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For this Question: Power Rangers have to trap a evil spirit monster into a cuboid shape box with dimensions A*B*C, where A, B and C are positive integers. The volume of the box should be exactly X cubic units (that is A*B*C should be equal to X). To allow minimum evil aura to leak from the box the surface area of the box should be minimized. Given X, find the minimum surface area of the optimal box.The first and the only line of input contains a single integer X.
Constraints:
1 <= X <= 1000000Print the minimum surface area of the optimal box.Sample Input 1
125
Sample Output 1
150
Explanation: Optimal dimensions are 5*5*5.
Sample Input 2
100
Sample Output 1
130
Explanation: Optimal dimensions are 5*4*5., I have written this Solution Code:
import java.io.*;
import java.util.*;
public class Main
{
public static void main(String[] args)throws Exception{ new Main().run();}
long mod=1000000000+7;
long tsa=Long.MAX_VALUE;
void solve() throws Exception
{
long X=nl();
div(X);
out.println(tsa);
}
long cal(long a,long b, long c)
{
return 2l*(a*b + b*c + c*a);
}
void div(long n)
{
for (long i=1; i*i<=n; i++)
{
if (n%i==0)
{
if (n/i == i)
{
all_div(i, i);
}
else
{
all_div(i, n/i);
all_div(n/i, i);
}
}
}
}
void all_div(long n , long alag)
{
ArrayList<Long> al = new ArrayList<>();
for (long i=1; i*i<=n; i++)
{
if (n%i==0)
{
if (n/i == i)
tsa=min(tsa,cal(i,i,alag));
else
{
tsa=min(tsa,cal(i,n/i,alag));
}
}
}
}
private byte[] buf=new byte[1024];
private int index;
private InputStream in;
private int total;
private SpaceCharFilter filter;
PrintWriter out;
int min(int... ar){int min=Integer.MAX_VALUE;for(int i:ar)min=Math.min(min, i);return min;}
long min(long... ar){long min=Long.MAX_VALUE;for(long i:ar)min=Math.min(min, i);return min;}
int max(int... ar) {int max=Integer.MIN_VALUE;for(int i:ar)max=Math.max(max, i);return max;}
long max(long... ar) {long max=Long.MIN_VALUE;for(long i:ar)max=Math.max(max, i);return max;}
void reverse(int a[]){for(int i=0;i<a.length>>1;i++){int tem=a[i];a[i]=a[a.length-1-i];a[a.length-1-i]=tem;}}
void reverse(long a[]){for(int i=0;i<a.length>>1;i++){long tem=a[i];a[i]=a[a.length-1-i];a[a.length-1-i]=tem;}}
String reverse(String s){StringBuilder sb=new StringBuilder(s);sb.reverse();return sb.toString();}
void shuffle(int a[]) {
ArrayList<Integer> al = new ArrayList<>();
for(int i=0;i<a.length;i++)
al.add(a[i]);
Collections.sort(al);
for(int i=0;i<a.length;i++)
a[i]=al.get(i);
}
long lcm(long a,long b)
{
return (a*b)/(gcd(a,b));
}
int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
long expo(long p,long q)
{
long z = 1;
while (q>0) {
if (q%2 == 1) {
z = (z * p)%mod;
}
p = (p*p)%mod;
q >>= 1;
}
return z;
}
void run()throws Exception
{
in=System.in; out = new PrintWriter(System.out);
solve();
out.flush();
}
private int scan()throws IOException
{
if(total<0)
throw new InputMismatchException();
if(index>=total)
{
index=0;
total=in.read(buf);
if(total<=0)
return -1;
}
return buf[index++];
}
private int ni() throws IOException
{
int c = scan();
while (isSpaceChar(c))
c = scan();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = scan();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = scan();
} while (!isSpaceChar(c));
return res * sgn;
}
private long nl() throws IOException
{
long num = 0;
int b;
boolean minus = false;
while ((b = scan()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
;
if (b == '-') {
minus = true;
b = scan();
}
while (true) {
if (b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = scan();
}
}
private double nd() throws IOException{
return Double.parseDouble(ns());
}
private String ns() throws IOException {
int c = scan();
while (isSpaceChar(c))
c = scan();
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c))
res.appendCodePoint(c);
c = scan();
} while (!isSpaceChar(c));
return res.toString();
}
private String nss() throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
return br.readLine();
}
private char nc() throws IOException
{
int c = scan();
while (isSpaceChar(c))
c = scan();
return (char) c;
}
private boolean isWhiteSpace(int n)
{
if(n==' '||n=='\n'||n=='\r'||n=='\t'||n==-1)
return true;
return false;
}
private boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return isWhiteSpace(c);
}
private interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Power Rangers have to trap a evil spirit monster into a cuboid shape box with dimensions A*B*C, where A, B and C are positive integers. The volume of the box should be exactly X cubic units (that is A*B*C should be equal to X). To allow minimum evil aura to leak from the box the surface area of the box should be minimized. Given X, find the minimum surface area of the optimal box.The first and the only line of input contains a single integer X.
Constraints:
1 <= X <= 1000000Print the minimum surface area of the optimal box.Sample Input 1
125
Sample Output 1
150
Explanation: Optimal dimensions are 5*5*5.
Sample Input 2
100
Sample Output 1
130
Explanation: Optimal dimensions are 5*4*5., I have written this Solution Code: m =int(input())
def print_factors(x):
factors=[]
for i in range(1, x + 1):
if x % i == 0:
factors.append(i)
return(factors)
area=[]
factors=print_factors(m)
for a in factors:
for b in factors:
for c in factors:
if(a*b*c==m):
area.append((2*a*b)+(2*b*c)+(2*a*c))
print(min(area)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Power Rangers have to trap a evil spirit monster into a cuboid shape box with dimensions A*B*C, where A, B and C are positive integers. The volume of the box should be exactly X cubic units (that is A*B*C should be equal to X). To allow minimum evil aura to leak from the box the surface area of the box should be minimized. Given X, find the minimum surface area of the optimal box.The first and the only line of input contains a single integer X.
Constraints:
1 <= X <= 1000000Print the minimum surface area of the optimal box.Sample Input 1
125
Sample Output 1
150
Explanation: Optimal dimensions are 5*5*5.
Sample Input 2
100
Sample Output 1
130
Explanation: Optimal dimensions are 5*4*5., I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int x;
cin >> x;
int ans = 6*x*x;
for (int i=1;i*i*i<=x;++i)
if (x%i==0)
for (int j=i;j*j<=x/i;++j)
if (x/i % j==0) {
int k=x/i/j;
int cur=0;
cur=i*j+i*k+k*j;
cur*=2;
ans=min(ans,cur);
}
cout<<ans;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a rule in ludo that a token can only be unlocked when either a 1 or 6 shown in the die. Given the die number N, Your task is to check whether the token can be unlocked or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>ludo()</b> that takes integers N as argument.
Constraints:-
1 <= N <= 6Return 1 if the token can be unlocked else return 0.Sample Input:-
1
Sample Output:-
1
Sample Input:-
2
Sample Output:-
0, I have written this Solution Code: int ludo(int N){
return (N==1||N==6);
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a rule in ludo that a token can only be unlocked when either a 1 or 6 shown in the die. Given the die number N, Your task is to check whether the token can be unlocked or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>ludo()</b> that takes integers N as argument.
Constraints:-
1 <= N <= 6Return 1 if the token can be unlocked else return 0.Sample Input:-
1
Sample Output:-
1
Sample Input:-
2
Sample Output:-
0, I have written this Solution Code: def ludo(N):
if N==1 or N==6:
return 1
return 0
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a rule in ludo that a token can only be unlocked when either a 1 or 6 shown in the die. Given the die number N, Your task is to check whether the token can be unlocked or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>ludo()</b> that takes integers N as argument.
Constraints:-
1 <= N <= 6Return 1 if the token can be unlocked else return 0.Sample Input:-
1
Sample Output:-
1
Sample Input:-
2
Sample Output:-
0, I have written this Solution Code:
int ludo(int N){
return (N==1||N==6);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a rule in ludo that a token can only be unlocked when either a 1 or 6 shown in the die. Given the die number N, Your task is to check whether the token can be unlocked or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>ludo()</b> that takes integers N as argument.
Constraints:-
1 <= N <= 6Return 1 if the token can be unlocked else return 0.Sample Input:-
1
Sample Output:-
1
Sample Input:-
2
Sample Output:-
0, I have written this Solution Code: static int ludo(int N){
if(N==1 || N==6){return 1;}
return 0;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying to open a lock for that she needs a passcode. Sara knows that the passcode contains N digits in it and the digits at even indices(0- indexing) are even and the digits at odd indices are {2, 3, 5, 7}.
Now Sara wants to know how many different passcodes are possible so that she can open her lock.The input contains a single integer N.
Constraints:-
1 <= N <= 10^15Print the number of possible passcodes.
Note:-
Since the answer can be quite large so print your ans modulo 10^9+7.Sample Input:-
1
Sample Output:-
5
Explanation:-
Possible answer:- 0,2,4,6,8
Sample Input:-
4
Sample Output:-
400, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.math.*;
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
long n = sc.nextLong();
BigInteger five = new BigInteger("5");
BigInteger four = new BigInteger("4");
BigInteger m = new BigInteger("1000000007");
long fourPower = n/2;
long fivePow = n - fourPower;
BigInteger res = five.modPow(BigInteger.valueOf(fivePow),m);
BigInteger res2 = four.modPow(BigInteger.valueOf(fourPower),m);
long ans = (res.longValue()* res2.longValue())%mod;
System.out.print(ans);
}
static long mod = 1000000007;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying to open a lock for that she needs a passcode. Sara knows that the passcode contains N digits in it and the digits at even indices(0- indexing) are even and the digits at odd indices are {2, 3, 5, 7}.
Now Sara wants to know how many different passcodes are possible so that she can open her lock.The input contains a single integer N.
Constraints:-
1 <= N <= 10^15Print the number of possible passcodes.
Note:-
Since the answer can be quite large so print your ans modulo 10^9+7.Sample Input:-
1
Sample Output:-
5
Explanation:-
Possible answer:- 0,2,4,6,8
Sample Input:-
4
Sample Output:-
400, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
long long fast_pow(int base, long long n,int M=1000000007)
{
if(n==0)
return 1;
if(n==1)
return base;
long long halfn=fast_pow(base,n/2);
if(n%2==0)
return ( halfn * halfn ) % M;
else
return ( ( ( halfn * halfn ) % M ) * base ) % M;
}
int solve(long long n) {
const int M = 1000000007;
return (int)(fast_pow(5,(n+1)/2)*fast_pow(4,(n/2))%M);
}
int main(){
long long n;
cin>>n;
cout<<solve(n);
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find out whether it is divisible by 3.The first and the only line of input contains the number N.
<b>Constraints</b>
1 <= N <= 10^100000 (N may consist of 100001 digits).
<b>No usual datatype will be able to input such large number.</b>Output "Yes" if the number is divisible by 3, else output "No".Sample Input 1
14
Sample Output 1
No
Sample Input 2
1234567890123456789012345678901234567890
Sample Output 2
Yes
Explanation: In the first sample case, the number is not divisible by 3, while in the second sample case, it is divisible by 3. (We know how weird this explanation is, but ok)., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define int long long
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define MOD 1000000007
#define INF 1000000000000000007LL
const int N = 100005;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
string s; cin>>s;
int sum=0;
For(i, 0, s.length()){
sum += (s[i]-'0');
}
if(sum%3==0)
cout<<"Yes";
else
cout<<"No";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find out whether it is divisible by 3.The first and the only line of input contains the number N.
