i++) s[i] = a[i - 1] - '0';
vector<int> w(n + 1, -1);
for (int i = 1; i <= n; i++) {
if (s[i] == 0) {
if (i + x <= n) {
w[i + x] = 0;
}
if (i > x) {
w[i - x] = 0;
}
if (i + x > n && i <= x) {
w[i] = 0;
}
}
}
bool poss = true;
for (int i = 1; i <= n; i++) {
if (s[i] == 1) {
if (i + x > n && i <= x) {
poss = false;
break;
}
if (i + x <= n && i <= x) {
if (w[i + x] == 0) {
poss = false;
break;
} else {
w[i + x] = 1;
}
} else if (i + x > n && i > x) {
if (w[i - x] == 0) {
poss = false;
break;
} else {
w[i - x] = 1;
}
} else if (i + x <= n && i > x) {
if (w[i + x] == 0 && w[i - x] == 0) {
poss = false;
break;
} else {
if (w[i + x] == -1) w[i + x] = 1;
if (w[i - x] == -1) w[i - x] = 1;
}
}
}
}
if (poss == false) {
cout << -1 << "\n";
continue;
}
for (int i = 1; i <= n; i++) {
cout << w[i];
}
cout << "\n";
}
return 0;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long p, f, s, a, sw, aw, ps = 0, pa = 0, fs = 0, fa = 0;
int main() {
int t;
cin >> t;
while (t--) {
string a;
cin >> a;
int n = a.length(), x;
cin >> x;
vector<int> s(n + 1, -1);
for (int i = 1; i <= n; i++) s[i] = a[i - 1] - '0';
vector<int> w(n + 1, -1);
for (int i = 1; i <= n; i++) {
if (s[i] == 0) {
if (i + x <= n) {
w[i + x] = 0;
}
if (i > x) {
w[i - x] = 0;
}
if (i + x > n && i <= x) {
w[i] = 0;
}
}
}
bool poss = true;
for (int i = 1; i <= n; i++) {
if (s[i] == 1) {
if (i + x > n && i <= x) {
poss = false;
break;
}
if (i + x <= n && i <= x) {
if (w[i + x] == 0) {
poss = false;
break;
} else {
w[i + x] = 1;
}
} else if (i + x > n && i > x) {
if (w[i - x] == 0) {
poss = false;
break;
} else {
w[i - x] = 1;
}
} else if (i + x <= n && i > x) {
if (w[i + x] == 0 && w[i - x] == 0) {
poss = false;
break;
} else {
if (w[i + x] == -1) w[i + x] = 1;
if (w[i - x] == -1) w[i - x] = 1;
}
}
}
}
if (poss == false) {
cout << -1 << "\n";
continue;
}
for (int i = 1; i <= n; i++) {
cout << w[i];
}
cout << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
const char* comma = strchr(names + 1, ',');
cerr.write(names, comma - names) << " : " << arg1 << " | ";
__f(comma + 1, args...);
}
int inf = 0x3f3f3f3f;
long long infl = 0x3f3f3f3f3f3f3f3fLL;
long double infd = 1.0 / 0.0;
const long long MOD = 1e9 + 7;
const long double pi = 2 * acos(0.0);
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int tt = 1;
cin >> tt;
while (tt--) {
string s;
cin >> s;
int x;
cin >> x;
int n = (int)s.size();
string tmp = string(n, '1');
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) tmp[i - x] = '0';
if (i + x < n) tmp[i + x] = '0';
}
}
string s1 = string(n, '0');
string w = tmp;
for (int i = 0; i < n; i++) {
if (i + x < n && w[i + x] == '1') s1[i] = '1';
if (i - x >= 0 && w[i - x] == '1') s1[i] = '1';
}
if (s1 != s) {
cout << -1 << '\n';
continue;
}
cout << tmp << '\n';
}
return 0;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
const char* comma = strchr(names + 1, ',');
cerr.write(names, comma - names) << " : " << arg1 << " | ";
__f(comma + 1, args...);
}
int inf = 0x3f3f3f3f;
long long infl = 0x3f3f3f3f3f3f3f3fLL;
long double infd = 1.0 / 0.0;
const long long MOD = 1e9 + 7;
const long double pi = 2 * acos(0.0);
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int tt = 1;
cin >> tt;
while (tt--) {
string s;
cin >> s;
int x;
cin >> x;
int n = (int)s.size();
string tmp = string(n, '1');
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) tmp[i - x] = '0';
if (i + x < n) tmp[i + x] = '0';
}
}
string s1 = string(n, '0');
string w = tmp;
for (int i = 0; i < n; i++) {
if (i + x < n && w[i + x] == '1') s1[i] = '1';
if (i - x >= 0 && w[i - x] == '1') s1[i] = '1';
}
if (s1 != s) {
cout << -1 << '\n';
continue;
}
cout << tmp << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int x;
cin >> x;
int n = s.length();
string t = "";
for (int i = 0; i < n; i++) t += "1";
for (int i = 0; i < n; i++)
if (s[i] == '0') {
if (i - x >= 0) t[i - x] = '0';
if (i + x < n) t[i + x] = '0';
}
bool can = true;
for (int i = 0; i < n; i++)
if (s[i] == '1') {
bool ok = false;
ok |= i - x >= 0 && t[i - x] == '1';
ok |= i + x < n && t[i + x] == '1';
can &= ok;
}
if (can) {
cout << t << "\n";
} else {
cout << "-1\n";
}
}
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int x;
cin >> x;
int n = s.length();
string t = "";
for (int i = 0; i < n; i++) t += "1";
for (int i = 0; i < n; i++)
if (s[i] == '0') {
if (i - x >= 0) t[i - x] = '0';
if (i + x < n) t[i + x] = '0';
}
bool can = true;
for (int i = 0; i < n; i++)
if (s[i] == '1') {
bool ok = false;
ok |= i - x >= 0 && t[i - x] == '1';
ok |= i + x < n && t[i + x] == '1';
can &= ok;
}
if (can) {
cout << t << "\n";
} else {
cout << "-1\n";
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n, x, no = 0;
string s;
cin >> s >> x;
n = s.length();
char w[n];
for (int a = 0; a < n; a++) w[a] = 'x';
for (int a = 0; a < n; a++) {
if (s[a] == '0') {
if ((a + x) < n) w[a + x] = '0';
if ((a - x) >= 0) w[a - x] = '0';
}
}
for (int a = 0; a < n; a++) {
if (s[a] == '1') {
int l = 0;
if ((a + x) < n && w[a + x] != '0') {
w[a + x] = '1';
l++;
}
if ((a - x) >= 0 && w[a - x] != '0') {
w[a - x] = '1';
l++;
}
if (l == 0) no++;
}
}
if (no > 0)
cout << "-1\n";
else {
for (int a = 0; a < n; a++) {
if (w[a] == 'x')
cout << "1";
else
cout << w[a];
}
cout << '\n';
}
}
}
| ### Prompt
Create a solution in cpp for the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n, x, no = 0;
string s;
cin >> s >> x;
n = s.length();
char w[n];
for (int a = 0; a < n; a++) w[a] = 'x';
for (int a = 0; a < n; a++) {
if (s[a] == '0') {
if ((a + x) < n) w[a + x] = '0';
if ((a - x) >= 0) w[a - x] = '0';
}
}
for (int a = 0; a < n; a++) {
if (s[a] == '1') {
int l = 0;
if ((a + x) < n && w[a + x] != '0') {
w[a + x] = '1';
l++;
}
if ((a - x) >= 0 && w[a - x] != '0') {
w[a - x] = '1';
l++;
}
if (l == 0) no++;
}
}
if (no > 0)
cout << "-1\n";
else {
for (int a = 0; a < n; a++) {
if (w[a] == 'x')
cout << "1";
else
cout << w[a];
}
cout << '\n';
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
char w[100005], s[100005];
void solve() {
scanf("%s", s);
int x;
scanf("%d", &x);
int n = strlen(s);
for (int i = 0; i < n; i++) w[i] = '2';
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
if (i - x >= 0 && w[i - x] == '1' || i + x < n && w[i + x] == '1')
continue;
else {
if (i - x >= 0 && w[i - x] == '2')
w[i - x] = '1';
else if (i + x < n && w[i + x] == '2')
w[i + x] = '1';
else {
printf("-1\n");
return;
}
}
} else {
if (i - x < 0 && i + x >= n)
continue;
else if (i + x >= n && w[i - x] == '0')
continue;
else if (i - x < 0 && w[i + x] == '0')
continue;
else {
if (i - x >= 0) {
if (w[i - x] == '1') {
puts("-1");
return;
} else
w[i - x] = '0';
}
if (i + x < n) {
if (w[i + x] == '1') {
puts("-1");
return;
} else
w[i + x] = '0';
}
}
}
}
for (int i = 0; i < n; i++) {
printf("%c", w[i] == '2' ? '0' : w[i]);
}
printf("\n");
}
int main() {
int T;
scanf("%d", &T);
while (T--) solve();
return 0;
}
| ### Prompt
Generate a cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
char w[100005], s[100005];
void solve() {
scanf("%s", s);
int x;
scanf("%d", &x);
int n = strlen(s);
for (int i = 0; i < n; i++) w[i] = '2';
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
if (i - x >= 0 && w[i - x] == '1' || i + x < n && w[i + x] == '1')
continue;
else {
if (i - x >= 0 && w[i - x] == '2')
w[i - x] = '1';
else if (i + x < n && w[i + x] == '2')
w[i + x] = '1';
else {
printf("-1\n");
return;
}
}
} else {
if (i - x < 0 && i + x >= n)
continue;
else if (i + x >= n && w[i - x] == '0')
continue;
else if (i - x < 0 && w[i + x] == '0')
continue;
else {
if (i - x >= 0) {
if (w[i - x] == '1') {
puts("-1");
return;
} else
w[i - x] = '0';
}
if (i + x < n) {
if (w[i + x] == '1') {
puts("-1");
return;
} else
w[i + x] = '0';
}
}
}
}
for (int i = 0; i < n; i++) {
printf("%c", w[i] == '2' ? '0' : w[i]);
}
printf("\n");
}
int main() {
int T;
scanf("%d", &T);
while (T--) solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void solve() {
int64_t n, x, i;
string str;
cin >> str;
cin >> x;
string ans = "";
for (int64_t i = 0; i < str.size(); i++) {
ans += "1";
}
for (int64_t i = 0; i < str.size(); i++) {
if (str[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < str.size()) ans[i + x] = '0';
}
}
for (int64_t i = 0; i < ans.size(); i++) {
bool flag = false;
if (str[i] == '1') {
if (i - x >= 0 and ans[i - x] == '1') flag = true;
if (i + x < str.size() and ans[i + x] == '1') flag = true;
if (!flag) {
cout << "-1" << '\n';
return;
}
}
}
cout << ans << '\n';
}
signed main() {
std::ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int64_t tt;
cin >> tt;
while (tt--) solve();
cerr << " Execution : " << (1.0 * clock()) / CLOCKS_PER_SEC << "s \n";
return 0;
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
int64_t n, x, i;
string str;
cin >> str;
cin >> x;
string ans = "";
for (int64_t i = 0; i < str.size(); i++) {
ans += "1";
}
for (int64_t i = 0; i < str.size(); i++) {
if (str[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < str.size()) ans[i + x] = '0';
}
}
for (int64_t i = 0; i < ans.size(); i++) {
bool flag = false;
if (str[i] == '1') {
if (i - x >= 0 and ans[i - x] == '1') flag = true;
if (i + x < str.size() and ans[i + x] == '1') flag = true;
if (!flag) {
cout << "-1" << '\n';
return;
}
}
}
cout << ans << '\n';
}
signed main() {
std::ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int64_t tt;
cin >> tt;
while (tt--) solve();
cerr << " Execution : " << (1.0 * clock()) / CLOCKS_PER_SEC << "s \n";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <typename P, typename Q>
P CeilDiv(P _dividend, Q _divider) {
return static_cast<P>((_dividend - 1LL + _divider) / _divider);
}
template <typename Tp>
inline void outarr(Tp _begin, Tp _end, const char* _delim = " ") {
for (Tp current = _begin; current != _end; ++current) {
std::cout << *current << _delim;
}
std::cout << '\n';
}
using ll = long long;
using pii = std::pair<int, int>;
constexpr int INF = 0x3f3f3f3f;
constexpr int MOD = static_cast<const int>(1e9 + 7);
string Gen(const string& s, int x) {
string res(static_cast<int>((s).size()), '0');
for (int i = 0; i < (static_cast<int>((s).size())); ++i) {
if (i - x >= 0 && s[i - x] == '1') {
res[i] = '1';
}
if (i + x < static_cast<int>((s).size()) && s[i + x] == '1') {
res[i] = '1';
}
}
return res;