<b>Constraints</b>
1 <= N <= 10^100000 (N may consist of 100001 digits).
<b>No usual datatype will be able to input such large number.</b>Output "Yes" if the number is divisible by 3, else output "No".Sample Input 1
14
Sample Output 1
No
Sample Input 2
1234567890123456789012345678901234567890
Sample Output 2
Yes
Explanation: In the first sample case, the number is not divisible by 3, while in the second sample case, it is divisible by 3. (We know how weird this explanation is, but ok)., I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
String S = sc.next();
int sum=0;
for(int i=0;i<S.length();i++){
sum+=S.charAt(i)-'0';
}
if(sum%3==0){
System.out.print("Yes");
}
else{
System.out.print("No");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find out whether it is divisible by 3.The first and the only line of input contains the number N.
<b>Constraints</b>
1 <= N <= 10^100000 (N may consist of 100001 digits).
<b>No usual datatype will be able to input such large number.</b>Output "Yes" if the number is divisible by 3, else output "No".Sample Input 1
14
Sample Output 1
No
Sample Input 2
1234567890123456789012345678901234567890
Sample Output 2
Yes
Explanation: In the first sample case, the number is not divisible by 3, while in the second sample case, it is divisible by 3. (We know how weird this explanation is, but ok)., I have written this Solution Code: n = int(input())
if n%3==0:
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given integer N, super primes are those integers from 1 to N whose multiples (other than themselves) do no exist in the range [1, N].
Your task is to generate all super primes <= N in sorted order.
</b>Note: Super primes are not related to primes in any way.</b><b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SuperPrime()</b> that takes the integer N as a parameter.
Constraints:-
2 < = N <= 1000Printing will be done by the driver code you just need to complete the generator function.Sample Input:-
5
Sample Output:-
3 4 5
Sample Input:-
4
Sample Output:-
3 4, I have written this Solution Code: public static void SuperPrimes(int n){
int x = n/2+1;
for(int i=x ; i<=n ; i++){
out.printf("%d ",i);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given integer N, super primes are those integers from 1 to N whose multiples (other than themselves) do no exist in the range [1, N].
Your task is to generate all super primes <= N in sorted order.
</b>Note: Super primes are not related to primes in any way.</b><b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SuperPrime()</b> that takes the integer N as a parameter.
Constraints:-
2 < = N <= 1000Printing will be done by the driver code you just need to complete the generator function.Sample Input:-
5
Sample Output:-
3 4 5
Sample Input:-
4
Sample Output:-
3 4, I have written this Solution Code: def SuperPrimes(N):
for i in range (int(N/2)+1,N+1):
yield i
return
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted doubly linked list containing n nodes. Your task is to remove duplicate nodes from the given list.
Example 1:
Input
1<->2<->2-<->3<->3<->4
Output:
1<->2<->3<->4
Example 2:
Input
1<->1<->1<->1
Output
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>deleteDuplicates()</b> that takes head node as parameter.
Constraints:
1 <=N <= 10000
1 <= Node. data<= 2*10000Return the head of the modified list.Sample Input:-
6
1 2 2 3 3 4
Sample Output:-
1 2 3 4
Sample Input:-
4
1 1 1 1
Sample Output:-
1, I have written this Solution Code: public static Node deleteDuplicates(Node head)
{
/* if list is empty */
if (head== null)
return head;
Node current = head;
while (current.next != null)
{
/* Compare current node with next node */
if (current.val == current.next.val)
/* delete the node pointed to by
' current->next' */
deleteNode(head, current.next);
/* else simply move to the next node */
else
current = current.next;
}
return head;
}
/* Function to delete a node in a Doubly Linked List.
head_ref --> pointer to head node pointer.
del --> pointer to node to be deleted. */
public static void deleteNode(Node head, Node del)
{
/* base case */
if(head==null || del==null)
{
return ;
}
/* If node to be deleted is head node */
if(head==del)
{
head=del.next;
}
/* Change next only if node to be deleted
is NOT the last node */
if(del.next!=null)
{
del.next.prev=del.prev;
}
/* Change prev only if node to be deleted
is NOT the first node */
if (del.prev != null)
del.prev.next = del.next;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Shivansh wants to give chapo to Arpit, but only if he is able to solve the following problem.
As you are a dear friend of Arpit, you decide to help Arpit solve the problem.
Given an array Arr of N numbers. You have to process Q queries on that array. Queries can be of following types :
> add x to index y
> given y compute sum of (Arr[i]*(y%i)) for i from 1 to N.First Line of input contains two space separated integers N and Q.
Second Line contains N space separated integers Arr[1] Arr[2] ..... Arr[N]
Next Q lines describe one query each and can be one of the two type:
> 1 i X - Add X to index i.
> 2 Y - compute sum of (Arr[i]*(Y%i)) for i from 1 to N.
Constraints
1 <= N,Q <= 100000
1 <= Arr[i], X, Y <= 100000
1 <= i <= NFor each second type query output one integer per line - answer for this query.Sample input
4 3
3 2 3 4
1 2 1
2 10
2 12
Sample output
11
0
Explaination
after 1st query array : 3 3 3 4
for second query ans = (3*(10%1)) +(3*(10%2))+(3*(10%3)) +(4*(10%4)) = 11
for third query ans = (3*(12%1)) +(3*(12%2))+(3*(12%3)) +(4*(12%4)) = 0
, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define se second
#define fi first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
int seg[300000];
int n=100001;
void up(int i, int value) {
int in=i;
for (seg[in += n] += i*value; in > 1; in >>= 1) seg[in>>1] = seg[in] + seg[in^1];
}
int query(int l, int r) {
++r;
int res = 0;
for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
if (l&1) res += seg[l++];
if (r&1) res += seg[--r];
}
return res;
}
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int q;
cin>>q;
int tot=0;
for(int i=1;i<=n;++i)
{
int d;
cin>>d;
tot+=d;
up(i,d);
}
while(q)
{
--q;
char c;
cin>>c;
if(c=='1')
{
int i,x;
cin>>i>>x;
up(i,x);
tot+=x;
}
if(c=='2')
{
int y;
cin>>y;
int total=tot*y;
int consumed=0;
int l=1,r=1;
while(1)
{
consumed+=(y/l)*(query(l,r));
if(r==y)
break;
l=r+1;
r=y/(y/l);
}
cout<<total-consumed<<"\n";
}
}
#ifdef ANIKET_GOYAL
cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sting S of lowercase English character's. Your task is to find the minimum length subsequence which contains all characters from 'a' to 'z' in increasing order (i. e "abcdefghijklmnopqrstuvwxyz").Input contains a single string S.
Constraints:-
1 <= |S| <= 100000
Note:- Sting will contain only lowercase english letters.Print the minimum length of such subsequence. If No such subsequence exist print -1.Sample Input:-
aaaabcdefghijklmnopqrstuvwxyabcdzzz
Sample Output:-
30
Sample Input:-
abcd
Sample Output:-
-1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner in = new Scanner(System.in);
String str = in.next();
System.out.print(checkSubstring(str));
}
static int checkSubstring(String str){
int n = str.length();
if(n < 26){
return -1;
}
Deque<Integer> indexOfA = new LinkedList<>();
for(int i=0; i<n; i++){
if(str.charAt(i)-'a' == 0){
if(!indexOfA.isEmpty() && indexOfA.peekLast() == i-1){
indexOfA.removeLast();
}
indexOfA.addLast(i);
}
}
int res = n+1;
while(!indexOfA.isEmpty()){
int holder = -1;
int start = 0, end = 0;
int i = indexOfA.removeFirst();
for(;i<n; i++){
if(str.charAt(i)-'a' == holder+1){
holder++;
}
if(holder == 0 && str.charAt(i)-'a' == 0){
start = i;
}
if(holder == 25){
end = i;
break;
}
}
res = Math.min(res,holder==25?end-start+1:n+1);
}
return res==n+1?-1:res;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sting S of lowercase English character's. Your task is to find the minimum length subsequence which contains all characters from 'a' to 'z' in increasing order (i. e "abcdefghijklmnopqrstuvwxyz").Input contains a single string S.
Constraints:-
1 <= |S| <= 100000
Note:- Sting will contain only lowercase english letters.Print the minimum length of such subsequence. If No such subsequence exist print -1.Sample Input:-
aaaabcdefghijklmnopqrstuvwxyabcdzzz
Sample Output:-
30
Sample Input:-
abcd
Sample Output:-
-1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 101
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int solve(string s)
{
int n = s.length();
queue<int> q[26];
for(int i=0;i<n;i++){
q[s[i]-'a'].push(i);
}
int ans = INT_MAX;
while(!q[0].empty()){
int start = q[0].front();
q[0].pop();
int cur = start;
bool allchar=true;
for(int i=1;i<26;i++){
while(q[i].size()!=0 && q[i].front()<cur){
q[i].pop();
}
if(q[i].size()==0){
allchar=false;
break;
}
cur=q[i].front();
}
if(allchar==false){break;}
ans=min(ans,cur-start+1);
}
if(ans==INT_MAX){return -1;}
return ans;
}
signed main() {
int t;
//cin>>t;
t=1;
//vector<string> v;
while(t--){
string s;
cin>>s;
int r=solve(s);
cout<<r<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N candies, i<sup>th</sup> of them costing p<sub>i</sub>. You have M amount of money with you. Find the maximum number of candies you can buy.The first line of the input contains two integers N and M.
The second line of the input contains N space seperated integers.
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= M <= 10<sup>14</sup>
1 <= p<sub>i</sub> <= 10<sup>9</sup>Print the maximum number of candies you can buy.Sample Input:
4 7
3 1 4 2
Sample Output:
3, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n, m;
cin >> n >> m;
vector<int> p(n);
for(auto &i : p) cin >> i;
sort(p.begin(), p.end());
int cur = 0, ans = 0;
for(int i = 0; i < n; i++){
if(cur + p[i] > m){
break;
}
cur += p[i];
ans++;
}
cout << ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string s, find the minimum number of changes required to it so that all substrings of the string become distinct.
Note: length of string is atmost 26.The first line contains an integer T, number of test cases. For each testcase there is only one line containing s atmost 26 characters.
1 <= T <= 100
1 <= |s| <= 26For each testcase in new line, print the minimum number of changes to the string.Sample Input:
3
aab
aebaecedabbee
ab
Sample Output:
1
8
0
Explanation:
Testcase 1: If we change one instance of 'a' to any character from 'c' to 'z', we get all distinct substrings.
Testcase 2: We need to change 2 a's, 2 b's and 4 e's to get distinct substrings.
Testcase 3: As no change is required hence 0., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws Exception {
InputStreamReader ir = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(ir);
int n = Integer.parseInt(br.readLine());
for(int i=0; i<n; i++) {
int[] arr = new int[26];
int ans = 0;
String s = br.readLine();
int l = s.length();
for (int j=0; j<l; j++) {
int c = s.charAt(j) - 97;
arr[c]++;
}
for (int k=0; k<26; k++) {
if(arr[k]>1) {
int val = arr[k];
ans += val-1;
}
}
System.out.println(ans);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string s, find the minimum number of changes required to it so that all substrings of the string become distinct.
Note: length of string is atmost 26.The first line contains an integer T, number of test cases. For each testcase there is only one line containing s atmost 26 characters.
1 <= T <= 100
1 <= |s| <= 26For each testcase in new line, print the minimum number of changes to the string.Sample Input:
3
aab
aebaecedabbee
ab
Sample Output:
1
8
0
Explanation:
Testcase 1: If we change one instance of 'a' to any character from 'c' to 'z', we get all distinct substrings.
Testcase 2: We need to change 2 a's, 2 b's and 4 e's to get distinct substrings.
Testcase 3: As no change is required hence 0., I have written this Solution Code:
t = int(input())
for _ in range(t):
a = input();
d1 = {}
c = 0
for x in a:
if x in d1:
c+=1;
else:
d1[x]=1
print(c), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string s, find the minimum number of changes required to it so that all substrings of the string become distinct.
Note: length of string is atmost 26.The first line contains an integer T, number of test cases. For each testcase there is only one line containing s atmost 26 characters.
1 <= T <= 100
1 <= |s| <= 26For each testcase in new line, print the minimum number of changes to the string.Sample Input:
3
aab
aebaecedabbee
ab
Sample Output:
1
8
0
Explanation:
Testcase 1: If we change one instance of 'a' to any character from 'c' to 'z', we get all distinct substrings.
Testcase 2: We need to change 2 a's, 2 b's and 4 e's to get distinct substrings.
Testcase 3: As no change is required hence 0., I have written this Solution Code: #include<bits/stdc++.h>
#define int long long
#define ld long double
#define ll long long
#define pb push_back
#define endl '\n'
#define pi pair<int,int>
#define vi vector<int>
#define all(a) (a).begin(),(a).end()
#define fi first
#define se second
#define sz(x) (int)x.size()
#define hell 1000000007
#define rep(i,a,b) for(int i=a;i<b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define lbnd lower_bound
#define ubnd upper_bound
#define bs binary_search
#define mp make_pair
using namespace std;
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
void solve(){
string s; cin >> s;
int a[26] = {0};
for(int i = 0; i < (int)s.length(); i++)
a[s[i]-'a']++;
int ans = 0;
for(int i = 0; i < 26; i++){
if(a[i])
ans += (a[i]-1);
}
cout << ans << endl;
}
void testcases(){
int tt = 1;
cin >> tt;
while(tt--){
solve();
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms: ";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ronaldo has challenged Messi to beat his team in the upcoming “Friendly” fixture. But both these legends are tired of the original football rules, so they decided to play Football with a twist. Each player is assigned an ID, from 1 to 10^6. Initially Ronaldo has the ball in his posession. Given a sequence of N passes, help Messi Know which ID player has the ball, after N passes.
Note: Passes are of two types :
i) P ID, which means the player currently having the ball passes it to Player with identity ID.
ii) B, which means the player currently having the ball passes it back to the Player he got the Pass From.
Also, It is guaranteed that the given order of passes will be a valid order.The first line of input contains number of testcases T. The 1st line of each testcase, contains two integers, N and ID1, denoting the total number of passes and the ID of Ronaldo. Respectively. Each of the next N lines, contains the information of a Pass which can be either of the above two types, i.e:
1) P ID
2) B
Constraints
1 <= T <= 100
1 <= N <= 10^5
1 <= IDs <= 10^6
Sum of N for every test case is less than 10^6For each testcase you need to print the ID of the Player having the ball after N passes.Input :
2
3 1
P 13
P 14
B
5 1
P 12
P 13
P 14
B
B
Output :
13
14
Explanation :
Testcase 1: Initially, the ball is with Ronaldo, having ID 1.
In the first pass, he passes the ball to Player with ID 13.
In the Second Pass, the player currently having the ball, ie Player with ID 13, passes it to player with ID 14.
In the last pass the player currently having the ball, ie Player with ID 14 passed the ball back to the player from whom he got the Pass, i.e the ball is passed back to Player with Player ID 13, as Player ID 13 had passed the ball to player ID 14 in the previous pass.
Testcase 2: Initially, the ball is with Ronaldo, having ID 1.
In the first pass he passes the ball to player with ID 12.
In the second pass, the second player passes the ball to palyer with ID 13, again the player with ID 13 passes the ball to player with ID 14.
Now, player with ID 14 passed back the ball to 13, and again 13 passes back the ball to 14., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine().trim());
while (t-- > 0) {
String[] line = br.readLine().trim().split(" ");
int n = Integer.parseInt(line[0]);
int curr = Integer.parseInt(line[1]);
int prev = curr;
while (n-- > 0) {
String[] currLine = br.readLine().trim().split(" ");
if (currLine[0].equals("P")) {
prev = curr;
curr = Integer.parseInt(currLine[1]);
}
if (currLine[0].equals("B")) {
int temp = curr;
curr = prev;
prev = temp;
}
}
System.out.println(curr);
System.gc();
}
br.close();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ronaldo has challenged Messi to beat his team in the upcoming “Friendly” fixture. But both these legends are tired of the original football rules, so they decided to play Football with a twist. Each player is assigned an ID, from 1 to 10^6. Initially Ronaldo has the ball in his posession. Given a sequence of N passes, help Messi Know which ID player has the ball, after N passes.
Note: Passes are of two types :
i) P ID, which means the player currently having the ball passes it to Player with identity ID.
ii) B, which means the player currently having the ball passes it back to the Player he got the Pass From.
Also, It is guaranteed that the given order of passes will be a valid order.The first line of input contains number of testcases T. The 1st line of each testcase, contains two integers, N and ID1, denoting the total number of passes and the ID of Ronaldo. Respectively. Each of the next N lines, contains the information of a Pass which can be either of the above two types, i.e:
1) P ID
2) B
Constraints
1 <= T <= 100
1 <= N <= 10^5
1 <= IDs <= 10^6
Sum of N for every test case is less than 10^6For each testcase you need to print the ID of the Player having the ball after N passes.Input :
2
3 1
P 13
P 14
B
5 1
P 12
P 13
P 14
B
B
Output :
13
14
Explanation :
Testcase 1: Initially, the ball is with Ronaldo, having ID 1.
In the first pass, he passes the ball to Player with ID 13.
In the Second Pass, the player currently having the ball, ie Player with ID 13, passes it to player with ID 14.
In the last pass the player currently having the ball, ie Player with ID 14 passed the ball back to the player from whom he got the Pass, i.e the ball is passed back to Player with Player ID 13, as Player ID 13 had passed the ball to player ID 14 in the previous pass.
Testcase 2: Initially, the ball is with Ronaldo, having ID 1.
In the first pass he passes the ball to player with ID 12.
In the second pass, the second player passes the ball to palyer with ID 13, again the player with ID 13 passes the ball to player with ID 14.
Now, player with ID 14 passed back the ball to 13, and again 13 passes back the ball to 14., I have written this Solution Code: for i in range(int(input())):
N, ID = map(int,input().split())
pre = 0
for i in range(N):
arr = input().split()
if len(arr)==2:
pre,ID = ID,arr[1]
else:
ID,pre = pre,ID
print(ID) if pre else print(0), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ronaldo has challenged Messi to beat his team in the upcoming “Friendly” fixture. But both these legends are tired of the original football rules, so they decided to play Football with a twist. Each player is assigned an ID, from 1 to 10^6. Initially Ronaldo has the ball in his posession. Given a sequence of N passes, help Messi Know which ID player has the ball, after N passes.
Note: Passes are of two types :
i) P ID, which means the player currently having the ball passes it to Player with identity ID.
ii) B, which means the player currently having the ball passes it back to the Player he got the Pass From.
Also, It is guaranteed that the given order of passes will be a valid order.The first line of input contains number of testcases T. The 1st line of each testcase, contains two integers, N and ID1, denoting the total number of passes and the ID of Ronaldo. Respectively. Each of the next N lines, contains the information of a Pass which can be either of the above two types, i.e:
1) P ID
2) B
Constraints
1 <= T <= 100
1 <= N <= 10^5
1 <= IDs <= 10^6
Sum of N for every test case is less than 10^6For each testcase you need to print the ID of the Player having the ball after N passes.Input :
2
3 1
P 13
P 14
B
5 1
P 12
P 13
P 14
B
B
Output :
13
14
Explanation :
Testcase 1: Initially, the ball is with Ronaldo, having ID 1.
In the first pass, he passes the ball to Player with ID 13.
In the Second Pass, the player currently having the ball, ie Player with ID 13, passes it to player with ID 14.
In the last pass the player currently having the ball, ie Player with ID 14 passed the ball back to the player from whom he got the Pass, i.e the ball is passed back to Player with Player ID 13, as Player ID 13 had passed the ball to player ID 14 in the previous pass.
Testcase 2: Initially, the ball is with Ronaldo, having ID 1.
In the first pass he passes the ball to player with ID 12.
In the second pass, the second player passes the ball to palyer with ID 13, again the player with ID 13 passes the ball to player with ID 14.
Now, player with ID 14 passed back the ball to 13, and again 13 passes back the ball to 14., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n, id; cin >> n >> id;
int pre = 0, cur = id;
for(int i = 1; i <= n; i++){
char c; cin >> c;
if(c == 'P'){
int x; cin >> x;
pre = cur;
cur = x;
}
else{
swap(pre, cur);
}
}
cout << cur << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alexa has a drink that has A kilocalories of energy per 100 milliliters. How many kilocalories of energy does B milliliters of this drink have?The input consists of two space separated integers A and B.
<b>Constraints</b>
0 ≤ A, B ≤1000
All values in input are integers.Print a number representing the answer up to 2 decimal places.
Your output is considered correct when its absolute or relative error from our answer is at most 10<sup>−2</sup>.<b>Sample Input 1</b>
45 200
<b>Sample Output 1</b>
90.00
<b>Sample Input 2</b>
37 450
<b>Sample Output 2</b>
166.50
<b>Sample Input 3</b>
50 0
<b>Sample Output 3</b>
0.00, I have written this Solution Code: import java.util.Scanner;
public class Main {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
double a, b;
a = sc.nextDouble();
b = sc.nextDouble();
double val = a*b/100.0;
System.out.printf("%.2f", val);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alexa has a drink that has A kilocalories of energy per 100 milliliters. How many kilocalories of energy does B milliliters of this drink have?The input consists of two space separated integers A and B.
<b>Constraints</b>
0 ≤ A, B ≤1000
All values in input are integers.Print a number representing the answer up to 2 decimal places.
Your output is considered correct when its absolute or relative error from our answer is at most 10<sup>−2</sup>.<b>Sample Input 1</b>
45 200
<b>Sample Output 1</b>
90.00
<b>Sample Input 2</b>
37 450
<b>Sample Output 2</b>
166.50
<b>Sample Input 3</b>
50 0
<b>Sample Output 3</b>
0.00, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main() {
int A, B;
cin >> A >> B;
float soln = (float)(A * B) / 100.0;
printf("%0.2f\n", soln);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alexa has a drink that has A kilocalories of energy per 100 milliliters. How many kilocalories of energy does B milliliters of this drink have?The input consists of two space separated integers A and B.