}
int main() {
std::ios_base::sync_with_stdio(false);
std::cin.tie(nullptr);
int test_cases;
cin >> test_cases;
while (test_cases--) {
string s;
int x;
cin >> s >> x;
string ans(static_cast<int>((s).size()), '1');
for (int i = 0; i < (static_cast<int>((s).size())); ++i) {
if (s[i] == '1') {
continue;
}
if (i - x >= 0) {
ans[i - x] = '0';
}
if (i + x < static_cast<int>((s).size())) {
ans[i + x] = '0';
}
}
if (s != Gen(ans, x)) {
cout << -1 << "\n";
} else {
cout << ans << "\n";
}
}
return 0;
}
| ### Prompt
Develop a solution in CPP to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename P, typename Q>
P CeilDiv(P _dividend, Q _divider) {
return static_cast<P>((_dividend - 1LL + _divider) / _divider);
}
template <typename Tp>
inline void outarr(Tp _begin, Tp _end, const char* _delim = " ") {
for (Tp current = _begin; current != _end; ++current) {
std::cout << *current << _delim;
}
std::cout << '\n';
}
using ll = long long;
using pii = std::pair<int, int>;
constexpr int INF = 0x3f3f3f3f;
constexpr int MOD = static_cast<const int>(1e9 + 7);
string Gen(const string& s, int x) {
string res(static_cast<int>((s).size()), '0');
for (int i = 0; i < (static_cast<int>((s).size())); ++i) {
if (i - x >= 0 && s[i - x] == '1') {
res[i] = '1';
}
if (i + x < static_cast<int>((s).size()) && s[i + x] == '1') {
res[i] = '1';
}
}
return res;
}
int main() {
std::ios_base::sync_with_stdio(false);
std::cin.tie(nullptr);
int test_cases;
cin >> test_cases;
while (test_cases--) {
string s;
int x;
cin >> s >> x;
string ans(static_cast<int>((s).size()), '1');
for (int i = 0; i < (static_cast<int>((s).size())); ++i) {
if (s[i] == '1') {
continue;
}
if (i - x >= 0) {
ans[i - x] = '0';
}
if (i + x < static_cast<int>((s).size())) {
ans[i + x] = '0';
}
}
if (s != Gen(ans, x)) {
cout << -1 << "\n";
} else {
cout << ans << "\n";
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
const int MOD = 1e9 + 7;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
cout.tie(NULL);
long long int t;
cin >> t;
while (t--) {
string s;
cin >> s;
long long int x;
cin >> x;
string res;
for (int i = 0; i < s.size(); i++) res += '#';
for (int i = 0; i < s.size(); i++) {
if (s[i] == '0') {
if (i - x >= 0) res[i - x] = '0';
if (i + x < s.size()) res[i + x] = '0';
}
}
bool fl = 0;
for (int i = 0; i < s.size(); i++) {
if (s[i] == '1') {
if (i - x >= 0 && res[i - x] != '0')
res[i - x] = '1';
else if (i + x < s.size() && res[i + x] != '0') {
res[i + x] = '1';
} else {
fl = 1;
break;
}
}
}
if (fl == 1) {
cout << -1 << '\n';
continue;
}
for (int i = 0; i < s.size(); i++) {
if (res[i] == '#') res[i] = '0';
}
cout << res << '\n';
}
return 0;
}
| ### Prompt
Create a solution in CPP for the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
const int MOD = 1e9 + 7;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
cout.tie(NULL);
long long int t;
cin >> t;
while (t--) {
string s;
cin >> s;
long long int x;
cin >> x;
string res;
for (int i = 0; i < s.size(); i++) res += '#';
for (int i = 0; i < s.size(); i++) {
if (s[i] == '0') {
if (i - x >= 0) res[i - x] = '0';
if (i + x < s.size()) res[i + x] = '0';
}
}
bool fl = 0;
for (int i = 0; i < s.size(); i++) {
if (s[i] == '1') {
if (i - x >= 0 && res[i - x] != '0')
res[i - x] = '1';
else if (i + x < s.size() && res[i + x] != '0') {
res[i + x] = '1';
} else {
fl = 1;
break;
}
}
}
if (fl == 1) {
cout << -1 << '\n';
continue;
}
for (int i = 0; i < s.size(); i++) {
if (res[i] == '#') res[i] = '0';
}
cout << res << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
const int alldx[] = {1, -1, 0, 0, 1, -1, 1, -1};
const int alldy[] = {0, 0, 1, -1, 1, -1, -1, 1};
const int dx[] = {1, -1, 0, 0};
const int dy[] = {0, 0, 1, -1};
struct matrix {
int a[2][2];
matrix operator*(matrix &b) {
matrix product;
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
product.a[i][j] = 0;
}
}
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++)
for (int k = 0; k < 2; k++) product.a[i][k] += (a[i][j] * b.a[j][k]);
return product;
}
matrix operator+(matrix &b) {
matrix sum;
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++) sum.a[i][j] = 0;
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++) sum.a[i][j] += a[i][j] + b.a[i][j];
return sum;
}
matrix operator-(matrix &b) {
matrix difference;
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++) difference.a[i][j] = 0;
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++) difference.a[i][j] += a[i][j] - b.a[i][j];
return difference;
}
};
using namespace std;
void task() {
string s;
cin >> s;
int k;
cin >> k;
int n = s.size();
string ans = "";
for (int i = 0; i < n; i++) {
ans += '1';
}
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i + k < n) {
ans[i + k] = '0';
}
if (i - k >= 0) {
ans[i - k] = '0';
}
}
}
bool pos = true;
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
if (((i + k < n) && ans[i + k] == '1') ||
(i - k >= 0 && ans[i - k] == '1')) {
pos = true;
} else {
pos = false;
break;
}
}
}
if (!pos) {
cout << "-1\n";
} else {
cout << ans << endl;
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int TT;
cin >> TT;
while (TT--) {
task();
}
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
const int alldx[] = {1, -1, 0, 0, 1, -1, 1, -1};
const int alldy[] = {0, 0, 1, -1, 1, -1, -1, 1};
const int dx[] = {1, -1, 0, 0};
const int dy[] = {0, 0, 1, -1};
struct matrix {
int a[2][2];
matrix operator*(matrix &b) {
matrix product;
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
product.a[i][j] = 0;
}
}
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++)
for (int k = 0; k < 2; k++) product.a[i][k] += (a[i][j] * b.a[j][k]);
return product;
}
matrix operator+(matrix &b) {
matrix sum;
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++) sum.a[i][j] = 0;
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++) sum.a[i][j] += a[i][j] + b.a[i][j];
return sum;
}
matrix operator-(matrix &b) {
matrix difference;
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++) difference.a[i][j] = 0;
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++) difference.a[i][j] += a[i][j] - b.a[i][j];
return difference;
}
};
using namespace std;
void task() {
string s;
cin >> s;
int k;
cin >> k;
int n = s.size();
string ans = "";
for (int i = 0; i < n; i++) {
ans += '1';
}
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i + k < n) {
ans[i + k] = '0';
}
if (i - k >= 0) {
ans[i - k] = '0';
}
}
}
bool pos = true;
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
if (((i + k < n) && ans[i + k] == '1') ||
(i - k >= 0 && ans[i - k] == '1')) {
pos = true;
} else {
pos = false;
break;
}
}
}
if (!pos) {
cout << "-1\n";
} else {
cout << ans << endl;
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int TT;
cin >> TT;
while (TT--) {
task();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int x;
cin >> x;
int n = s.size();
vector<char> ans(n, '1');
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i + x < n) ans[i + x] = '0';
if (i - x >= 0) ans[i - x] = '0';
}
}
bool flag = true;
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
bool pos = 0;
if (i + x < n && ans[i + x] == '1') pos = 1;
if (i - x >= 0 && ans[i - x] == '1') pos = 1;
if (!pos) flag = false;
}
}
if (flag)
for (char i : ans) cout << i;
else
cout << -1;
cout << "\n";
}
}
| ### Prompt
Generate a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int x;
cin >> x;
int n = s.size();
vector<char> ans(n, '1');
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i + x < n) ans[i + x] = '0';
if (i - x >= 0) ans[i - x] = '0';
}
}
bool flag = true;
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
bool pos = 0;
if (i + x < n && ans[i + x] == '1') pos = 1;
if (i - x >= 0 && ans[i - x] == '1') pos = 1;
if (!pos) flag = false;
}
}
if (flag)
for (char i : ans) cout << i;
else
cout << -1;
cout << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int mod = (int)1e9 + 7;
class apurvakb {
public:
apurvakb() {
string s;
cin >> s;
int x;
cin >> x;
int n = s.size();
string str = string(n, '1');
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x > -1) str[i - x] = '0';
if (i + x < n) str[i + x] = '0';
}
}
for (int i = 0; i < n; i++) {
bool fl = false;
fl = fl or (i - x > -1 and str[i - x] == '1');
fl = fl or (i + x < n and str[i + x] == '1');
if ((s[i] - '0') != fl) {
cout << -1 << endl;
return;
}
}
cout << str << endl;
}
};
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int T = 1;
cin >> T;
int cas = 1;
while (T--) {
apurvakb obj;
}
return 0;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = (int)1e9 + 7;
class apurvakb {
public:
apurvakb() {
string s;
cin >> s;
int x;
cin >> x;
int n = s.size();
string str = string(n, '1');
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x > -1) str[i - x] = '0';
if (i + x < n) str[i + x] = '0';
}
}
for (int i = 0; i < n; i++) {
bool fl = false;
fl = fl or (i - x > -1 and str[i - x] == '1');
fl = fl or (i + x < n and str[i + x] == '1');
if ((s[i] - '0') != fl) {
cout << -1 << endl;
return;
}
}
cout << str << endl;
}
};
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int T = 1;
cin >> T;
int cas = 1;
while (T--) {
apurvakb obj;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const long long mod = 1e9 + 7;
const int mxN = 1e5 + 9;
char s[mxN];
void solve() {
string res;
int x;
cin >> res >> x;
int n = res.length();
for (int i = 0; i < n; i++) s[i] = '1';
for (int i = 0; i < n; i++) {
if (res[i] == '0') {
if (i + x < n) s[i + x] = '0';
if (i - x >= 0) s[i - x] = '0';
}
}
for (int i = 0; i < n; i++) {
if (res[i] == '1') {
bool ok1 = false;
bool ok2 = false;
if (i + x < n && s[i + x] == '1') ok1 = true;
if (i - x >= 0 && s[i - x] == '1') ok2 = true;
if (ok1 || ok2) continue;
cout << -1 << "\n";
return;
}
}
for (int i = 0; i < n; i++) cout << s[i];
cout << "\n";
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while (t--) {
solve();
}
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const long long mod = 1e9 + 7;
const int mxN = 1e5 + 9;
char s[mxN];
void solve() {
string res;
int x;
cin >> res >> x;
int n = res.length();
for (int i = 0; i < n; i++) s[i] = '1';
for (int i = 0; i < n; i++) {
if (res[i] == '0') {
if (i + x < n) s[i + x] = '0';
if (i - x >= 0) s[i - x] = '0';
}
}
for (int i = 0; i < n; i++) {
if (res[i] == '1') {