<b>Constraints</b>
0 ≤ A, B ≤1000
All values in input are integers.Print a number representing the answer up to 2 decimal places.
Your output is considered correct when its absolute or relative error from our answer is at most 10<sup>−2</sup>.<b>Sample Input 1</b>
45 200
<b>Sample Output 1</b>
90.00
<b>Sample Input 2</b>
37 450
<b>Sample Output 2</b>
166.50
<b>Sample Input 3</b>
50 0
<b>Sample Output 3</b>
0.00, I have written this Solution Code: a, b = map(int, input().split())
ans = a*b/100.0
print("%.2f" %ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to check whether the given number is prime or notThe input contains a single integer N.
Constraints:-
1 <= N <= 100000000000Print "YES" If the given number is prime else print "NO".Sample Input:-
2
Sample Output:-
YES
Sample Input:-
4
Sample Output:-
NO, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
long n = sc.nextLong();
int p=(int)Math.sqrt(n);
for(int i=2;i<=p;i++){
if(n%i==0){System.out.print("NO");return;}
}
System.out.print("YES");
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to check whether the given number is prime or notThe input contains a single integer N.
Constraints:-
1 <= N <= 100000000000Print "YES" If the given number is prime else print "NO".Sample Input:-
2
Sample Output:-
YES
Sample Input:-
4
Sample Output:-
NO, I have written this Solution Code: import math
def isprime(A):
if A == 1:
return False
sqrt = int(math.sqrt(A))
for i in range(2,sqrt+1):
if A%i == 0:
return False
return True
inp = int(input())
if isprime(inp):
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to check whether the given number is prime or notThe input contains a single integer N.
Constraints:-
1 <= N <= 100000000000Print "YES" If the given number is prime else print "NO".Sample Input:-
2
Sample Output:-
YES
Sample Input:-
4
Sample Output:-
NO, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
long n;
cin>>n;
long x = sqrt(n);
for(int i=2;i<=x;i++){
if(n%i==0){cout<<"NO";return 0;}
}
cout<<"YES";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a 2X2 square matrix. You need to find the determinant of the matrix.The input contains two-line, each line contain two integers separated by spaces.
Each element of the matrix is from 1 to 100.Output a single integer, the determinant of the matrix.Sample Input
4 5
2 3
Sample Output
2
Sample Input
2 10
10 40
Sample Output
-20, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc=new Scanner(System.in);
int[][]arr=new int[2][2];
for(int i=0;i<2;i++){
for(int j=0;j<2;j++){
arr[i][j]=sc.nextInt();
}
}
int determinent=(arr[0][0]*arr[1][1])-(arr[1][0]*arr[0][1]);
System.out.print(determinent);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a 2X2 square matrix. You need to find the determinant of the matrix.The input contains two-line, each line contain two integers separated by spaces.
Each element of the matrix is from 1 to 100.Output a single integer, the determinant of the matrix.Sample Input
4 5
2 3
Sample Output
2
Sample Input
2 10
10 40
Sample Output
-20, I have written this Solution Code: arr1=list(map(int,input().split()))
arr2=list(map(int,input().split()))
lis=[]
a11=arr1[0]
a12=arr1[1]
a21=arr2[0]
a22=arr2[1]
det=(a11*a22)-(a12*a21)
print(det), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a 2X2 square matrix. You need to find the determinant of the matrix.The input contains two-line, each line contain two integers separated by spaces.
Each element of the matrix is from 1 to 100.Output a single integer, the determinant of the matrix.Sample Input
4 5
2 3
Sample Output
2
Sample Input
2 10
10 40
Sample Output
-20, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int a,b,c,d;
cin>>a>>b>>c>>d;
cout<<a*d-b*c;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given of array of N integers, find its number of increasing subsequences modulo (10^9 + 7)
A subsequence of the given array is the array formed after removing zero or more elements from it.
It is said to be increasing when every element is greater than its previous one.First line of input contains T, the number of test cases. Each test case is of two lines.
First line of each test case contains N, the number of elements of the array.
Next line contains N integers Ai, denoting elements of the array
1 <= T <= 50
1 <= N <= 2000
1 <= Ai <= 10^9Output T lines denoting the number of increasing subsequence for every test case.
Print the answer modulo (10^9 + 7)Sample input 1:
2
5
1 5 2 6 1
3
8 4 2
Sample output 1:
12
3
Explanation:
All increasing subsequence for test 1 are:
{1}
{5}
{2}
{6}
{1}
{1, 5}
{1, 2}
{1. 6}
{2, 6}
{5, 6}
{1, 5, 6}
{1, 2, 6}, I have written this Solution Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in)) ;
int test = Integer.parseInt(read.readLine()) ;
while(test-- > 0) {
int n = Integer.parseInt(read.readLine()) ;
String[] str = read.readLine().trim().split(" ") ;
long[] arr = new long[n] ;
for(int i=0; i<n; i++) {
arr[i] = Long.parseLong(str[i]) ;
}
long sum = 0L ;
long[] count = new long[n] ;
for(int i=0; i<n; i++) {
count[i] = 1L ;
for(int j=0; j<i; j++) {
if(arr[i] > arr[j]) {
count[i] = (count[i] + count[j]) % 1000000007 ;
}
}
sum = (sum + count[i]) % 1000000007;
}
System.out.println(sum);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given of array of N integers, find its number of increasing subsequences modulo (10^9 + 7)
A subsequence of the given array is the array formed after removing zero or more elements from it.
It is said to be increasing when every element is greater than its previous one.First line of input contains T, the number of test cases. Each test case is of two lines.
First line of each test case contains N, the number of elements of the array.
Next line contains N integers Ai, denoting elements of the array
1 <= T <= 50
1 <= N <= 2000
1 <= Ai <= 10^9Output T lines denoting the number of increasing subsequence for every test case.
Print the answer modulo (10^9 + 7)Sample input 1:
2
5
1 5 2 6 1
3
8 4 2
Sample output 1:
12
3
Explanation:
All increasing subsequence for test 1 are:
{1}
{5}
{2}
{6}
{1}
{1, 5}
{1, 2}
{1. 6}
{2, 6}
{5, 6}
{1, 5, 6}
{1, 2, 6}, I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e3+ 5;
const int mod = 1e9 + 7;
const int inf = 1e18 + 9;
int a[N], dp[N];
void solve(){
int n; cin >> n;
for(int i = 1; i <= n; i++){
cin >> a[i];
dp[i] = 0;
}
dp[0] = 1;
int ans = 0;
for(int i = 1; i <= n; i++){
for(int j = 0; j < i; j++){
if(a[i] > a[j])
(dp[i] += dp[j]) %= mod;
}
(ans += dp[i]) %= mod;
}
cout << ans << endl;
}
void testcases(){
int tt = 1;
cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N. You need to print the sum of the first N natural numbers.The first line of input contains a single integer T, the next T lines contains a single integer N.
Constraints:
1 < = T < = 100
1 < = N < = 100Print the sum of first N natural numbers for each test case in a new line.Sample Input:
2
5
10
Sample Output:
15
55, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
while(t-->0){
int n=Integer.parseInt(br.readLine());
System.out.println(n*(n+1)/2);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N. You need to print the sum of the first N natural numbers.The first line of input contains a single integer T, the next T lines contains a single integer N.
Constraints:
1 < = T < = 100
1 < = N < = 100Print the sum of first N natural numbers for each test case in a new line.Sample Input:
2
5
10
Sample Output:
15
55, I have written this Solution Code: for t in range(int(input())):
n = int(input())
print(n*(n+1)//2), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N. You need to print the sum of the first N natural numbers.The first line of input contains a single integer T, the next T lines contains a single integer N.
Constraints:
1 < = T < = 100
1 < = N < = 100Print the sum of first N natural numbers for each test case in a new line.Sample Input:
2
5
10
Sample Output:
15
55, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
cout<<(n*(n+1))/2<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: static int isPresent(long arr[], int n, long k)
{
int left = 0;
int right = n-1;
int res = -1;
while(left<=right){
int mid = (left+right)/2;
if(arr[mid] == k){
res = 1;
break;
}else if(arr[mid] < k){
left = mid + 1;
}else{
right = mid - 1;
}
}
return res;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
unordered_map<long long,int> m;
long k;
cin>>k;
long long a;
for(int i=0;i<n;i++){
cin>>a;
m[a]++;
}
if(m.find(k)!=m.end()){
cout<<1<<endl;
}
else{
cout<<-1<<endl;
}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: def binary_search(arr, low, high, x):
if high >= low:
mid = (high + low) // 2
if arr[mid] == x:
return 1
elif arr[mid] > x:
return binary_search(arr, low, mid - 1, x)
else:
return binary_search(arr, mid + 1, high, x)
else:
return -1
def position(n,arr,x):
return binary_search(arr,0,n-1,x)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: // arr is they array to search from
// x is target
function binSearch(arr, x) {
// write code here
// do not console.log
// return the 1 or -1
let l = 0;
let r = arr.length - 1;
let mid;
while (r >= l) {
mid = l + Math.floor((r - l) / 2);
// If the element is present at the middle
// itself
if (arr[mid] == x)
return 1;
// If element is smaller than mid, then
// it can only be present in left subarray
if (arr[mid] > x)
r = mid - 1;
// Else the element can only be present
// in right subarray
else
l = mid + 1;
}
// We reach here when element is not
// present in array
return -1;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. The task is to generate and print all binary numbers with decimal values from 1 to N.
Try solving this using a queue.The only line of input contains a single integer N.
1 <= N <= 50000Print all binary numbers with decimal values from 1 to N in a single line.Sample Input:
5
Sample Output:
1 10 11 100 101, I have written this Solution Code: def DecimalToBinary(num):
return "{0:b}".format(int(num))
n = int(input())
for i in range(1,n+1):
print(DecimalToBinary(i),end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. The task is to generate and print all binary numbers with decimal values from 1 to N.
Try solving this using a queue.The only line of input contains a single integer N.
1 <= N <= 50000Print all binary numbers with decimal values from 1 to N in a single line.Sample Input:
5
Sample Output:
1 10 11 100 101, I have written this Solution Code: import java.util.*;
import java.math.*;
import java.io.*;
class Main{
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
StringBuilder s=new StringBuilder("");
for(int i=1;i<=n;i++){
s.append(Integer.toBinaryString(i)+" ");
}
System.out.print(s);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. The task is to generate and print all binary numbers with decimal values from 1 to N.
Try solving this using a queue.The only line of input contains a single integer N.
1 <= N <= 50000Print all binary numbers with decimal values from 1 to N in a single line.Sample Input:
5
Sample Output:
1 10 11 100 101, I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 500 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
void solve(){
int n; cin >> n;
queue<string> q;
q.push("1");
while(n--){
string x = q.front();
q.pop();
cout << x << " ";
q.push(x + "0");
q.push(x + "1");
}
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given the following pseudocode:
code :
res = a
for i = 1 to k
if i is odd
res = res & b
else
res = res | b
You are also given the values of a, b and k. Find the value of res.First line contains of single integer t denoting number of test cases.
Each test cases consists of 3 lines where each line consists of a single integer denoting the values of a, b and k respectively.