bool ok1 = false;
bool ok2 = false;
if (i + x < n && s[i + x] == '1') ok1 = true;
if (i - x >= 0 && s[i - x] == '1') ok2 = true;
if (ok1 || ok2) continue;
cout << -1 << "\n";
return;
}
}
for (int i = 0; i < n; i++) cout << s[i];
cout << "\n";
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while (t--) {
solve();
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long t;
scanf("%ld", &t);
while (t--) {
int x, i, n;
string w, s;
cin >> s >> x;
n = s.length();
w = string(n, '1');
for (i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x < n) w[i + x] = '0';
}
}
for (i = 0; i < n; i++) {
if ((i - x >= 0 && w[i - x] == '1') || (i + x < n && w[i + x] == '1')) {
if (s[i] != '1') break;
} else if (s[i] != '0')
break;
}
if (i < n)
cout << -1;
else
cout << w;
cout << endl;
}
return 0;
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long t;
scanf("%ld", &t);
while (t--) {
int x, i, n;
string w, s;
cin >> s >> x;
n = s.length();
w = string(n, '1');
for (i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x < n) w[i + x] = '0';
}
}
for (i = 0; i < n; i++) {
if ((i - x >= 0 && w[i - x] == '1') || (i + x < n && w[i + x] == '1')) {
if (s[i] != '1') break;
} else if (s[i] != '0')
break;
}
if (i < n)
cout << -1;
else
cout << w;
cout << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
long long int t;
cin >> t;
while (t--) {
string s;
cin >> s;
long long int x;
cin >> x;
string temp = "";
long long int n = s.length();
for (long long int i = 0; i < s.length(); ++i) {
temp += '1';
}
for (long long int i = 0; i < s.length(); ++i) {
if (s[i] == '0') {
if (i + x < n) temp[i + x] = '0';
if (i - x >= 0) temp[i - x] = '0';
}
}
string ss;
long long int flag = 0;
for (long long int i = 0; i < n; ++i) {
ss += '2';
}
for (long long int i = 0; i < n; ++i) {
if (i + x < n) {
if (temp[i + x] == '1') ss[i] = '1';
}
if (i - x >= 0) {
if (temp[i - x] == '1') ss[i] = '1';
}
}
for (long long int i = 0; i < n; ++i) {
if (ss[i] == '2') ss[i] = '0';
}
if (ss == s)
cout << temp << "\n";
else
cout << -1 << "\n";
}
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
long long int t;
cin >> t;
while (t--) {
string s;
cin >> s;
long long int x;
cin >> x;
string temp = "";
long long int n = s.length();
for (long long int i = 0; i < s.length(); ++i) {
temp += '1';
}
for (long long int i = 0; i < s.length(); ++i) {
if (s[i] == '0') {
if (i + x < n) temp[i + x] = '0';
if (i - x >= 0) temp[i - x] = '0';
}
}
string ss;
long long int flag = 0;
for (long long int i = 0; i < n; ++i) {
ss += '2';
}
for (long long int i = 0; i < n; ++i) {
if (i + x < n) {
if (temp[i + x] == '1') ss[i] = '1';
}
if (i - x >= 0) {
if (temp[i - x] == '1') ss[i] = '1';
}
}
for (long long int i = 0; i < n; ++i) {
if (ss[i] == '2') ss[i] = '0';
}
if (ss == s)
cout << temp << "\n";
else
cout << -1 << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while (t--) {
string str;
cin >> str;
int x;
cin >> x;
int n = str.size();
vector<char> v(n, '$');
for (int i = 0; i < n; i++) {
if (str[i] == '0') {
if (i - x >= 0) v[i - x] = '0';
if (i + x < n) v[i + x] = '0';
}
}
for (int i = 0; i < n; i++)
if (v[i] == '$') v[i] = '1';
string s = "";
for (int i = 0; i < n; i++) {
if (i - x >= 0 && v[i - x] == '1') {
s += '1';
continue;
}
if (i + x < n && v[i + x] == '1') {
s += '1';
continue;
}
s += '0';
}
if (s == str) {
for (int i = 0; i < n; i++) cout << v[i];
cout << endl;
} else {
cout << -1 << endl;
}
}
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while (t--) {
string str;
cin >> str;
int x;
cin >> x;
int n = str.size();
vector<char> v(n, '$');
for (int i = 0; i < n; i++) {
if (str[i] == '0') {
if (i - x >= 0) v[i - x] = '0';
if (i + x < n) v[i + x] = '0';
}
}
for (int i = 0; i < n; i++)
if (v[i] == '$') v[i] = '1';
string s = "";
for (int i = 0; i < n; i++) {
if (i - x >= 0 && v[i - x] == '1') {
s += '1';
continue;
}
if (i + x < n && v[i + x] == '1') {
s += '1';
continue;
}
s += '0';
}
if (s == str) {
for (int i = 0; i < n; i++) cout << v[i];
cout << endl;
} else {
cout << -1 << endl;
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1000000007;
void __print(long long int x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
cerr << '{';
__print(x.first);
cerr << ',';
__print(x.second);
cerr << '}';
}
template <typename T>
void __print(const T &x) {
long long int f = 0;
cerr << '{';
for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
__print(t);
if (sizeof...(v)) cerr << ", ";
_print(v...);
}
void solve() {
string s;
cin >> s;
long long int x;
cin >> x;
long long int n = s.length();
string ans(n, '1');
for (long long int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) {
ans[i - x] = '0';
}
if (i + x < n) {
ans[i + x] = '0';
}
}
};
bool check = true;
for (long long int i = 0; i < n; i++) {
if (s[i] == '1') {
bool l = false;
bool r = false;
if (i - x >= 0 && ans[i - x] == '1') {
l = true;
}
if (i + x < n && ans[i + x] == '1') {
r = true;
}
if (l == false && r == false) {
check = false;
break;
}
}
}
if (!check) {
cout << -1 << endl;
} else {
cout << ans << endl;
}
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long int t;
cin >> t;
while (t--) {
solve();
}
}
| ### Prompt
Generate a CPP solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1000000007;
void __print(long long int x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
cerr << '{';
__print(x.first);
cerr << ',';
__print(x.second);
cerr << '}';
}
template <typename T>
void __print(const T &x) {
long long int f = 0;
cerr << '{';
for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
__print(t);
if (sizeof...(v)) cerr << ", ";
_print(v...);
}
void solve() {
string s;
cin >> s;
long long int x;
cin >> x;
long long int n = s.length();
string ans(n, '1');
for (long long int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) {
ans[i - x] = '0';
}
if (i + x < n) {
ans[i + x] = '0';
}
}
};
bool check = true;
for (long long int i = 0; i < n; i++) {
if (s[i] == '1') {
bool l = false;
bool r = false;
if (i - x >= 0 && ans[i - x] == '1') {
l = true;
}
if (i + x < n && ans[i + x] == '1') {
r = true;
}
if (l == false && r == false) {
check = false;
break;
}
}
}
if (!check) {
cout << -1 << endl;
} else {
cout << ans << endl;
}
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long int t;
cin >> t;
while (t--) {
solve();
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
void solve() {
string a;
int x;
cin >> a >> x;
int n = a.length();
vector<int> o(n, -1);
for (int i = 0; i < n; i++) {
if (i - x < 0 && i + x < n) {
if (o[i + x] == -1) o[i + x] = a[i] - '0';
if (o[i + x] != a[i] - '0') {
cout << "-1";
goto fail;
}
} else if (i + x >= n && i - x >= 0) {
if (o[i - x] == -1) o[i - x] = a[i] - '0';
if (o[i - x] != a[i] - '0') {
cout << "-1";
goto fail;
}
} else if (i + x >= n && i - x < 0) {
if (a[i] == '1') {
cout << "-1";
goto fail;
}
} else if (i - x >= 0 && i + x < n) {
if (a[i] == '0') {
if (o[i + x] == -1) o[i + x] = 0;
if (o[i - x] == -1) o[i - x] = 0;
if (o[i - x] == 1) {
cout << "-1";
goto fail;
}
if (o[i + x] == 1) {
cout << "-1";
goto fail;
}
}
if (a[i] == '1') {
if (o[i - x] == -1) o[i - x] = 1;
if (o[i - x] == 0) o[i + x] = 1;
}
}
}
for (int i = 0; i < n; i++) {
if (o[i] == -1) o[i] = 1;
cout << o[i];
}
fail:
cout << "\n";
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int q = 1;
cin >> q;
while (q--) {
solve();
}
return 0;
}
| ### Prompt
Generate a cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
string a;
int x;
cin >> a >> x;
int n = a.length();
vector<int> o(n, -1);
for (int i = 0; i < n; i++) {
if (i - x < 0 && i + x < n) {
if (o[i + x] == -1) o[i + x] = a[i] - '0';
if (o[i + x] != a[i] - '0') {
cout << "-1";
goto fail;
}
} else if (i + x >= n && i - x >= 0) {
if (o[i - x] == -1) o[i - x] = a[i] - '0';
if (o[i - x] != a[i] - '0') {
cout << "-1";
goto fail;
}
} else if (i + x >= n && i - x < 0) {
if (a[i] == '1') {
cout << "-1";
goto fail;
}
} else if (i - x >= 0 && i + x < n) {
if (a[i] == '0') {
if (o[i + x] == -1) o[i + x] = 0;
if (o[i - x] == -1) o[i - x] = 0;
if (o[i - x] == 1) {
cout << "-1";
goto fail;
}
if (o[i + x] == 1) {
cout << "-1";
goto fail;
}
}
if (a[i] == '1') {
if (o[i - x] == -1) o[i - x] = 1;
if (o[i - x] == 0) o[i + x] = 1;
}
}
}
for (int i = 0; i < n; i++) {
if (o[i] == -1) o[i] = 1;
cout << o[i];
}
fail:
cout << "\n";
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int q = 1;
cin >> q;
while (q--) {
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int t, x, sum, flag = 1;
char a[200000];
bool b[200000];
int main() {
bool flag = false;
cin >> t;
while (t--) {
flag = false;
int j;
cin >> a >> x;
j = strlen(a);
memset(b, true, sizeof(b));
for (int i = 0; i < j; i++) {
if (i - x >= 0 && a[i] == '0') b[i - x] = false;
if (i + x < j && a[i] == '0') b[i + x] = false;
}
int y = 1;
for (int i = 0; i < j; i++) {
if (a[i] == '0') continue;
flag = false;
if ((b[i - x] == true && i - x >= 0) || (b[i + x] == true && i + x < j))
flag = true;
if (!flag) {
y = 0;
break;
}
}
if (y == 1) {
for (int i = 0; i < j; i++) {
if (b[i] == true)
cout << "1";
else
cout << "0";
}
} else
cout << "-1";
cout << endl;
}
return 0;
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int t, x, sum, flag = 1;
char a[200000];
bool b[200000];
int main() {
bool flag = false;
cin >> t;
while (t--) {
flag = false;
int j;
cin >> a >> x;
j = strlen(a);
memset(b, true, sizeof(b));
for (int i = 0; i < j; i++) {
if (i - x >= 0 && a[i] == '0') b[i - x] = false;
if (i + x < j && a[i] == '0') b[i + x] = false;
}
int y = 1;
for (int i = 0; i < j; i++) {
if (a[i] == '0') continue;
flag = false;
if ((b[i - x] == true && i - x >= 0) || (b[i + x] == true && i + x < j))
flag = true;
if (!flag) {
y = 0;
break;
}
}
if (y == 1) {
for (int i = 0; i < j; i++) {
if (b[i] == true)
cout << "1";
else
cout << "0";
}
} else
cout << "-1";
cout << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
if (a == 0) return b;
return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
void func() {
string s;
cin >> s;
int n, x;
cin >> x;
n = s.size();
std::vector<int> w(n, 1);
for (int i = 0; i < n; i++) {
if (s.at(i) == '0') {
if (i >= x) w[i - x] = 0;
if (i + x < n) w[i + x] = 0;
}
}
string s1 = "";
for (int i = 0; i < n; i++) {
if (i >= x && w[i - x] == 1)
s1 += '1';
else if (i + x < n && w[i + x] == 1)
s1 += '1';
else
s1 += '0';
}
for (int i = 0; i < n; i++) {
if (s1.at(i) != s.at(i)) {
cout << -1 << endl;
return;
}
}
for (int i = 0; i < n; i++) cout << w[i];
cout << endl;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long t;
cin >> t;
for (long long i = 0; i < t; i++) func();
return 0;
}
| ### Prompt
Please create a solution in Cpp to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
if (a == 0) return b;
return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
void func() {
string s;
cin >> s;
int n, x;
cin >> x;
n = s.size();
std::vector<int> w(n, 1);
for (int i = 0; i < n; i++) {
if (s.at(i) == '0') {
if (i >= x) w[i - x] = 0;
if (i + x < n) w[i + x] = 0;
}
}
string s1 = "";
for (int i = 0; i < n; i++) {
if (i >= x && w[i - x] == 1)
s1 += '1';
else if (i + x < n && w[i + x] == 1)
s1 += '1';
else
s1 += '0';
}
for (int i = 0; i < n; i++) {
if (s1.at(i) != s.at(i)) {
cout << -1 << endl;
return;
}
}
for (int i = 0; i < n; i++) cout << w[i];
cout << endl;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long t;
cin >> t;
for (long long i = 0; i < t; i++) func();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void solve() {
string a;
int x;
cin >> a >> x;
int n = a.length();
vector<int> o(n, -1);
for (int i = 0; i < n; i++) {
if (i - x < 0 && i + x < n)
o[i + x] = a[i] - '0';
else if (i + x >= n && i - x >= 0) {
if (o[i - x] == -1) o[i - x] = a[i] - '0';
if (o[i - x] != a[i] - '0') {
cout << "-1";
goto fail;
}
} else if (i + x >= n && i - x < 0) {
if (a[i] == '1') {
cout << "-1";
goto fail;
}
} else {
if (a[i] == '0') {
o[i + x] = 0;
if (o[i - x] == -1) o[i - x] = 0;
if (o[i - x] == 1) {
cout << "-1";
goto fail;
}
}
if (a[i] == '1') {
if (o[i - x] == -1) o[i - x] = 1;
if (o[i - x] == 0) o[i + x] = 1;
}
}
}
for (int i = 0; i < n; i++) {
if (o[i] == -1) o[i] = 1;
cout << o[i];
}
fail:
cout << "\n";
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int q = 1;
cin >> q;
while (q--) {
solve();
}
return 0;
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
string a;
int x;
cin >> a >> x;
int n = a.length();
vector<int> o(n, -1);
for (int i = 0; i < n; i++) {
if (i - x < 0 && i + x < n)
o[i + x] = a[i] - '0';
else if (i + x >= n && i - x >= 0) {
if (o[i - x] == -1) o[i - x] = a[i] - '0';
if (o[i - x] != a[i] - '0') {
cout << "-1";
goto fail;
}
} else if (i + x >= n && i - x < 0) {
if (a[i] == '1') {
cout << "-1";
goto fail;
}
} else {
if (a[i] == '0') {
o[i + x] = 0;
if (o[i - x] == -1) o[i - x] = 0;
if (o[i - x] == 1) {
cout << "-1";
goto fail;
}
}
if (a[i] == '1') {
if (o[i - x] == -1) o[i - x] = 1;
if (o[i - x] == 0) o[i + x] = 1;
}
}
}
for (int i = 0; i < n; i++) {
if (o[i] == -1) o[i] = 1;
cout << o[i];
}
fail:
cout << "\n";
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int q = 1;
cin >> q;
while (q--) {
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
int t;
void solve() {
int x;
string s;
cin >> s >> x;
int a = s.size();
string st(a, '1');
for (int j = 0; j < a; j++) {
if (s[j] == '0') {
if (j - x >= 0) st[j - x] = '0';
if (j + x < a) st[j + x] = '0';
}
}
for (int j = 0; j < a; j++) {
bool b = false;
b = b || (j - x >= 0 && st[j - x] == '1');
b = b || (j + x < a && st[j + x] == '1');
if (s[j] != b + '0') {
cout << -1 << endl;
return;
}
}
cout << st << endl;
}
int main() {
ios::sync_with_stdio(false), cin.tie(0);
cin >> t;
for (int i = 0; i < t; i++) solve();
return 0;
}
| ### Prompt
Create a solution in cpp for the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
int t;
void solve() {
int x;
string s;
cin >> s >> x;
int a = s.size();
string st(a, '1');
for (int j = 0; j < a; j++) {
if (s[j] == '0') {
if (j - x >= 0) st[j - x] = '0';
if (j + x < a) st[j + x] = '0';
}
}
for (int j = 0; j < a; j++) {
bool b = false;
b = b || (j - x >= 0 && st[j - x] == '1');
b = b || (j + x < a && st[j + x] == '1');
if (s[j] != b + '0') {
cout << -1 << endl;
return;
}
}
cout << st << endl;
}
int main() {
ios::sync_with_stdio(false), cin.tie(0);
cin >> t;
for (int i = 0; i < t; i++) solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int MAXN = 1e5 + 5;
ll x;
string s;
int ans[MAXN];
void init() {
;
;
}
void solve(int casei) {
ll n = s.length();
s = "#" + s;
for (int i = 1; i <= n; ++i) {
ans[i] = -1;
}
for (int i = 1; i <= n; ++i) {
if (s[i] == '0') {
if (i + x <= n) ans[i + x] = 0;
if (i - x >= 1) ans[i - x] = 0;
}
}
for (int i = 1; i <= n; ++i) {
if (s[i] == '1') {
bool flg = false;
if (i + x <= n && ans[i + x] != 0) ans[i + x] = 1, flg = true;
if (i - x >= 1 && ans[i - x] != 0) ans[i - x] = 1, flg = true;
if (flg == false) {
cout << "-1\n";
return;
}
}
}
for (int i = 1; i <= n; ++i) {
if (ans[i] == -1) ans[i] = 1;
}
for (int i = 1; i <= n; ++i) cout << ans[i];
cout << endl;
}
int main() {
int Case = 1;
ios::sync_with_stdio(false);
cin.tie(0);
cin >> Case;
for (int casei = 1; casei <= Case; ++casei) {
cin >> s >> x;
init();
solve(casei);
}
return 0;
}
| ### Prompt
Generate a cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int MAXN = 1e5 + 5;
ll x;
string s;
int ans[MAXN];
void init() {
;
;
}
void solve(int casei) {
ll n = s.length();
s = "#" + s;
for (int i = 1; i <= n; ++i) {
ans[i] = -1;
}
for (int i = 1; i <= n; ++i) {
if (s[i] == '0') {
if (i + x <= n) ans[i + x] = 0;
if (i - x >= 1) ans[i - x] = 0;
}
}
for (int i = 1; i <= n; ++i) {
if (s[i] == '1') {
bool flg = false;
if (i + x <= n && ans[i + x] != 0) ans[i + x] = 1, flg = true;
if (i - x >= 1 && ans[i - x] != 0) ans[i - x] = 1, flg = true;
if (flg == false) {
cout << "-1\n";
return;
}
}
}
for (int i = 1; i <= n; ++i) {
if (ans[i] == -1) ans[i] = 1;
}
for (int i = 1; i <= n; ++i) cout << ans[i];
cout << endl;
}
int main() {
int Case = 1;
ios::sync_with_stdio(false);
cin.tie(0);
cin >> Case;
for (int casei = 1; casei <= Case; ++casei) {
cin >> s >> x;
init();
solve(casei);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
bool compare(const pair<int, int> &i, const pair<int, int> &j) {
return (i.second == j.second) ? (i.first < j.first) : (i.second < j.second);
}
const long long mod = 1e9 + 7;
int main() {
long long t;
cin >> t;
while (t--) {
string s;
cin >> s;
long long x;
cin >> x;
std::vector<char> w(s.size(), 'k');
bool poss = true;
for (int i = (0); i < (s.size()); i++) {
if (s[i] == '0') {
if (i - x >= 0) {
w[i - x] = '0';
}
if (i + x < s.size()) {
w[i + x] = '0';
}
}
}
for (int i = (0); i < (s.size()); i++) {
if (s[i] == '1') {
if (i - x >= 0) {
if (w[i - x] != '0') {
w[i - x] = '1';
} else if (i + x < s.size() and w[i + x] == 'k') {
w[i + x] = '1';
} else {
poss = false;
}
} else if (i + x < s.size() and w[i + x] == 'k') {
w[i + x] = '1';
} else
poss = false;
}
}
if (poss) {
for (int i = (0); i < (w.size()); i++) {
if (w[i] != 'k')
cout << w[i];
else
cout << '1';
}
} else {
cout << -1;
}
cout << "\n";
}
}
| ### Prompt
In cpp, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
bool compare(const pair<int, int> &i, const pair<int, int> &j) {
return (i.second == j.second) ? (i.first < j.first) : (i.second < j.second);
}
const long long mod = 1e9 + 7;
int main() {
long long t;
cin >> t;
while (t--) {
string s;
cin >> s;
long long x;
cin >> x;
std::vector<char> w(s.size(), 'k');
bool poss = true;
for (int i = (0); i < (s.size()); i++) {
if (s[i] == '0') {
if (i - x >= 0) {
w[i - x] = '0';
}
if (i + x < s.size()) {
w[i + x] = '0';
}
}
}
for (int i = (0); i < (s.size()); i++) {
if (s[i] == '1') {
if (i - x >= 0) {
if (w[i - x] != '0') {
w[i - x] = '1';
} else if (i + x < s.size() and w[i + x] == 'k') {
w[i + x] = '1';
} else {
poss = false;
}
} else if (i + x < s.size() and w[i + x] == 'k') {
w[i + x] = '1';
} else
poss = false;
}
}
if (poss) {
for (int i = (0); i < (w.size()); i++) {
if (w[i] != 'k')
cout << w[i];
else
cout << '1';
}
} else {
cout << -1;
}
cout << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long int t;
cin >> t;
while (t--) {
string s;
long int x;
cin >> s >> x;
string w = "";
long int n = s.length();
for (long int i = 0; i < n; i++) w += "1";
for (long int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x < n) w[i + x] = '0';
}
}
string ss = "";
for (long int i = 0; i < n; i++) {
bool pp = 0;
if (i - x >= 0 && w[i - x] == '1') pp = 1;
if (i + x < n && w[i + x] == '1') pp = 1;
if (pp)
ss += '1';
else
ss += '0';
}
if (ss == s)
cout << w << endl;
else
cout << "-1" << endl;
}
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long int t;
cin >> t;
while (t--) {
string s;
long int x;
cin >> s >> x;
string w = "";
long int n = s.length();
for (long int i = 0; i < n; i++) w += "1";
for (long int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x < n) w[i + x] = '0';
}
}
string ss = "";
for (long int i = 0; i < n; i++) {
bool pp = 0;
if (i - x >= 0 && w[i - x] == '1') pp = 1;
if (i + x < n && w[i + x] == '1') pp = 1;
if (pp)
ss += '1';
else
ss += '0';
}
if (ss == s)
cout << w << endl;
else
cout << "-1" << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const double pii = 2 * pi;
const double eps = 1e-6;
const long long MOD = 1e9 + 7;
void solve() {
int n, x;
string s;
cin >> s;
cin >> x;
n = s.size();
vector<int> ans(n, 1);
for (int i = 0; i < n; i++)
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = 0;
if (i + x < n) ans[i + x] = 0;
}
for (int i = 0; i < n; i++)
if (s[i] == '1')
if (!((i - x >= 0 && ans[i - x] == 1) ||
(i + x < n && ans[i + x] == 1))) {
cout << "-1\n";
return;
}
for (int i = 0; i < n; i++) cout << ans[i];
cout << "\n";
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(NULL);
std::cout.tie(NULL);
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
| ### Prompt
Generate a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const double pii = 2 * pi;
const double eps = 1e-6;
const long long MOD = 1e9 + 7;
void solve() {
int n, x;
string s;
cin >> s;
cin >> x;
n = s.size();
vector<int> ans(n, 1);
for (int i = 0; i < n; i++)
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = 0;
if (i + x < n) ans[i + x] = 0;
}
for (int i = 0; i < n; i++)
if (s[i] == '1')
if (!((i - x >= 0 && ans[i - x] == 1) ||
(i + x < n && ans[i + x] == 1))) {
cout << "-1\n";
return;
}
for (int i = 0; i < n; i++) cout << ans[i];
cout << "\n";
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(NULL);
std::cout.tie(NULL);
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