<b>Constraints</b>
1<= T <= 1e5
1<= a, b, k <= 1e18Print the final value of res for each test case in a new lineSample Input :
1
4 5 1
Sample Output :
4, I have written this Solution Code: def solve(a,b,k):
res=a
for i in range(1,k):
if(i%2==1):
res= res & b
else:
res= res | b
return res
t=int(input())
for i in range(t):
a,b,k=list(map(int, input().split()))
if k==1:
print(a&b)
else:
print(b), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given the following pseudocode:
code :
res = a
for i = 1 to k
if i is odd
res = res & b
else
res = res | b
You are also given the values of a, b and k. Find the value of res.First line contains of single integer t denoting number of test cases.
Each test cases consists of 3 lines where each line consists of a single integer denoting the values of a, b and k respectively.
<b>Constraints</b>
1<= T <= 1e5
1<= a, b, k <= 1e18Print the final value of res for each test case in a new lineSample Input :
1
4 5 1
Sample Output :
4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int tt;
cin >> tt;
while (tt--) {
int a, b, k;
cin >> a >> b >> k;
cout << ((k == 1) ? a & b : b) << "\n";
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given the following pseudocode:
code :
res = a
for i = 1 to k
if i is odd
res = res & b
else
res = res | b
You are also given the values of a, b and k. Find the value of res.First line contains of single integer t denoting number of test cases.
Each test cases consists of 3 lines where each line consists of a single integer denoting the values of a, b and k respectively.
<b>Constraints</b>
1<= T <= 1e5
1<= a, b, k <= 1e18Print the final value of res for each test case in a new lineSample Input :
1
4 5 1
Sample Output :
4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
for(int j=0;j<t;j++){
String[] arr = br.readLine().split(" ");
long a = Long.parseLong(arr[0]);
long b = Long.parseLong(arr[1]);
long k = Long.parseLong(arr[2]);
System.out.println((k==1)?a&b:b);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While Solo was enjoying her daddy's party, her little sister Tono gave her the following problem to solve:
Given an array A of N integers, find the number of occurrences of the maximum integer in the array.
As Solo is small and enjoying the party, please help her solve this problem.The first line of the input contains an integer N, the size of the array A.
The next line contains N space separated integers, the elements of the array A.
Constraints
1 <= N <= 100
1 <= A[i] <= 100Output a single integer, the number of occurrences of the maximum integer in the array A.Sample Input
5
1 2 3 2 1
Sample Output
1
Explanation: The maximum integer is 3 and it occurs once in the array.
Sample Input
5
5 5 5 5 5
Sample Output
5, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
int[] arr=new int[n];
String[] s=br.readLine().trim().split(" ");
for(int i=0;i<arr.length;i++)
arr[i]=Integer.parseInt(s[i]);
int max=Integer.MIN_VALUE,c=0;
for(int i=0;i<arr.length;i++){
if(max==arr[i])
c++;
if(max<arr[i]){
max=arr[i];
c=1;
}
}
System.out.println(c);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While Solo was enjoying her daddy's party, her little sister Tono gave her the following problem to solve:
Given an array A of N integers, find the number of occurrences of the maximum integer in the array.
As Solo is small and enjoying the party, please help her solve this problem.The first line of the input contains an integer N, the size of the array A.
The next line contains N space separated integers, the elements of the array A.
Constraints
1 <= N <= 100
1 <= A[i] <= 100Output a single integer, the number of occurrences of the maximum integer in the array A.Sample Input
5
1 2 3 2 1
Sample Output
1
Explanation: The maximum integer is 3 and it occurs once in the array.
Sample Input
5
5 5 5 5 5
Sample Output
5, I have written this Solution Code: n=int(input())
a=list(map(int,input().split()))
print(a.count(max(a))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While Solo was enjoying her daddy's party, her little sister Tono gave her the following problem to solve:
Given an array A of N integers, find the number of occurrences of the maximum integer in the array.
As Solo is small and enjoying the party, please help her solve this problem.The first line of the input contains an integer N, the size of the array A.
The next line contains N space separated integers, the elements of the array A.
Constraints
1 <= N <= 100
1 <= A[i] <= 100Output a single integer, the number of occurrences of the maximum integer in the array A.Sample Input
5
1 2 3 2 1
Sample Output
1
Explanation: The maximum integer is 3 and it occurs once in the array.
Sample Input
5
5 5 5 5 5
Sample Output
5, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
vector<int> a(101, 0);
For(i, 0, n){
int x; cin>>x;
a[x]++;
}
for(int i=100; i>=1; i--){
if(a[i]){
cout<<a[i];
return;
}
}
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N (N is always even), you need to find exactly (N/2) numbers where each number represents the pair-wise sum of consecutive elements of the array A.
In simple terms print (A[1]+A[2]), (A[3]+A[4]), ..., (A[N-1]+A[N]).The first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N(even number) denoting the number of elements in the array. The next line contains N (white-space separated) integers.
Constraints
1 <= N <= 100
1 <= A[I] <= 1000000000For each test case, output N/2 elements representing the pairwise sum of adjacent elements in the array.Input:-1
4
1 2 6 4
output-1
3 10
input-2
10
1 2 3 4 5 6 0 7 8 9
output-2
3 7 11 7 17
Explanation(might now be the optimal solution):
Testcase 1:
Follow the below steps:-
Step 1: [1 2 6 4]
Step 2: (1 2) and (6 4)
Step 3: 3 10, I have written this Solution Code: n = int(input())
all_no = input().split(' ')
i = 0
joined_str = ''
while(i < n-1):
if(i == 0):
joined_str = str(int(all_no[i]) + int(all_no[i+1]))
else:
joined_str = joined_str + ' ' + str(int(all_no[i]) + int(all_no[i+1]))
i = i + 2
print(joined_str), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N (N is always even), you need to find exactly (N/2) numbers where each number represents the pair-wise sum of consecutive elements of the array A.
In simple terms print (A[1]+A[2]), (A[3]+A[4]), ..., (A[N-1]+A[N]).The first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N(even number) denoting the number of elements in the array. The next line contains N (white-space separated) integers.
Constraints
1 <= N <= 100
1 <= A[I] <= 1000000000For each test case, output N/2 elements representing the pairwise sum of adjacent elements in the array.Input:-1
4
1 2 6 4
output-1
3 10
input-2
10
1 2 3 4 5 6 0 7 8 9
output-2
3 7 11 7 17
Explanation(might now be the optimal solution):
Testcase 1:
Follow the below steps:-
Step 1: [1 2 6 4]
Step 2: (1 2) and (6 4)
Step 3: 3 10, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n=sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int t;
for(int i=0;i<n;i+=2){
System.out.print(a[i]+a[i+1]+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N (N is always even), you need to find exactly (N/2) numbers where each number represents the pair-wise sum of consecutive elements of the array A.
In simple terms print (A[1]+A[2]), (A[3]+A[4]), ..., (A[N-1]+A[N]).The first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N(even number) denoting the number of elements in the array. The next line contains N (white-space separated) integers.
Constraints
1 <= N <= 100
1 <= A[I] <= 1000000000For each test case, output N/2 elements representing the pairwise sum of adjacent elements in the array.Input:-1
4
1 2 6 4
output-1
3 10
input-2
10
1 2 3 4 5 6 0 7 8 9
output-2
3 7 11 7 17
Explanation(might now be the optimal solution):
Testcase 1:
Follow the below steps:-
Step 1: [1 2 6 4]
Step 2: (1 2) and (6 4)
Step 3: 3 10, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
long a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
for(int i=0;i<n;i+=2){
cout<<a[i]+a[i+1]<<" ";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a linked list consisting of N nodes and an integer K, your task is to reverse every alternate K nodes.
In other words , you have to reverse first k nodes , then skip the next k nodes , then reverse next k nodes and so on .
NOTE: if there are not K nodes to reverse then reverse all the nodes left (See example for better understanding)<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>ReverseAlternateK()</b> that takes head node and K as parameter.
Constraints:
1 <=k<=N<=10000Return the head of the modified linked list.Sample Input:-
8 3
1 2 3 4 5 6 7 8
Sample Output:-
3 2 1 4 5 6 8 7
Explanation:
[{1 , 2 ,3 } , {4, 5 , 6} , {7 , 8}]
Reverse 1st segment.
Skip the 2nd segment.
Reverse the 3rd segment. , I have written this Solution Code: public static Node ReverseAlternateK(Node head,int k){
Node current = head;
Node next = null, prev = null;
int count = 0;
while (current != null && count < k) {
next = current.next;
current.next = prev;
prev = current;
current = next;
count++;
}
if (head != null) {
head.next = current;
}
count = 0;
while (count < k - 1 && current != null) {
current = current.next;
count++;
}
if (current != null) {
current.next = ReverseAlternateK(current.next, k);
}
return prev;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find the number of prime numbers before N (Including that number too).The first line of the input contains the number of test cases T.
Next T lines contain the value N.
<b>Constraints</b>
1 <= T <= 1e5
1 <= N <= 1e5Print the number of primes number before that number.Sample Input 1:
2
3
11
Sample Output 1:
2
5, I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.StringTokenizer;
class Main
{
static int [] booleanArray(int num)
{
boolean [] bool = new boolean[num+1];
int [] count = new int [num+1];
bool[0] = true;
bool[1] = true;
for(int i=2; i*i<=num; i++)
{
if(bool[i]==false)
{
for(int j=i*i; j<=num; j+=i)
bool[j] = true;
}
}
int counter = 0;
for(int i=1; i<=num; i++)
{
if(bool[i]==false)
{
counter = counter+1;
count[i] = counter;
}
else
{
count[i] = counter;
}
}
return count;
}
public static void main (String[] args) throws IOException {
InputStreamReader ak = new InputStreamReader(System.in);
BufferedReader hk = new BufferedReader(ak);
int[] v = booleanArray(100000);
int t = Integer.parseInt(hk.readLine());
for (int i = 1; i <= t; i++) {
int a = Integer.parseInt(hk.readLine());
System.out.println(v[a]);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find the number of prime numbers before N (Including that number too).The first line of the input contains the number of test cases T.
Next T lines contain the value N.
<b>Constraints</b>
1 <= T <= 1e5
1 <= N <= 1e5Print the number of primes number before that number.Sample Input 1:
2
3
11
Sample Output 1:
2
5, I have written this Solution Code: m=100001
prime=[True for i in range(m)]
p=2
while(p*p<=m):
if prime[p]:
for i in range(p*p,m,p):
prime[i]=False
p+=1
c=[0 for i in range(m)]
c[2]=1
for i in range(3,m):
if prime[i]:
c[i]=c[i-1]+1
else:
c[i]=c[i-1]
t=int(input())
while t>0:
n=int(input())
print(c[n])
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find the number of prime numbers before N (Including that number too).The first line of the input contains the number of test cases T.
Next T lines contain the value N.
<b>Constraints</b>
1 <= T <= 1e5
1 <= N <= 1e5Print the number of primes number before that number.Sample Input 1:
2
3
11
Sample Output 1:
2
5, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
vector<bool> sieve(int n) {
vector<bool> is_prime(n + 1, true);
is_prime[0] = is_prime[1] = false;
for (int i = 2; i * i <= n; i++) {
if (is_prime[i]) {
for (int j = i * i; j <= n; j += i)
is_prime[j] = false;
}
}
return is_prime;
}
int main() {
vector<bool> prime = sieve(1e5 + 1);
vector<int> prefix(1e5 + 1, 0);
for (int i = 1; i <= 1e5; i++) {
if (prime[i]) {
prefix[i] = prefix[i - 1] + 1;
} else {
prefix[i] = prefix[i - 1];
}
}
int tt;
cin >> tt;
while (tt--) {
int n;
cin >> n;
cout << prefix[n] << "\n";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A graph is given with N nodes and M edges. The first visit to a vertex i will give 2<sup>i-1</sup> -1 score. Initially, we are at node 1. Then calculate the maximum score that we can get if we are only allowed to move to an adjacent vertex at most K times.First line contains 3 integers separated by space denoting N (1<=N<=60) , M (1<=M<=N*(N-1)/2) , and K (1<=K<=15), the number of nodes, edges, and the maximum move allowed respectively.
Then, M lines follow with each line containing two integers u and v representing edge between vertex u and v (u≠v).Output maximum score in first line.Input:
4 3 1
1 2
1 3
1 4
Output:
7
Explanation:
Alice can move to node 2 or node 3 or node 4. Here node 4 has (2^3 - 1 = 7)coins., I have written this Solution Code: //*** Author: ShlokG ***//
#include <bits/stdc++.h>
using namespace std;
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
typedef long long int ll;
typedef unsigned long long int ull;
#define pb push_back
#define endl "\n"
const ll MAX=1e18+5;
ll mod=1e9+7;
ll mod1=998244353;
const int infi = 1e9;
const ll INF=1e18;
void precompute(){
}
void TEST_CASE(){
int n,m,k;
cin >> n >> m >> k;
int mat[100][100];
int dp[1<<17][17];
vector<int> v;
for(int i=0 ; i<100 ; i++) for(int j=0 ; j<100 ; j++) mat[i][j]=1000;
for(int i=0 ; i<100 ; i++) mat[i][i]=0;
for(int i=0 ; i<m ; i++){
int x,y;
cin >> x >> y;
mat[x-1][y-1]=mat[y-1][x-1]=1;
}
for(int i=0 ; i<n ; i++) for(int j=0 ; j<n ; j++) for(int l=0 ; l<n ; l++) mat[j][l] = min (mat [j] [l], mat [j] [i] + mat [i] [l ]);
v.push_back (0);
ll ret = 0 ;
for(int i=n-1 ; i>0 ; i--) {
if(v.size()>=k+1) break;
v.push_back(i);
memset(dp,0x3f,sizeof(dp));
dp[1][0]=0;
for(int mask=1 ; mask<(1<<v.size()) ; mask++){
for(int x=0 ; x<v.size() ; x++) if(mask&(1<<x)) {
for(int y=0 ; y<v.size() ; y++) if(x!=y && (mask&(1<<y))) dp[mask][x] = min(dp[mask][x], dp[mask^(1<<x)][y] + mat[v[x]][v[y]]);
}
}
int ok = 0 ;
for(int j=0 ; j<v.size() ; j++) if(dp[(1<<v.size())-1][j]<=k) ok++;
if (ok == 0 ) v.pop_back();
else ret += ( 1LL << i) -1 ;
}
cout << ret << endl;
}
signed main(){
fast;
//freopen ("INPUT.txt","r",stdin);
//freopen ("OUTPUT.txt","w",stdout);
int test=1,TEST=1;
precompute();
//cin >> test;
while(test--){
//cout << "Case #" << TEST++ << ": ";
TEST_CASE();
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For an integer N, your task is to calculate sum of first N natural numbers.<b>User Task:</b>
Since this will be a functional problem, you don't have to worry about input. You just have to complete the function <b>sum()</b> which takes the integer N as a parameter.
Constraints:
1 <= N < = 100000000Print the sum of first N natural numbers.Sample Input:-
5
Sample Output:-
15
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code:
static void sum(int N){
long x=N;
x=x*(x+1);
x=x/2;
System.out.print(x);
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N pizza delivery people, they need to deliver pizzas to M customers in the town. The town is a straight line, and all the above N + M people are at distinct positions in it. The initial position of i-th pizza delivery person is P<sub>i</sub> and the initial position i-th person to whom pizza should be delivered is Q<sub>i</sub>.
Every second, every pizza delivery person can move one unit to either the left or right, or remain stationary. Multiple pizza delivery people are permitted to occupy the same position. When a pizza delivery person occupies the same position as a customer, they can spend 0 seconds to deliver the pizza to the person.
Find the minimum time (in seconds) required to deliver pizzas to all the customers.The first line of the input contains two integers N and M.
The second line of the input contains N integers, the initial positions of the pizza delivery people.
The third and the final line of input contains M integers, the initial positions of the customers.
Constraints
1 <= N <= 200000
1 <= M <= 200000
1 <= P<sub>i</sub>, Q<sub>i</sub> <= 500,000,000Output a single integer, the minimum time required to deliver all pizzas.Sample Input
2 2
10 20
13 16
Sample Output
4
Explanation: The 1st pizza delivery person delivers to 1st customer in 3 sec while the 2nd delivery person delivers the pizza to 2nd customer in 4 sec.
Sample Input
2 2
10 20
17 18
Sample Output
3
Explanation: The 2nd delivery person delivers to both customers.
Sample Input
3 3
20 60 30
25 10 35
Sample Output
15
Sample Input
2 3
20 80
10 21 81
Sample Output
12, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static int dboys,cusM;
static long[]posOfDevBoy,posOfCust;
public static void main (String[] args) {
FasterIo sc = new FasterIo();
dboys= sc.nextInt();
cusM= sc.nextInt();
posOfDevBoy = new long[dboys];
posOfCust = new long[cusM];
for(int i=0;i<dboys;i++){
posOfDevBoy[i] = sc.nextLong();
}
for(int i=0;i<cusM;i++){
posOfCust[i] = sc.nextLong();
}
sc.close();
Arrays.sort(posOfDevBoy);
Arrays.sort(posOfCust);
System.out.println(minTimeToDeliverPizzas());
}
public static long minTimeToDeliverPizzas(){
long low = 0,high = Long.MAX_VALUE,mid = 0;
long minTime = Long.MAX_VALUE;
while(low<=high){
mid = low+(high-low )/2;
if(canAllCustomerReachable(mid)){
minTime = mid;
high = mid-1;
}else{
low = mid+1;
}
}
return minTime;
}
public static boolean canAllCustomerReachable(long totalTime){
int k =0;
int remainingCus = cusM;
long dist=0;
for(int i=0;i<dboys;i++){
long timeForCustK = Math.abs(posOfDevBoy[i]-posOfCust[k]);
if(posOfCust[k]<posOfDevBoy[i])
{
if(totalTime<timeForCustK) return false;
dist = Math.max((totalTime-2 * timeForCustK),(totalTime-timeForCustK)/2);
}
else dist = totalTime;
while(posOfDevBoy[i]+dist >= posOfCust[k]){
if(k>=cusM-1){
return true;
}
k++;
}
}
return k>=cusM-1;
}
}
class FasterIo{
BufferedReader br;
StringTokenizer st;
public FasterIo()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt(){
return Integer.parseInt(next());
}
long nextLong(){
return Long.parseLong(next());
}
public void close(){
try{
br.close();
}
catch(IOException e){
e.printStackTrace();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N pizza delivery people, they need to deliver pizzas to M customers in the town. The town is a straight line, and all the above N + M people are at distinct positions in it. The initial position of i-th pizza delivery person is P<sub>i</sub> and the initial position i-th person to whom pizza should be delivered is Q<sub>i</sub>.
Every second, every pizza delivery person can move one unit to either the left or right, or remain stationary. Multiple pizza delivery people are permitted to occupy the same position. When a pizza delivery person occupies the same position as a customer, they can spend 0 seconds to deliver the pizza to the person.
Find the minimum time (in seconds) required to deliver pizzas to all the customers.The first line of the input contains two integers N and M.
The second line of the input contains N integers, the initial positions of the pizza delivery people.
The third and the final line of input contains M integers, the initial positions of the customers.
Constraints
1 <= N <= 200000
1 <= M <= 200000
1 <= P<sub>i</sub>, Q<sub>i</sub> <= 500,000,000Output a single integer, the minimum time required to deliver all pizzas.Sample Input
2 2
10 20
13 16
Sample Output
4
Explanation: The 1st pizza delivery person delivers to 1st customer in 3 sec while the 2nd delivery person delivers the pizza to 2nd customer in 4 sec.
Sample Input
2 2
10 20
17 18
Sample Output
3
Explanation: The 2nd delivery person delivers to both customers.
Sample Input
3 3
20 60 30
25 10 35
Sample Output
15
Sample Input
2 3
20 80
10 21 81
Sample Output
12, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
// #define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
int n, m, k, s;
vector<int> p, q;
bool ch(int x){
int j = 0;
For(i, 0, n){
int cur = x;
if(p[i] - q[j] > cur)
return false;
if(q[j] < p[i]){
int d = p[i]-q[j];
cur = max(cur-2*d, (cur-d)/2);
for(; j<m; j++){
if(q[j] > p[i])
break;
}
}
for(; j<m; j++){
int d = q[j]-p[i];
if(d > cur)
break;
}
if(j == m){
return true;
}
}
return false;
}
void solve(){
cin>>n>>m;
p.resize(n);
q.resize(m);
For(i, 0, n){
cin>>p[i];
}
For(i, 0, m){
cin>>q[i];
}
sort(all(p));
sort(all(q));
int low = 0;
int high = 1000000001;
while(high > low){
int mid = (low+high)>>1;
if(ch(mid))
high = mid;
else
low = mid + 1;
}
cout<<low;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
for(int i=1; i<=t; i++){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of size N. You have to print the maximum in every <b>K-sized subarray</b> from the left to the right in the array A.
More formally, you have to print N - K + 1 integers X<sub>1</sub>, X<sub>2</sub>, ..., X<sub>N-K+1</sub> such that X<sub>i</sub> (1 <= i <= N - K + 1) is the maximum element in the subarray A<sub>i</sub>, A<sub>i+1</sub>, ..., A<sub>i+K-1</sub>.1. The first line contains an integer N, denoting the size of the array.
2. The second line has N space-separated integers of the array A.
3. The third line contains integer K, denoting the size of the sliding window
<b>Constraints :</b>
1 ≤ N ≤ 10<sup>5</sup>
-10<sup>4</sup> ≤ A[i] ≤ 10<sup>4</sup>
1 ≤ K ≤ NPrint the max of K numbers for each position of sliding windowSample Input:-
8
1 3 -1 -3 5 3 6 7
3
Sample Output:-
3 3 5 5 6 7
Explanation:-
Window position Max
- - - -
[1 3 -1] -3 5 3 6 7 3
1 [3 -1 -3] 5 3 6 7 3
1 3 [-1 -3 5] 3 6 7 5
1 3 -1 [-3 5 3] 6 7 5
1 3 -1 -3 [5 3 6] 7 6
1 3 -1 -3 5 [3 6 7] 7
Sample Input:-
1
1
1
Sample Output:-
1 , I have written this Solution Code: import java.io.*;
import java.util.*;
class Main
{
public static void main(String args[])throws Exception
{
BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
StringBuilder sb=new StringBuilder();
int n=Integer.parseInt(bu.readLine());
String s[]=bu.readLine().split(" ");
int a[]=new int[n],i;
PriorityQueue<int[]> pq=new PriorityQueue<>(new Comparator<int[]>() {
@Override
public int compare(int[] o1, int[] o2) {
if(o1[0]<o2[0]) return 1;
else return -1;
}});
for(i=0;i<n;i++) a[i]=Integer.parseInt(s[i]);
int k=Integer.parseInt(bu.readLine());
for(i=0;i<k;i++) pq.add(new int[]{a[i],i});
sb.append(pq.peek()[0]+" ");
for(i=k;i<n;i++)
{
pq.add(new int[]{a[i],i});
while(!pq.isEmpty() && pq.peek()[1]<=i-k) pq.poll();
sb.append(pq.peek()[0]+" ");
}
System.out.println(sb);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of size N. You have to print the maximum in every <b>K-sized subarray</b> from the left to the right in the array A.