ll a[200004], b[200005];
ll fnc(ll n) {
ll cnt;
return cnt;
}
int main() {
int t;
cin >> t;
while (t--) {
ll n, m, sum = 0, cnt = 0;
string s, s1;
map<ll, ll> mp;
cin >> s >> n;
for (int i = 0; i < s.size(); i++) {
if (s[i] == '0') {
ll x = i + 1;
x += n;
mp[x] = 1;
if (x - 2 * n > 0) mp[x - 2 * n] = 1;
}
}
for (int i = 0; i < s.size(); i++) {
if (s[i] == '1') {
ll x = i + 1;
if ((x + n <= s.size() and mp[x + n] == 0 ||
x - n > 0 and mp[x - n] == 0)) {
} else {
cnt = -1;
break;
}
}
}
if (cnt == -1)
cout << -1 << endl;
else {
ll x = s.size();
for (int i = 1; i <= x; i++) {
if (mp[i] == 1)
cout << 0;
else
cout << 1;
}
cout << endl;
}
}
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
ll a[200004], b[200005];
ll fnc(ll n) {
ll cnt;
return cnt;
}
int main() {
int t;
cin >> t;
while (t--) {
ll n, m, sum = 0, cnt = 0;
string s, s1;
map<ll, ll> mp;
cin >> s >> n;
for (int i = 0; i < s.size(); i++) {
if (s[i] == '0') {
ll x = i + 1;
x += n;
mp[x] = 1;
if (x - 2 * n > 0) mp[x - 2 * n] = 1;
}
}
for (int i = 0; i < s.size(); i++) {
if (s[i] == '1') {
ll x = i + 1;
if ((x + n <= s.size() and mp[x + n] == 0 ||
x - n > 0 and mp[x - n] == 0)) {
} else {
cnt = -1;
break;
}
}
}
if (cnt == -1)
cout << -1 << endl;
else {
ll x = s.size();
for (int i = 1; i <= x; i++) {
if (mp[i] == 1)
cout << 0;
else
cout << 1;
}
cout << endl;
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <class T>
T nrem(T x, T y) {
return ((x % y) + y) % y;
}
const long long M = 1e9 + 7;
long long mod(long long x) { return ((x % M + M) % M); }
void in() {}
void solve() {
string s;
cin >> s;
int len = s.length();
int x;
cin >> x;
string ans = string(len, '1');
for (int i = 0; i <= len - 1; ++i) {
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < len) ans[i + x] = '0';
}
}
string w = string(len, '0');
for (int i = 0; i <= len - 1; ++i) {
if (i - x >= 0 and ans[i - x] == '1')
w[i] = '1';
else if (i + x <= len - 1 && ans[i + x] == '1')
w[i] = '1';
else
w[i] = '0';
}
if (w == s)
cout << ans << "\n";
else
cout << -1 << "\n";
}
int main() {
in();
int t;
cin >> t;
while (t--) {
solve();
}
}
| ### Prompt
Develop a solution in CPP to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
T nrem(T x, T y) {
return ((x % y) + y) % y;
}
const long long M = 1e9 + 7;
long long mod(long long x) { return ((x % M + M) % M); }
void in() {}
void solve() {
string s;
cin >> s;
int len = s.length();
int x;
cin >> x;
string ans = string(len, '1');
for (int i = 0; i <= len - 1; ++i) {
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < len) ans[i + x] = '0';
}
}
string w = string(len, '0');
for (int i = 0; i <= len - 1; ++i) {
if (i - x >= 0 and ans[i - x] == '1')
w[i] = '1';
else if (i + x <= len - 1 && ans[i + x] == '1')
w[i] = '1';
else
w[i] = '0';
}
if (w == s)
cout << ans << "\n";
else
cout << -1 << "\n";
}
int main() {
in();
int t;
cin >> t;
while (t--) {
solve();
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while (t--) {
int x;
string s;
cin >> s >> x;
int n = s.length();
char w[n + 2];
for (int i = 0; i < n + 2; i++) {
w[i] = '@';
}
bool canchange[n + 2];
for (int i = 0; i < n + 2; i++) {
canchange[i] = true;
}
int i;
for (i = 0; i < n; i++) {
if (i - x < 0 && i + x >= n) {
if (s[i] == '1') {
break;
}
continue;
} else if (i - x < 0) {
if (!canchange[i + x] && (w[i + x] != '@' && w[i + x] != s[i])) {
break;
}
w[i + x] = s[i];
canchange[i + x] = false;
} else if (i + x >= n) {
if (!canchange[i - x] && (w[i - x] != '@' && w[i - x] != s[i])) {
break;
}
w[i - x] = s[i];
canchange[i - x] = false;
} else {
if (s[i] == '1') {
if (w[i - x] == '@') {
w[i - x] = '1';
canchange[i - x] = false;
} else if (w[i - x] == '0') {
if (!canchange[i - x]) {
if (!canchange[i + x] && w[i + x] == '0') {
break;
}
w[i + x] = '1';
canchange[i + x] = false;
} else {
w[i - x] = '1';
canchange[i - x] = false;
w[i + x] = '0';
}
} else {
w[i + x] = '0';
}
} else {
if (w[i - x] != '0') {
if (!canchange[i - x]) {
break;
} else {
w[i - x] = '0';
canchange[i - x] = false;
}
}
if (w[i + x] != '0') {
if (!canchange[i + x]) {
break;
} else {
w[i + x] = '0';
canchange[i + x] = false;
}
}
}
}
}
if (i < n) {
cout << -1 << endl;
continue;
}
for (i = 0; i < n; i++) {
if (w[i] != '@') {
cout << w[i];
} else {
cout << '1';
}
}
cout << endl;
}
}
| ### Prompt
In cpp, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while (t--) {
int x;
string s;
cin >> s >> x;
int n = s.length();
char w[n + 2];
for (int i = 0; i < n + 2; i++) {
w[i] = '@';
}
bool canchange[n + 2];
for (int i = 0; i < n + 2; i++) {
canchange[i] = true;
}
int i;
for (i = 0; i < n; i++) {
if (i - x < 0 && i + x >= n) {
if (s[i] == '1') {
break;
}
continue;
} else if (i - x < 0) {
if (!canchange[i + x] && (w[i + x] != '@' && w[i + x] != s[i])) {
break;
}
w[i + x] = s[i];
canchange[i + x] = false;
} else if (i + x >= n) {
if (!canchange[i - x] && (w[i - x] != '@' && w[i - x] != s[i])) {
break;
}
w[i - x] = s[i];
canchange[i - x] = false;
} else {
if (s[i] == '1') {
if (w[i - x] == '@') {
w[i - x] = '1';
canchange[i - x] = false;
} else if (w[i - x] == '0') {
if (!canchange[i - x]) {
if (!canchange[i + x] && w[i + x] == '0') {
break;
}
w[i + x] = '1';
canchange[i + x] = false;
} else {
w[i - x] = '1';
canchange[i - x] = false;
w[i + x] = '0';
}
} else {
w[i + x] = '0';
}
} else {
if (w[i - x] != '0') {
if (!canchange[i - x]) {
break;
} else {
w[i - x] = '0';
canchange[i - x] = false;
}
}
if (w[i + x] != '0') {
if (!canchange[i + x]) {
break;
} else {
w[i + x] = '0';
canchange[i + x] = false;
}
}
}
}
}
if (i < n) {
cout << -1 << endl;
continue;
}
for (i = 0; i < n; i++) {
if (w[i] != '@') {
cout << w[i];
} else {
cout << '1';
}
}
cout << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
bool vis[100100];
string s;
bool process(string s, int x, string Res) {
for (int i = 0; i < s.size(); i++) {
char curr = '0';
if ((i - x >= 0 && s[i - x] == '1') ||
(i + x < s.size() && s[i + x] == '1'))
curr = '1';
if (curr != Res[i]) return false;
}
return true;
}
void Solve() {
cin >> s;
int x;
cin >> x;
for (int i = 0; i < s.size(); i++) {
if (s[i] == '0') {
if (i - x >= 0) vis[i - x] = true;
if (i + x < s.size()) vis[i + x] = true;
}
}
string Ans;
for (int i = 0; i < s.size(); i++) Ans += (vis[i] ? '0' : '1');
if (process(Ans, x, s))
cout << Ans;
else
cout << -1;
cout << '\n';
}
void Clear() {
for (int i = 0; i < s.size(); i++) vis[i] = false;
}
int main() {
int T = 1;
cin >> T;
while (T--) {
Solve();
Clear();
}
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
bool vis[100100];
string s;
bool process(string s, int x, string Res) {
for (int i = 0; i < s.size(); i++) {
char curr = '0';
if ((i - x >= 0 && s[i - x] == '1') ||
(i + x < s.size() && s[i + x] == '1'))
curr = '1';
if (curr != Res[i]) return false;
}
return true;
}
void Solve() {
cin >> s;
int x;
cin >> x;
for (int i = 0; i < s.size(); i++) {
if (s[i] == '0') {
if (i - x >= 0) vis[i - x] = true;
if (i + x < s.size()) vis[i + x] = true;
}
}
string Ans;
for (int i = 0; i < s.size(); i++) Ans += (vis[i] ? '0' : '1');
if (process(Ans, x, s))
cout << Ans;
else
cout << -1;
cout << '\n';
}
void Clear() {
for (int i = 0; i < s.size(); i++) vis[i] = false;
}
int main() {
int T = 1;
cin >> T;
while (T--) {
Solve();
Clear();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t, x, n;
string s, ans;
cin >> t;
while (t--) {
cin >> s >> x;
n = s.length();
ans = string(n, '1');
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < n) ans[i + x] = '0';
}
}
bool isone;
bool ispossible = true;
for (int i = 0; i < n; i++) {
isone = false;
if (i - x >= 0 && ans[i - x] == '1') isone = true;
if (i + x < n && ans[i + x] == '1') isone = true;
if (isone ^ (s[i] - '0')) {
cout << -1 << endl;
ispossible = false;
break;
}
}
if (ispossible) cout << ans << endl;
}
return 0;
}
| ### Prompt
Create a solution in CPP for the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t, x, n;
string s, ans;
cin >> t;
while (t--) {
cin >> s >> x;
n = s.length();
ans = string(n, '1');
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < n) ans[i + x] = '0';
}
}
bool isone;
bool ispossible = true;
for (int i = 0; i < n; i++) {
isone = false;
if (i - x >= 0 && ans[i - x] == '1') isone = true;
if (i + x < n && ans[i + x] == '1') isone = true;
if (isone ^ (s[i] - '0')) {
cout << -1 << endl;
ispossible = false;
break;
}
}
if (ispossible) cout << ans << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
priority_queue<pair<long long, long long>, vector<pair<long long, long long> >,
greater<pair<long long, long long> > >
pq;
priority_queue<long long, vector<long long>, greater<long long> > pq1;
long long result(long long a, long long b, long long p) {
long long ans = 1;
if (b == (-1)) b = p - 2;
while (b) {
if (b & 1) {
ans = (ans * a) % p;
}
a = (a * a) % p;
b >>= 1;
}
return ans;
}
long long T = 1, n, m, k;
int main() {
if (fopen("input.txt", "r"))
freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout);
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> T;
while (T--) {
string s, w;
cin >> s;
w = s;
long long x;
cin >> x;
n = s.size();
for (int i = 0; i < n; i++) w[i] = '1';
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x < n) w[i + x] = '0';
}
}
string p = "";
for (int i = 0; i < n; i++) {
if ((i - x >= 0 && w[i - x] == '1') || (i + x < n && w[i + x] == '1')) {
p.push_back('1');
} else {
p.push_back('0');
}
}
if (p == s) {
cout << w << "\n";
} else {
cout << "-1\n";
}
}
return 0;
}
| ### Prompt
Your challenge is to write a cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
priority_queue<pair<long long, long long>, vector<pair<long long, long long> >,
greater<pair<long long, long long> > >
pq;
priority_queue<long long, vector<long long>, greater<long long> > pq1;
long long result(long long a, long long b, long long p) {
long long ans = 1;
if (b == (-1)) b = p - 2;
while (b) {
if (b & 1) {
ans = (ans * a) % p;
}
a = (a * a) % p;
b >>= 1;
}
return ans;
}
long long T = 1, n, m, k;
int main() {
if (fopen("input.txt", "r"))
freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout);
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> T;
while (T--) {
string s, w;
cin >> s;
w = s;
long long x;
cin >> x;
n = s.size();
for (int i = 0; i < n; i++) w[i] = '1';
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x < n) w[i + x] = '0';
}
}
string p = "";
for (int i = 0; i < n; i++) {
if ((i - x >= 0 && w[i - x] == '1') || (i + x < n && w[i + x] == '1')) {
p.push_back('1');
} else {
p.push_back('0');
}
}
if (p == s) {
cout << w << "\n";
} else {
cout << "-1\n";
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
bool prime[100005];
void Sieve(int n) {
memset(prime, true, sizeof(prime));
for (int p = 2; p * p <= n; p++) {
if (prime[p] == true) {
for (int i = p * p; i <= n; i += p) prime[i] = false;
}
}
}
void solve() {
string s;
int x;
cin >> s >> x;
string w = string(s.length(), '1');
for (int i = 0; i < s.length(); i++) {
if (s[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x < s.length()) w[i + x] = '0';
}
}
int f = 1;
for (int i = 0; i < s.length(); i++) {
if (s[i] == '1') {
if (i - x >= 0 && w[i - x] == '1') continue;
if (i + x < s.length() && w[i + x] == '1') continue;
f = -1;
}
}
if (f == -1)
cout << -1 << '\n';
else {
cout << w << '\n';
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t = 1;
cin >> t;
while (t--) {
solve();
}
}
| ### Prompt
In cpp, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
bool prime[100005];
void Sieve(int n) {
memset(prime, true, sizeof(prime));
for (int p = 2; p * p <= n; p++) {
if (prime[p] == true) {
for (int i = p * p; i <= n; i += p) prime[i] = false;
}
}
}
void solve() {
string s;
int x;
cin >> s >> x;
string w = string(s.length(), '1');
for (int i = 0; i < s.length(); i++) {
if (s[i] == '0') {
if (i - x >= 0) w[i - x] = '0';
if (i + x < s.length()) w[i + x] = '0';
}
}
int f = 1;
for (int i = 0; i < s.length(); i++) {
if (s[i] == '1') {
if (i - x >= 0 && w[i - x] == '1') continue;
if (i + x < s.length() && w[i + x] == '1') continue;
f = -1;
}
}
if (f == -1)
cout << -1 << '\n';
else {
cout << w << '\n';
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t = 1;
cin >> t;
while (t--) {
solve();
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
int32_t main() {
long long t;
cin >> t;
while (t--) {
string s;
long long x;
cin >> s >> x;
string c, p;
long long n = s.size();
for (long long i = 0; i < n; i++) {
c.push_back('0');
p.push_back('1');
}
for (long long i = 0; i < n; i++) {
if (s[i] == '0') {
if (i + x < n) {
p[i + x] = '0';
}
if (i - x >= 0) {
p[i - x] = '0';
}
}
}
for (long long i = 0; i < n; i++) {
if (p[i] == '1') {
if (i + x < n) {
c[i + x] = '1';
}
if (i - x >= 0) {
c[i - x] = '1';
}
}
}
if (s == c) {
cout << p << endl;
} else
cout << -1 << endl;
}
}
| ### Prompt
Please create a solution in CPP to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
int32_t main() {
long long t;
cin >> t;
while (t--) {
string s;
long long x;
cin >> s >> x;
string c, p;
long long n = s.size();
for (long long i = 0; i < n; i++) {
c.push_back('0');
p.push_back('1');
}
for (long long i = 0; i < n; i++) {
if (s[i] == '0') {
if (i + x < n) {
p[i + x] = '0';
}
if (i - x >= 0) {
p[i - x] = '0';
}
}
}
for (long long i = 0; i < n; i++) {
if (p[i] == '1') {
if (i + x < n) {
c[i + x] = '1';
}
if (i - x >= 0) {
c[i - x] = '1';
}
}
}
if (s == c) {
cout << p << endl;
} else
cout << -1 << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void drop(const T &x) {
cout << x << '\n';
exit(0);
}
void solve() {
string s;
cin >> s;
int x;
cin >> x;
int n = int(s.size());
string w(n, '1');
for (int i = 0; i < n; ++i) {
if (s[i] == '0') {
if (i - x >= 0) {
w[i - x] = '0';
}
if (i + x < n) {
w[i + x] = '0';
}
}
}
bool ok = true;
for (int i = 0; i < n; ++i) {
if (s[i] == '1') {
if (i - x >= 0 && w[i - x] == '1') continue;
if (i + x < n && w[i + x] == '1') continue;
ok = false;
}
}
if (!ok) {
cout << -1 << '\n';
} else {
cout << w << '\n';
}
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(0);
int T = 1;
cin >> T;
while (T--) solve();
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void drop(const T &x) {
cout << x << '\n';
exit(0);
}
void solve() {
string s;
cin >> s;
int x;
cin >> x;
int n = int(s.size());
string w(n, '1');
for (int i = 0; i < n; ++i) {
if (s[i] == '0') {
if (i - x >= 0) {
w[i - x] = '0';
}
if (i + x < n) {
w[i + x] = '0';
}
}
}
bool ok = true;
for (int i = 0; i < n; ++i) {
if (s[i] == '1') {
if (i - x >= 0 && w[i - x] == '1') continue;
if (i + x < n && w[i + x] == '1') continue;
ok = false;
}
}
if (!ok) {
cout << -1 << '\n';
} else {
cout << w << '\n';
}
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(0);
int T = 1;
cin >> T;
while (T--) solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
int test, cs = 1;
cin >> test;
while (test--) {
string s, res;
cin >> s;
int x, i, j, n = (int)(s.size());
cin >> x;
res.resize(n, '1');
for (i = 0; i < n; i++) {
if (i + x < n && s[i + x] == '0') res[i] = '0';
if (i - x >= 0 && s[i - x] == '0') res[i] = '0';
}
bool ok = true;
for (i = 0; i <= n; i++) {
if (s[i] == '1') {
if ((i + x < n && res[i + x] == '1') ||
(i - x >= 0 && res[i - x] == '1'))
ok = true;
else {
ok = false;
break;
}
}
}
if (ok)
cout << res << endl;
else
cout << -1 << endl;
}
}
| ### Prompt
Generate a CPP solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
int test, cs = 1;
cin >> test;
while (test--) {
string s, res;
cin >> s;
int x, i, j, n = (int)(s.size());
cin >> x;
res.resize(n, '1');
for (i = 0; i < n; i++) {
if (i + x < n && s[i + x] == '0') res[i] = '0';
if (i - x >= 0 && s[i - x] == '0') res[i] = '0';
}
bool ok = true;
for (i = 0; i <= n; i++) {
if (s[i] == '1') {
if ((i + x < n && res[i + x] == '1') ||
(i - x >= 0 && res[i - x] == '1'))
ok = true;
else {
ok = false;
break;
}
}
}
if (ok)
cout << res << endl;
else
cout << -1 << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int x;
cin >> x;
int n = s.size();
vector<char> w(s.size(), '1');
for (int i = 0; i < s.size(); i++) {
if (s[i] == '1') continue;
if (i - x >= 0) w[i - x] = '0';
if (i + x < n) w[i + x] = '0';
}
int flag = 0;
for (int i = 0; i < s.size(); i++) {
if (s[i] == '0') continue;
if (i - x >= 0 && s[i] == w[i - x]) continue;
if (i + x < n && s[i] == w[i + x]) continue;
flag = 1;
}
if (flag)
cout << -1;
else {
for (auto x : w) cout << x;
}
cout << endl;
}
}
| ### Prompt
In CPP, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
int x;
cin >> x;
int n = s.size();
vector<char> w(s.size(), '1');
for (int i = 0; i < s.size(); i++) {
if (s[i] == '1') continue;
if (i - x >= 0) w[i - x] = '0';
if (i + x < n) w[i + x] = '0';
}
int flag = 0;
for (int i = 0; i < s.size(); i++) {
if (s[i] == '0') continue;
if (i - x >= 0 && s[i] == w[i - x]) continue;
if (i + x < n && s[i] == w[i + x]) continue;
flag = 1;
}
if (flag)
cout << -1;
else {
for (auto x : w) cout << x;
}
cout << endl;
}
}
``` |
#include <bits/stdc++.h>
const int MAX = 1e5 + 7;
const int mod = 1e9 + 7;
using namespace std;
int n;
string s;
string check(string w) {
string tmp = w;
for (int i = 0; i < w.size(); ++i) {
if (i - n >= 0 && w[i - n] == '1' || i + n < w.size() && w[i + n] == '1')
tmp[i] = '1';
else
tmp[i] = '0';
}
return tmp;
}
string solve() {
int len = s.size();
string w = string(len, '1');
for (int i = 0; i < len; ++i) {
if (s[i] == '0') {
if (i + n < len) {
w[i + n] = '0';
}
if (i - n >= 0) {
w[i - n] = '0';
}
}
}
if (check(w) == s) return w;
return "-1";
}
int main() {
int t;
cin >> t;
while (t--) {
cin >> s >> n;
cout << solve() << endl;
}
}
| ### Prompt
Develop a solution in cpp to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
const int MAX = 1e5 + 7;
const int mod = 1e9 + 7;
using namespace std;
int n;
string s;
string check(string w) {
string tmp = w;
for (int i = 0; i < w.size(); ++i) {
if (i - n >= 0 && w[i - n] == '1' || i + n < w.size() && w[i + n] == '1')
tmp[i] = '1';
else
tmp[i] = '0';
}
return tmp;
}
string solve() {
int len = s.size();
string w = string(len, '1');
for (int i = 0; i < len; ++i) {
if (s[i] == '0') {
if (i + n < len) {
w[i + n] = '0';
}
if (i - n >= 0) {
w[i - n] = '0';
}
}
}
if (check(w) == s) return w;
return "-1";
}
int main() {
int t;
cin >> t;
while (t--) {
cin >> s >> n;
cout << solve() << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
int add(int a, int b) {
a += b;
if (a >= MOD) a -= MOD;
if (a < 0) a += MOD;
return a;
}
int mult(int a, int b) { return a * 1ll * b % MOD; }
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int q;
cin >> q;
while (q) {
--q;
string s;
cin >> s;
long long x;
cin >> x;
long long n = s.size();
vector<int> a(n), w(n);
for (long long i = 0; i < n; ++i) {
w[i] = 1;
if (s[i] == '1') {
a[i] = 1;
} else
a[i] = 0;
}
for (long long i = 0; i < n; ++i) {
if (a[i] == 0) {
if (i - x >= 0) {
w[i - x] = 0;
}
if (i + x < n) {
w[i + x] = 0;
}
}
}
bool g = true;
for (long long i = 0; i < n; ++i) {
if (a[i] == 1) {
bool f = false;
if (i - x >= 0) {
if (w[i - x] != 0) {
f = true;
}
}
if (i + x < n) {
if (w[i + x] != 0) {
f = true;
}
}
if (!f) {
g = false;
}
}
}
if (g) {
for (long long i = 0; i < n; ++i) {
cout << w[i];
}
cout << endl;
} else {
cout << -1 << endl;
}
}
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
int add(int a, int b) {
a += b;
if (a >= MOD) a -= MOD;
if (a < 0) a += MOD;
return a;
}
int mult(int a, int b) { return a * 1ll * b % MOD; }
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int q;
cin >> q;
while (q) {
--q;
string s;
cin >> s;
long long x;
cin >> x;
long long n = s.size();
vector<int> a(n), w(n);
for (long long i = 0; i < n; ++i) {
w[i] = 1;
if (s[i] == '1') {
a[i] = 1;
} else
a[i] = 0;
}
for (long long i = 0; i < n; ++i) {
if (a[i] == 0) {
if (i - x >= 0) {
w[i - x] = 0;
}
if (i + x < n) {
w[i + x] = 0;
}
}
}
bool g = true;
for (long long i = 0; i < n; ++i) {
if (a[i] == 1) {
bool f = false;
if (i - x >= 0) {
if (w[i - x] != 0) {
f = true;
}
}
if (i + x < n) {
if (w[i + x] != 0) {
f = true;
}
}
if (!f) {
g = false;
}
}
}
if (g) {
for (long long i = 0; i < n; ++i) {
cout << w[i];
}
cout << endl;
} else {
cout << -1 << endl;
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <class A>
void read(vector<A> &v);
template <class A, size_t S>
void read(array<A, S> &a);
template <class T>
void read(T &x) {
cin >> x;
}
void read(double &d) {
string t;
read(t);
d = stod(t);
}
void read(long double &d) {
string t;
read(t);
d = stold(t);
}
template <class H, class... T>
void read(H &h, T &...t) {
read(h);
read(t...);
}
template <class A>
void read(vector<A> &x) {