More formally, you have to print N - K + 1 integers X<sub>1</sub>, X<sub>2</sub>, ..., X<sub>N-K+1</sub> such that X<sub>i</sub> (1 <= i <= N - K + 1) is the maximum element in the subarray A<sub>i</sub>, A<sub>i+1</sub>, ..., A<sub>i+K-1</sub>.1. The first line contains an integer N, denoting the size of the array.
2. The second line has N space-separated integers of the array A.
3. The third line contains integer K, denoting the size of the sliding window
<b>Constraints :</b>
1 ≤ N ≤ 10<sup>5</sup>
-10<sup>4</sup> ≤ A[i] ≤ 10<sup>4</sup>
1 ≤ K ≤ NPrint the max of K numbers for each position of sliding windowSample Input:-
8
1 3 -1 -3 5 3 6 7
3
Sample Output:-
3 3 5 5 6 7
Explanation:-
Window position Max
- - - -
[1 3 -1] -3 5 3 6 7 3
1 [3 -1 -3] 5 3 6 7 3
1 3 [-1 -3 5] 3 6 7 5
1 3 -1 [-3 5 3] 6 7 5
1 3 -1 -3 [5 3 6] 7 6
1 3 -1 -3 5 [3 6 7] 7
Sample Input:-
1
1
1
Sample Output:-
1 , I have written this Solution Code: N = int(input())
A = [int(x) for x in input().split()]
K = int(input())
def print_max(a, n, k):
max_upto = [0 for i in range(n)]
s = []
s.append(0)
for i in range(1, n):
while (len(s) > 0 and a[s[-1]] < a[i]):
max_upto[s[-1]] = i - 1
del s[-1]
s.append(i)
while (len(s) > 0):
max_upto[s[-1]] = n - 1
del s[-1]
j = 0
for i in range(n - k + 1):
while (j < i or max_upto[j] < i + k - 1):
j += 1
print(a[j], end=" ")
print()
print_max(A, N, K), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of size N. You have to print the maximum in every <b>K-sized subarray</b> from the left to the right in the array A.
More formally, you have to print N - K + 1 integers X<sub>1</sub>, X<sub>2</sub>, ..., X<sub>N-K+1</sub> such that X<sub>i</sub> (1 <= i <= N - K + 1) is the maximum element in the subarray A<sub>i</sub>, A<sub>i+1</sub>, ..., A<sub>i+K-1</sub>.1. The first line contains an integer N, denoting the size of the array.
2. The second line has N space-separated integers of the array A.
3. The third line contains integer K, denoting the size of the sliding window
<b>Constraints :</b>
1 ≤ N ≤ 10<sup>5</sup>
-10<sup>4</sup> ≤ A[i] ≤ 10<sup>4</sup>
1 ≤ K ≤ NPrint the max of K numbers for each position of sliding windowSample Input:-
8
1 3 -1 -3 5 3 6 7
3
Sample Output:-
3 3 5 5 6 7
Explanation:-
Window position Max
- - - -
[1 3 -1] -3 5 3 6 7 3
1 [3 -1 -3] 5 3 6 7 3
1 3 [-1 -3 5] 3 6 7 5
1 3 -1 [-3 5 3] 6 7 5
1 3 -1 -3 [5 3 6] 7 6
1 3 -1 -3 5 [3 6 7] 7
Sample Input:-
1
1
1
Sample Output:-
1 , I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
vector<int> maxSwindow(vector<int>& arr, int k) {
vector<int> result;
deque<int> Q(k); // store the indices
// process the first window
for(int i = 0; i < k; i++) {
while(!Q.empty() and arr[i] >= arr[Q.back()])
Q.pop_back();
Q.push_back(i);
}
// process the remaining elements
for(int i = k; i < arr.size(); i++) {
// add the max of the current window
result.push_back(arr[Q.front()]);
// remove the elements going out of the window
while(!Q.empty() and Q.front() <= i - k)
Q.pop_front();
// remove the useless elements
while(!Q.empty() and arr[i] >= arr[Q.back()])
Q.pop_back();
// add the current element in the deque
Q.push_back(i);
}
result.push_back(arr[Q.front()]);
return result;
}
int main()
{
int k, n, m;
cin >> n;
vector<int> nums, res;
for(int i =0; i < n; i++){
cin >> m;
nums.push_back(m);
}
cin >> k;
res = maxSwindow(nums,k);
for(auto i = res.begin(); i!=res.end(); i++)
cout << *i << " ";
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono loves to do shopping. Today, she went to the market where there are N different types of products. She wants to buy exactly K of them at the minimum cost. Although she is super smart, she wants to check your smartness as well. Can you tell her the minimum cost required to buy exactly K products if she has already decided to buy product J?
<b>Note:</b> Tono does not buy the same product twice, and Tono will definitely buy product J (J is the <b>index</b> of the item).The first line of the input contains three integers, N, K, and J, denoting the number of products in the market, the number of products Tono needs to buy, and the product that Tono will definitely buy.
The next line contains N singly spaced integers, the cost of the N products C[1], C[2], ..., C[N].
<b>Constraints:</b>
1 <= N <= 200000
1 <= K <= N
1 <= J <= N
1 <= C[i] <= 1000
Output a single integer, the minimum amount Tono needs to pay.Sample Input 1:
5 3 4
1 2 3 4 5
Sample Output 1:
7
Sample Input 2:
5 1 3
2 4 3 1 1
Sample Output 2:
3
<b>Explanation:</b>
Tono needs to buy exactly 3 products, and she will definitely buy the 4th product. Thus, she will buy the 1st, 2nd, and the 4th product. The total cost she pays is 1+2+4=7.
, I have written this Solution Code: import java.util.*;
import java.io.*;
class Main {
public static void main(String[] args) throws IOException {
int n = io.nextInt(), k = io.nextInt(), j = io.nextInt() - 1;
int[] arr = new int[n];
for(int i = 0; i < n; i++) {
arr[i] = io.nextInt();
}
int cost = arr[j];
arr[j] = Integer.MAX_VALUE;
Arrays.sort(arr);
for(int i = 0; i < k - 1; i++) {
cost += arr[i];
}
io.println(cost);
io.close();
}
static IO io = new IO();
static class IO {
private byte[] buf;
private InputStream in;
private PrintWriter pw;
private int total, index;
public IO() {
buf = new byte[1024];
in = System.in;
pw = new PrintWriter(System.out);
}
public int next() throws IOException {
if(total < 0)
throw new InputMismatchException();
if(index >= total) {
index = 0;
total = in.read(buf);
if(total <= 0)
return -1;
}
return buf[index++];
}
public int nextInt() throws IOException {
int n = next(), integer = 0;
while(isWhiteSpace(n))
n = next();
int neg = 1;
if(n == '-') {
neg = -1;
n = next();
}
while(!isWhiteSpace(n)) {
if(n >= '0' && n <= '9') {
integer *= 10;
integer += n - '0';
n = next();
}
else
throw new InputMismatchException();
}
return neg * integer;
}
public int[] nextIntArray(int n) throws IOException {
int[] arr = new int[n];
for(int i = 0; i < n; i++)
arr[i] = nextInt();
return arr;
}
public long nextLong() throws IOException {
long integer = 0l;
int n = next();
while(isWhiteSpace(n))
n = next();
int neg = 1;
if(n == '-') {
neg = -1;
n = next();
}
while(!isWhiteSpace(n)) {
if(n >= '0' && n <= '9') {
integer *= 10;
integer += n - '0';
n = next();
}
else
throw new InputMismatchException();
}
return neg * integer;
}
public double nextDouble() throws IOException {
double doub = 0;
int n = next();
while(isWhiteSpace(n))
n = next();
int neg = 1;
if(n == '-') {
neg = -1;
n = next();
}
while(!isWhiteSpace(n) && n != '.') {
if(n >= '0' && n <= '9') {
doub *= 10;
doub += n - '0';
n = next();
}
else
throw new InputMismatchException();
}
if(n == '.') {
n = next();
double temp = 1;
while(!isWhiteSpace(n)) {
if(n >= '0' && n <= '9') {
temp /= 10;
doub += (n - '0') * temp;
n = next();
}
else
throw new InputMismatchException();
}
}
return doub * neg;
}
public String nextString() throws IOException {
StringBuilder sb = new StringBuilder();
int n = next();
while(isWhiteSpace(n))
n = next();
while(!isWhiteSpace(n)) {
sb.append((char)n);
n = next();
}
return sb.toString();
}
public String nextLine() throws IOException {
int n = next();
while(isWhiteSpace(n))
n = next();
StringBuilder sb = new StringBuilder();
while(!isEndOfLine(n)) {
sb.append((char)n);
n = next();
}
return sb.toString();
}
private boolean isWhiteSpace(int n) {
return n == ' ' || n == '\n' || n == '\r' || n == '\t' || n == -1;
}
private boolean isEndOfLine(int n) {
return n == '\n' || n == '\r' || n == -1;
}
public void print(Object obj) {
pw.print(obj);
}
public void println(Object... obj) {
if(obj.length == 1)
pw.println(obj[0]);
else {
for(Object o: obj)
pw.print(o + " ");
pw.println();
}
}
public void flush() throws IOException {
pw.flush();
}
public void close() throws IOException {
pw.close();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono loves to do shopping. Today, she went to the market where there are N different types of products. She wants to buy exactly K of them at the minimum cost. Although she is super smart, she wants to check your smartness as well. Can you tell her the minimum cost required to buy exactly K products if she has already decided to buy product J?
<b>Note:</b> Tono does not buy the same product twice, and Tono will definitely buy product J (J is the <b>index</b> of the item).The first line of the input contains three integers, N, K, and J, denoting the number of products in the market, the number of products Tono needs to buy, and the product that Tono will definitely buy.
The next line contains N singly spaced integers, the cost of the N products C[1], C[2], ..., C[N].
<b>Constraints:</b>
1 <= N <= 200000
1 <= K <= N
1 <= J <= N
1 <= C[i] <= 1000
Output a single integer, the minimum amount Tono needs to pay.Sample Input 1:
5 3 4
1 2 3 4 5
Sample Output 1:
7
Sample Input 2:
5 1 3
2 4 3 1 1
Sample Output 2:
3
<b>Explanation:</b>
Tono needs to buy exactly 3 products, and she will definitely buy the 4th product. Thus, she will buy the 1st, 2nd, and the 4th product. The total cost she pays is 1+2+4=7.