for (auto &a : x) read(a);
}
template <class A, size_t S>
void read(array<A, S> &x) {
for (auto &a : x) read(a);
}
string to_string(char c) { return string(1, c); }
string to_string(bool b) { return b ? "true" : "false"; }
string to_string(const char *s) { return string(s); }
string to_string(string s) { return s; }
string to_string(vector<bool> v) {
string res;
for (int i = (0); (1) > 0 ? i < ((int)(v).size()) : i > ((int)(v).size());
i += (1))
res += char('0' + v[i]);
return res;
}
template <size_t S>
string to_string(bitset<S> b) {
string res;
for (int i = (0); (1) > 0 ? i < (S) : i > (S); i += (1))
res += char('0' + b[i]);
return res;
}
template <class T>
string to_string(T v) {
bool f = 1;
string res;
for (auto &x : v) {
if (!f) res += ' ';
f = 0;
res += to_string(x);
}
return res;
}
template <class A>
void write(A x) {
cout << to_string(x);
}
template <class H, class... T>
void write(const H &h, const T &...t) {
write(h);
write(t...);
}
void print() { write("\n"); }
template <class H, class... T>
void print(const H &h, const T &...t) {
write(h);
if (sizeof...(t)) write(' ');
print(t...);
}
void DBG() { cerr << "]" << endl; }
template <class H, class... T>
void DBG(H h, T... t) {
cerr << to_string(h);
if (sizeof...(t)) cerr << ", ";
DBG(t...);
}
bool isLegal(int x, int n) { return x >= 0 && x < n; }
void solve() {
string str;
int k;
cin >> str >> k;
int n = str.length();
string res(n, '1');
for (int i = 0; i < n; i++) {
if (str[i] == '0') {
if (isLegal(i - k, n)) res[i - k] = '0';
if (isLegal(i + k, n)) res[i + k] = '0';
}
}
bool works = 1;
for (int i = 0; i < n; i++) {
if (str[i] == '1') {
if (isLegal(i - k, n) && res[i - k] == '1') continue;
if (isLegal(i + k, n) && res[i + k] == '1') continue;
works = 0;
}
}
if (works) {
cout << res << "\n";
} else {
cout << -1 << "\n";
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int t = 1;
read(t);
for (int i = (0); (1) > 0 ? i < (t) : i > (t); i += (1)) {
solve();
}
}
| ### Prompt
In CPP, your task is to solve the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <class A>
void read(vector<A> &v);
template <class A, size_t S>
void read(array<A, S> &a);
template <class T>
void read(T &x) {
cin >> x;
}
void read(double &d) {
string t;
read(t);
d = stod(t);
}
void read(long double &d) {
string t;
read(t);
d = stold(t);
}
template <class H, class... T>
void read(H &h, T &...t) {
read(h);
read(t...);
}
template <class A>
void read(vector<A> &x) {
for (auto &a : x) read(a);
}
template <class A, size_t S>
void read(array<A, S> &x) {
for (auto &a : x) read(a);
}
string to_string(char c) { return string(1, c); }
string to_string(bool b) { return b ? "true" : "false"; }
string to_string(const char *s) { return string(s); }
string to_string(string s) { return s; }
string to_string(vector<bool> v) {
string res;
for (int i = (0); (1) > 0 ? i < ((int)(v).size()) : i > ((int)(v).size());
i += (1))
res += char('0' + v[i]);
return res;
}
template <size_t S>
string to_string(bitset<S> b) {
string res;
for (int i = (0); (1) > 0 ? i < (S) : i > (S); i += (1))
res += char('0' + b[i]);
return res;
}
template <class T>
string to_string(T v) {
bool f = 1;
string res;
for (auto &x : v) {
if (!f) res += ' ';
f = 0;
res += to_string(x);
}
return res;
}
template <class A>
void write(A x) {
cout << to_string(x);
}
template <class H, class... T>
void write(const H &h, const T &...t) {
write(h);
write(t...);
}
void print() { write("\n"); }
template <class H, class... T>
void print(const H &h, const T &...t) {
write(h);
if (sizeof...(t)) write(' ');
print(t...);
}
void DBG() { cerr << "]" << endl; }
template <class H, class... T>
void DBG(H h, T... t) {
cerr << to_string(h);
if (sizeof...(t)) cerr << ", ";
DBG(t...);
}
bool isLegal(int x, int n) { return x >= 0 && x < n; }
void solve() {
string str;
int k;
cin >> str >> k;
int n = str.length();
string res(n, '1');
for (int i = 0; i < n; i++) {
if (str[i] == '0') {
if (isLegal(i - k, n)) res[i - k] = '0';
if (isLegal(i + k, n)) res[i + k] = '0';
}
}
bool works = 1;
for (int i = 0; i < n; i++) {
if (str[i] == '1') {
if (isLegal(i - k, n) && res[i - k] == '1') continue;
if (isLegal(i + k, n) && res[i + k] == '1') continue;
works = 0;
}
}
if (works) {
cout << res << "\n";
} else {
cout << -1 << "\n";
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int t = 1;
read(t);
for (int i = (0); (1) > 0 ? i < (t) : i > (t); i += (1)) {
solve();
}
}
``` |
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const long long mod1 = 998244353;
const long long mod = 1e9 + 7;
long long mod_mul(long long a, long long b) {
a = a % mod;
b = b % mod;
return (((a * b) % mod) + mod) % mod;
}
long long inv(long long i) {
if (i == 1) return 1;
return (mod - ((mod / i) * inv(mod % i)) % mod) % mod;
}
long long gcd(long long a, long long b) {
if (b == 0) return a;
return gcd(b, a % b);
}
long long pwr(long long a, long long b) {
a %= mod;
long long res = 1;
while (b > 0) {
if (b & 1) res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
long long findMinNumber(long long n) {
long long count = 0, ans = 1;
while (n % 2 == 0) {
count++;
n /= 2;
}
if (count % 2) ans *= 2;
for (int i = 3; i <= sqrt(n); i += 2) {
count = 0;
while (n % i == 0) {
count++;
n /= i;
}
if (count % 2) ans *= i;
}
if (n > 2) ans *= n;
return ans;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long t;
cin >> t;
while (t--) {
long long x, i, res1 = 0, res2 = 0, res = 0;
string s;
cin >> s;
cin >> x;
long long n = s.length();
char ans[n];
for (i = 0; i < n; i++) ans[i] = '2';
for (i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < n) ans[i + x] = '0';
}
}
for (i = 0; i < n; i++) {
res1 = 0;
res2 = 0;
if (s[i] == '1') {
if (i - x >= 0) {
if (ans[i - x] == '0') {
res1 = 1;
} else {
ans[i - x] = '1';
}
} else {
res1 = 1;
}
if (i + x < n) {
if (ans[i + x] == '0') {
res2 = 1;
} else {
ans[i + x] = '1';
}
} else {
res2 = 1;
}
if (res1 == 1 and res2 == 1) {
res = 1;
break;
}
if (i - x < 0 and i + x > n) {
res = 1;
break;
}
}
}
if (res == 1)
cout << -1 << '\n';
else {
for (i = 0; i < n; i++) {
if (ans[i] == '2')
cout << 1;
else
cout << ans[i];
}
cout << '\n';
}
}
return 0;
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const long long mod1 = 998244353;
const long long mod = 1e9 + 7;
long long mod_mul(long long a, long long b) {
a = a % mod;
b = b % mod;
return (((a * b) % mod) + mod) % mod;
}
long long inv(long long i) {
if (i == 1) return 1;
return (mod - ((mod / i) * inv(mod % i)) % mod) % mod;
}
long long gcd(long long a, long long b) {
if (b == 0) return a;
return gcd(b, a % b);
}
long long pwr(long long a, long long b) {
a %= mod;
long long res = 1;
while (b > 0) {
if (b & 1) res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
long long findMinNumber(long long n) {
long long count = 0, ans = 1;
while (n % 2 == 0) {
count++;
n /= 2;
}
if (count % 2) ans *= 2;
for (int i = 3; i <= sqrt(n); i += 2) {
count = 0;
while (n % i == 0) {
count++;
n /= i;
}
if (count % 2) ans *= i;
}
if (n > 2) ans *= n;
return ans;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long t;
cin >> t;
while (t--) {
long long x, i, res1 = 0, res2 = 0, res = 0;
string s;
cin >> s;
cin >> x;
long long n = s.length();
char ans[n];
for (i = 0; i < n; i++) ans[i] = '2';
for (i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) ans[i - x] = '0';
if (i + x < n) ans[i + x] = '0';
}
}
for (i = 0; i < n; i++) {
res1 = 0;
res2 = 0;
if (s[i] == '1') {
if (i - x >= 0) {
if (ans[i - x] == '0') {
res1 = 1;
} else {
ans[i - x] = '1';
}
} else {
res1 = 1;
}
if (i + x < n) {
if (ans[i + x] == '0') {
res2 = 1;
} else {
ans[i + x] = '1';
}
} else {
res2 = 1;
}
if (res1 == 1 and res2 == 1) {
res = 1;
break;
}
if (i - x < 0 and i + x > n) {
res = 1;
break;
}
}
}
if (res == 1)
cout << -1 << '\n';
else {
for (i = 0; i < n; i++) {
if (ans[i] == '2')
cout << 1;
else
cout << ans[i];
}
cout << '\n';
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, x, w[100003];
string s;
void in() {
cin >> s >> x;
n = s.length();
s = " " + s;
}
void xuly() {
memset(w, -1, sizeof w);
for (int i = 1; i <= n; i++)
if (s[i] == '0') {
if (i > x) w[i - x] = 0;
if (i + x <= n) w[i + x] = 0;
}
for (int i = 1; i <= n; i++)
if (s[i] == '1') {
bool check = false;
if (i > x)
if (w[i - x] != 0) check = true, w[i - x] = 1;
if (i + x <= n && w[i + x] != 0) check = true, w[i + x] = 1;
if (!check) {
cout << -1 << '\n';
return;
}
}
for (int i = 1; i <= n; i++) {
if (w[i] == -1) w[i] = 1;
cout << w[i];
}
cout << '\n';
}
int main() {
int T;
cin >> T;
while (T--) {
in();
xuly();
}
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, x, w[100003];
string s;
void in() {
cin >> s >> x;
n = s.length();
s = " " + s;
}
void xuly() {
memset(w, -1, sizeof w);
for (int i = 1; i <= n; i++)
if (s[i] == '0') {
if (i > x) w[i - x] = 0;
if (i + x <= n) w[i + x] = 0;
}
for (int i = 1; i <= n; i++)
if (s[i] == '1') {
bool check = false;
if (i > x)
if (w[i - x] != 0) check = true, w[i - x] = 1;
if (i + x <= n && w[i + x] != 0) check = true, w[i + x] = 1;
if (!check) {
cout << -1 << '\n';
return;
}
}
for (int i = 1; i <= n; i++) {
if (w[i] == -1) w[i] = 1;
cout << w[i];
}
cout << '\n';
}
int main() {
int T;
cin >> T;
while (T--) {
in();
xuly();
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
;
long long t;
cin >> t;
while (t--) {
long long n, i, j;
string s;
cin >> s;
long long x;
cin >> x;
n = s.length();
string w = s;
for (i = 0; i < n; i++) {
w[i] = '1';
}
for (i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) {
w[i - x] = '0';
}
if (i + x < n) {
w[i + x] = '0';
}
}
}
bool con = true;
for (i = 0; i < n; i++) {
if (s[i] == '1') {
long long c1 = 0;
long long c2 = 0;
if (i - x >= 0) {
c1 = w[i - x] - '0';
}
if (i + x < n) {
c2 = w[i + x] - '0';
}
long long num = c1 | c2;
if (num != 1) {
con = false;
break;
}
}
}
if (con)
cout << w << "\n";
else
cout << "-1 \n";
}
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
;
long long t;
cin >> t;
while (t--) {
long long n, i, j;
string s;
cin >> s;
long long x;
cin >> x;
n = s.length();
string w = s;
for (i = 0; i < n; i++) {
w[i] = '1';
}
for (i = 0; i < n; i++) {
if (s[i] == '0') {
if (i - x >= 0) {
w[i - x] = '0';
}
if (i + x < n) {
w[i + x] = '0';
}
}
}
bool con = true;
for (i = 0; i < n; i++) {
if (s[i] == '1') {
long long c1 = 0;
long long c2 = 0;
if (i - x >= 0) {
c1 = w[i - x] - '0';
}
if (i + x < n) {
c2 = w[i + x] - '0';
}
long long num = c1 | c2;
if (num != 1) {
con = false;
break;
}
}
}
if (con)
cout << w << "\n";
else
cout << "-1 \n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int mod = (int)1000000000 + 7;
const int MOD = (int)998244353;
const int INF = (int)1e18;
bool cmp(const pair<int, int>& a, const pair<int, int>& b) {
return a.second - a.first < b.second - b.first;
}
long long bit_ct1(long long x) {
long long ct = 0;
while (x > 0) {
x = x & (x - 1);
ct++;
}
return ct;
}
long long cc[27] = {0};
void solve() {
string s;
long long p;
cin >> s >> p;
long long ln = s.length();
vector<char> w(ln), ans(ln);
for (long long i = 0; i < ln; i++) {
w[i] = '1';
}
for (long long i = 0; i < ln; i++) {
if (s[i] == '0') {
if (i - p >= 0) w[i - p] = '0';
if (i + p < ln) w[i + p] = '0';
}
}
for (long long i = 0; i < ln; i++) {
if (i + p < ln && w[i + p] == '1') {
ans[i] = '1';
} else if (i - p >= 0 && w[i - p] == '1') {
ans[i] = '1';
} else {
ans[i] = '0';
}
}
for (long long i = 0; i < ln; i++) {
if (s[i] != ans[i]) {
cout << -1 << '\n';
return;
}
}
for (long long i = 0; i < ln; i++) {
cout << w[i];
}
cout << '\n';
return;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long t = 1;
cin >> t;
while (t--) {
solve();
}
clock_t timest = clock();
cerr << "[Executes in " << (double)(clock() - timest) / CLOCKS_PER_SEC << "s]"
<< '\n';
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = (int)1000000000 + 7;
const int MOD = (int)998244353;
const int INF = (int)1e18;
bool cmp(const pair<int, int>& a, const pair<int, int>& b) {
return a.second - a.first < b.second - b.first;
}
long long bit_ct1(long long x) {
long long ct = 0;
while (x > 0) {
x = x & (x - 1);
ct++;
}
return ct;
}
long long cc[27] = {0};
void solve() {
string s;
long long p;
cin >> s >> p;
long long ln = s.length();
vector<char> w(ln), ans(ln);
for (long long i = 0; i < ln; i++) {
w[i] = '1';
}
for (long long i = 0; i < ln; i++) {
if (s[i] == '0') {
if (i - p >= 0) w[i - p] = '0';
if (i + p < ln) w[i + p] = '0';
}
}
for (long long i = 0; i < ln; i++) {
if (i + p < ln && w[i + p] == '1') {
ans[i] = '1';
} else if (i - p >= 0 && w[i - p] == '1') {
ans[i] = '1';
} else {
ans[i] = '0';
}
}
for (long long i = 0; i < ln; i++) {
if (s[i] != ans[i]) {
cout << -1 << '\n';
return;
}
}
for (long long i = 0; i < ln; i++) {
cout << w[i];
}
cout << '\n';
return;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long t = 1;
cin >> t;
while (t--) {
solve();
}
clock_t timest = clock();
cerr << "[Executes in " << (double)(clock() - timest) / CLOCKS_PER_SEC << "s]"
<< '\n';
}
``` |
#include <bits/stdc++.h>
using namespace std;
int dx[4] = {-1, 0, 0, 1};
int dy[4] = {0, -1, 1, 0};
bool sortbysec(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);
}
long long a[10000005];
void sieve() {
for (int i = 1; i < 10000005; i++) {
for (int j = i; j <= 10000005; j += i) {
a[j]++;
}
}
}
long long n, w, h;
bool g(long long x) { return (x / w) * (x / h) >= n; }
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long test;
cin >> test;
while (test) {
int n, x, y;
string s, w = "";
cin >> s >> x;
n = s.length();
for (int i = 0; i < n; i++) {
w += "1";
}
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i + x < n) w[i + x] = '0';
if (i - x >= 0) w[i - x] = '0';
}
}
bool flag = true;
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
if ((i + x >= n || w[i + x] == '0') && (i - x < 0 || w[i - x] == '0')) {
flag = false;
}
}
}
if (!flag)
cout << "-1" << endl;
else
cout << w << endl;
test--;
}
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int dx[4] = {-1, 0, 0, 1};
int dy[4] = {0, -1, 1, 0};
bool sortbysec(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);
}
long long a[10000005];
void sieve() {
for (int i = 1; i < 10000005; i++) {
for (int j = i; j <= 10000005; j += i) {
a[j]++;
}
}
}
long long n, w, h;
bool g(long long x) { return (x / w) * (x / h) >= n; }
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long test;
cin >> test;
while (test) {
int n, x, y;
string s, w = "";
cin >> s >> x;
n = s.length();
for (int i = 0; i < n; i++) {
w += "1";
}
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
if (i + x < n) w[i + x] = '0';
if (i - x >= 0) w[i - x] = '0';
}
}
bool flag = true;
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
if ((i + x >= n || w[i + x] == '0') && (i - x < 0 || w[i - x] == '0')) {
flag = false;
}
}
}
if (!flag)
cout << "-1" << endl;
else
cout << w << endl;
test--;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t, n, i, flag, x, y;
vector<int> w;
cin >> t;
y = 0;
while (t--) {
y++;
string s;
cin >> s;
cin >> x;
i = 0;
while (s[i] != '\0') {
i++;
}
n = i;
w.clear();
w.resize(n, -1);
i = 0;
while (s[i] != '\0') {
if (s[i] == '0') {
if (i + x < n) w[i + x] = 0;
if (i - x > -1) w[i - x] = 0;
}
i++;
}
i = 0;
flag = 0;
while (s[i] != '\0') {
if (s[i] == '1') {
if (i + x < n && i - x > -1) {
if (w[i + x] == 0 && w[i - x] == 0) {
flag = 1;
break;
}
} else if (i + x < n) {
if (w[i + x] == 0) {
flag = 1;
break;
}
} else if (i - x > -1) {
if (w[i - x] == 0) {
flag = 1;
break;
}
} else {
flag = 1;
break;
}
}
i++;
}
if (flag == 0) {
for (i = 0; i < n; i++) {
if (w[i] == -1)
cout << 1;
else
cout << w[i];
}
cout << endl;
} else {
cout << "-1" << endl;
}
}
return 0;
}
| ### Prompt
Please formulate a CPP solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t, n, i, flag, x, y;
vector<int> w;
cin >> t;
y = 0;
while (t--) {
y++;
string s;
cin >> s;
cin >> x;
i = 0;
while (s[i] != '\0') {
i++;
}
n = i;
w.clear();
w.resize(n, -1);
i = 0;
while (s[i] != '\0') {
if (s[i] == '0') {
if (i + x < n) w[i + x] = 0;
if (i - x > -1) w[i - x] = 0;
}
i++;
}
i = 0;
flag = 0;
while (s[i] != '\0') {
if (s[i] == '1') {
if (i + x < n && i - x > -1) {
if (w[i + x] == 0 && w[i - x] == 0) {
flag = 1;
break;
}
} else if (i + x < n) {
if (w[i + x] == 0) {
flag = 1;
break;
}
} else if (i - x > -1) {
if (w[i - x] == 0) {
flag = 1;
break;
}
} else {
flag = 1;
break;
}
}
i++;
}
if (flag == 0) {
for (i = 0; i < n; i++) {
if (w[i] == -1)
cout << 1;
else
cout << w[i];
}
cout << endl;
} else {
cout << "-1" << endl;
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long min(long long a, long long b) { return (a < b) ? a : b; }
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long fp(long long a, long long b) {
if (b == 0) return 1;
long long x = fp(a, b / 2);
x = (x * x) % mod;
if (b & 1) x = (x * a) % mod;
return x;
}
const long long N = 2e5 + 5;
void solve() {
string s;
cin >> s;
long long n = s.size();
s = '#' + s;
long long x;
cin >> x;
string tmp = "#";
for (long long i = 1; i <= n; i++) tmp.push_back('1');
for (long long i = 1; i <= n; i++) {
if (i + x <= n && s[i] == '0') tmp[i + x] = '0';
if (i - x >= 1 && s[i] == '0') tmp[i - x] = '0';
}
long long f = 1;
for (long long i = 1; i <= n; i++) {
long long ok = 0;
if (i + x <= n && tmp[i + x] == '1') {
if (s[i] == '0') f = 0;
ok = 1;
}
if (i - x >= 1 && tmp[i - x] == '1') {
if (s[i] == '0') f = 0;
ok = 1;
}
if (ok == 0) {
if (s[i] == '1') f = 0;
}
}
if (f) {
cout << tmp.substr(1) << "\n";
} else {
cout << -1 << "\n";
}
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
srand(time(0));
long long t;
cin >> t;
for (long long i = 1; i <= t; i++) {
solve();
}
}
| ### Prompt
Develop a solution in CPP to the problem described below:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long min(long long a, long long b) { return (a < b) ? a : b; }
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long fp(long long a, long long b) {
if (b == 0) return 1;
long long x = fp(a, b / 2);
x = (x * x) % mod;
if (b & 1) x = (x * a) % mod;
return x;
}
const long long N = 2e5 + 5;
void solve() {
string s;
cin >> s;
long long n = s.size();
s = '#' + s;
long long x;
cin >> x;
string tmp = "#";
for (long long i = 1; i <= n; i++) tmp.push_back('1');
for (long long i = 1; i <= n; i++) {
if (i + x <= n && s[i] == '0') tmp[i + x] = '0';
if (i - x >= 1 && s[i] == '0') tmp[i - x] = '0';
}
long long f = 1;
for (long long i = 1; i <= n; i++) {
long long ok = 0;
if (i + x <= n && tmp[i + x] == '1') {
if (s[i] == '0') f = 0;
ok = 1;
}
if (i - x >= 1 && tmp[i - x] == '1') {
if (s[i] == '0') f = 0;
ok = 1;
}
if (ok == 0) {
if (s[i] == '1') f = 0;
}
}
if (f) {
cout << tmp.substr(1) << "\n";
} else {
cout << -1 << "\n";
}
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
srand(time(0));
long long t;
cin >> t;
for (long long i = 1; i <= t; i++) {
solve();
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long modx = 998244353;
const long long Nx = 1e5 + 2;
const long long inf = (long long)(1e19 + 10LL);
void solve() {
string str;
cin >> str;
long long x;
cin >> x;
long long n = (long long)(str.size());
vector<char> ans(n, '1');
for (long long i = 0; i < (long long)(str.size()); i++) {
if (str[i] == '0') {
if (i - x >= 0) {
ans[i - x] = '0';
}
if (i + x < n) {
ans[i + x] = '0';
}
}
}
string dup = "";
for (long long i = 0; i < (long long)(ans.size()); i++) {
if ((i - x >= 0 && ans[i - x] == '1') || (i + x < n && ans[i + x] == '1'))
dup = dup + "1";
else
dup = dup + "0";
}
if (str != dup) {
cout << "-1" << endl;
return;
}
for (auto it : ans) cout << it;
cout << endl;
return;
}
signed main() {
long long test = 1;
cin >> test;
while (test--) {
solve();
}
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long modx = 998244353;
const long long Nx = 1e5 + 2;
const long long inf = (long long)(1e19 + 10LL);
void solve() {
string str;
cin >> str;
long long x;
cin >> x;
long long n = (long long)(str.size());
vector<char> ans(n, '1');
for (long long i = 0; i < (long long)(str.size()); i++) {
if (str[i] == '0') {
if (i - x >= 0) {
ans[i - x] = '0';
}
if (i + x < n) {
ans[i + x] = '0';
}
}
}
string dup = "";
for (long long i = 0; i < (long long)(ans.size()); i++) {
if ((i - x >= 0 && ans[i - x] == '1') || (i + x < n && ans[i + x] == '1'))
dup = dup + "1";
else
dup = dup + "0";
}
if (str != dup) {
cout << "-1" << endl;
return;
}
for (auto it : ans) cout << it;
cout << endl;
return;
}
signed main() {
long long test = 1;
cin >> test;
while (test--) {
solve();
}
return 0;
}
``` |
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