, I have written this Solution Code:
a=input().split()
b=input().split()
for j in [a,b]:
for i in range(0,len(j)):
j[i]=int(j[i])
n,k,j=a[0],a[1],a[2]
c_j=b[j-1]
b.sort()
if b[k-1]<=c_j:
b[k-1]=c_j
sum=0
for i in range(0,k):
sum+=b[i]
print(sum), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono loves to do shopping. Today, she went to the market where there are N different types of products. She wants to buy exactly K of them at the minimum cost. Although she is super smart, she wants to check your smartness as well. Can you tell her the minimum cost required to buy exactly K products if she has already decided to buy product J?
<b>Note:</b> Tono does not buy the same product twice, and Tono will definitely buy product J (J is the <b>index</b> of the item).The first line of the input contains three integers, N, K, and J, denoting the number of products in the market, the number of products Tono needs to buy, and the product that Tono will definitely buy.
The next line contains N singly spaced integers, the cost of the N products C[1], C[2], ..., C[N].
<b>Constraints:</b>
1 <= N <= 200000
1 <= K <= N
1 <= J <= N
1 <= C[i] <= 1000
Output a single integer, the minimum amount Tono needs to pay.Sample Input 1:
5 3 4
1 2 3 4 5
Sample Output 1:
7
Sample Input 2:
5 1 3
2 4 3 1 1
Sample Output 2:
3
<b>Explanation:</b>
Tono needs to buy exactly 3 products, and she will definitely buy the 4th product. Thus, she will buy the 1st, 2nd, and the 4th product. The total cost she pays is 1+2+4=7.
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n, k, j; cin>>n>>k>>j;
vector<int> vect;
int ans = 0;
For(i, 1, n+1){
int a; cin>>a;
if(i!=j)
vect.pb(a);
else
ans += a;
}
sort(all(vect));
for(int i=0; i<k-1; i++){
ans += vect[i];
}
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a tree consisting of N nodes (from 1 to N) rooted at 1. You are given a node, your task is to calculate the sum of values of all nodes present at the same level that is of the given node.First line of input contains two space separated integers denoting the number of nodes N and the given node P. Next N-1 lines contains two space separated integers U and V, denoting an edge of the tree.
Last line of input contains N space separated integers denoting value of each node.
Constraints:-
2 <= N <= 10<sup>4</sup>
1 <= P <= N
1 <= U, V <= N
1 <= value[i] <= 10<sup>5</sup>Print the sum of all the nodes present at the same level as the node P.Sample Input:-
5 4
1 2
1 3
3 4
3 5
2 4 8 16 32
Sample Output:-
48
Explanation:-
1
/ \
2 3
/ \
4 5
Nodes present at same level as node 5 are:- 4 and 5, I have written this Solution Code: import java.io.*;
import java.util.*;
class Edge{
int s;
int d;
Edge(int s,int d){
this.s = s;
this.d = d;
}
}
class Tree{
LinkedList<Integer> node = new LinkedList<>();
}
class Main {
public static void print(Tree tree[]){
int i,n;
n = tree.length;
for(i=1;i<n;i++){
LinkedList mlist = tree[i].node;
System.out.print(i + " : ");
for(Object val : mlist)
System.out.print(val + " ");
System.out.println();
}
}
public static int calSum(Tree tree[],int data[],int p){
int i;
int n=data.length;
if(n == 0)
return 0;
if(n == 1)
return data[1];
int level[] = new int[n];
int visit[] = new int[n];
Queue<Integer> queue = new LinkedList<>();
queue.add(1);
visit[1] = 1;
while(queue.size() > 0){
int val = queue.peek();
queue.remove();
int size = tree[val].node.size();
for(i=0;i<size;i++){
int d = tree[val].node.get(i);
if(visit[d] == 0){
level[d] = level[val] + 1;
queue.add(d);
visit[d] = 1;
}
}
}
int sum = 0;
for(i=1;i<n;i++){
if(level[i] == level[p])
sum+=data[i];
}
return sum;
}
public static void main (String[] args) {
int i,n,p,s,d;
Scanner sc = new Scanner(System.in);
n = sc.nextInt();
p = sc.nextInt();
Tree tree[] = new Tree[n+1];
Edge edge[] = new Edge[n];
int data[] = new int[n+1];
for(i=1;i<=n;i++)
tree[i] = new Tree();
for(i=1;i<n;i++){
s = sc.nextInt();
d = sc.nextInt();
edge[i] = new Edge(s,d);
tree[s].node.add(d);
tree[d].node.add(s);
}
for(i=1;i<=n;i++)
data[i] = sc.nextInt();
int ans = calSum(tree,data,p);
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a tree consisting of N nodes (from 1 to N) rooted at 1. You are given a node, your task is to calculate the sum of values of all nodes present at the same level that is of the given node.First line of input contains two space separated integers denoting the number of nodes N and the given node P. Next N-1 lines contains two space separated integers U and V, denoting an edge of the tree.
Last line of input contains N space separated integers denoting value of each node.
Constraints:-
2 <= N <= 10<sup>4</sup>
1 <= P <= N
1 <= U, V <= N
1 <= value[i] <= 10<sup>5</sup>Print the sum of all the nodes present at the same level as the node P.Sample Input:-
5 4
1 2
1 3
3 4
3 5
2 4 8 16 32
Sample Output:-
48
Explanation:-
1
/ \
2 3
/ \
4 5
Nodes present at same level as node 5 are:- 4 and 5, I have written this Solution Code: from collections import deque
N,node = map(int,input().strip().split())
dic = {}
for i in range(N-1):
n1,n2 = map(int,input().strip().split())
if dic.get(n1):
dic[n1].append(n2)
else:
dic[n1] = [n2]
if dic.get(n2):
dic[n2].append(n1)
else:
dic[n2] = [n1]
arr = [int(i) for i in input().strip().split()]
visited = set()
queue = deque()
queue.append(1)
def find_sum(node,queue):
if node==1:
return arr[0]
flag = False
while(queue):
length = len(queue)
while(length):
nodes = queue.popleft()
visited.add(nodes)
if dic.get(nodes):
for i in dic[nodes]:
if i not in visited:
queue.append(i)
if i==node:
flag = True
length -= 1
#print(queue)
if flag:
sumi = 0
for i in queue:
sumi += arr[i-1]
return sumi
print(find_sum(node,queue)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a tree consisting of N nodes (from 1 to N) rooted at 1. You are given a node, your task is to calculate the sum of values of all nodes present at the same level that is of the given node.First line of input contains two space separated integers denoting the number of nodes N and the given node P. Next N-1 lines contains two space separated integers U and V, denoting an edge of the tree.
Last line of input contains N space separated integers denoting value of each node.
Constraints:-
2 <= N <= 10<sup>4</sup>
1 <= P <= N
1 <= U, V <= N
1 <= value[i] <= 10<sup>5</sup>Print the sum of all the nodes present at the same level as the node P.Sample Input:-
5 4
1 2
1 3
3 4
3 5
2 4 8 16 32
Sample Output:-
48
Explanation:-
1
/ \
2 3
/ \
4 5
Nodes present at same level as node 5 are:- 4 and 5, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1000001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
signed main()
{
int n;
cin>>n;
vector<int> v[n];
int P;
cin>>P;
int a,b;
for(int i=0;i<n-1;i++){
cin>>a>>b;
a--;b--;
v[a].EB(b);
v[b].EB(a);
}
P--;
int ans[n];
FOR(i,n){
cin>>ans[i];
}
queue<int> q,p;
q.push(0);
bool check[n];
FOR(i,n){check[i]=false;}
int sum=0;
check[0]=true;
if(P==0){
out(ans[0]);
return 0;
}
while(q.size()!=0){
bool win=false;
int s=q.size();
for(int i=0;i<s;i++){
a=q.front();
//out(a);
for(int j=0;j<v[a].size();j++){
if(check[v[a][j]]==false){
if(v[a][j]==P){win=true;}
p.push(v[a][j]);
check[v[a][j]]=true;
}
}
q.pop();
}
while(p.size()!=0){
q.push(p.front());
p.pop();
}
if(win){
a=q.size();
//out(a);
FOR(i,a){
sum+=ans[q.front()];
q.pop();
}
break;
}
}
out(sum);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joffrey has issued orders for Stark's execution. Now, you are trying to predict if Stark will survive or not. You are confident that Stark's survival depends on the N<sup>th</sup> fibonacci number's parity. If it is odd you will predict that Stark lives but if it is even you will predict that Stark dies. Given N, print your prediction.
Fibonacci series is a series where,
Fibonacci(1) = 0
Fibonacci(2) = 1
Fibonacci(i) = Fibonacci(i-1) + Fibonacci(i-2); for i > 2Input Contains a single integer N.
Constraints:
1 <= N <= 1000000Print "Alive" if Nth Fibonacci number is odd and print "Dead" if Nth Fibonacci number is even.Sample Input 1
3
Sample Output 1
Alive
Explanation: Fibonacci(3) = 1 which is odd.
Sample Input 2
4
Sample Output 1
Dead
Explanation: Fibonacci(4) = 2 which is even., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
InputStreamReader is = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(is);
int n=Integer.parseInt(br.readLine().trim());
if(n%3==1)
System.out.println("Dead");
else
System.out.println("Alive");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joffrey has issued orders for Stark's execution. Now, you are trying to predict if Stark will survive or not. You are confident that Stark's survival depends on the N<sup>th</sup> fibonacci number's parity. If it is odd you will predict that Stark lives but if it is even you will predict that Stark dies. Given N, print your prediction.
Fibonacci series is a series where,
Fibonacci(1) = 0
Fibonacci(2) = 1
Fibonacci(i) = Fibonacci(i-1) + Fibonacci(i-2); for i > 2Input Contains a single integer N.
Constraints:
1 <= N <= 1000000Print "Alive" if Nth Fibonacci number is odd and print "Dead" if Nth Fibonacci number is even.Sample Input 1
3
Sample Output 1
Alive
Explanation: Fibonacci(3) = 1 which is odd.
Sample Input 2
4
Sample Output 1
Dead
Explanation: Fibonacci(4) = 2 which is even., I have written this Solution Code: n = int(input().strip())
if n%3 == 1:
print("Dead")
else:
print("Alive"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joffrey has issued orders for Stark's execution. Now, you are trying to predict if Stark will survive or not. You are confident that Stark's survival depends on the N<sup>th</sup> fibonacci number's parity. If it is odd you will predict that Stark lives but if it is even you will predict that Stark dies. Given N, print your prediction.
Fibonacci series is a series where,
Fibonacci(1) = 0
Fibonacci(2) = 1
Fibonacci(i) = Fibonacci(i-1) + Fibonacci(i-2); for i > 2Input Contains a single integer N.
Constraints:
1 <= N <= 1000000Print "Alive" if Nth Fibonacci number is odd and print "Dead" if Nth Fibonacci number is even.Sample Input 1
3
Sample Output 1
Alive
Explanation: Fibonacci(3) = 1 which is odd.
Sample Input 2
4
Sample Output 1
Dead
Explanation: Fibonacci(4) = 2 which is even., I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
if(n%3==1)
cout<<"Dead";
else
cout<<"Alive";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array <strong>arr[]</strong> of size <strong>N</strong> and an element <strong>X</strong>. The task is to find and print the indices of the given element if it is present in array if not then print “<strong>Not found</strong>” without quotes.
<strong>Note</strong>: The elements may be present more than once.The first line of input contains T, denoting the number of test cases.
The first line of each test case contains N and X, N is the size of array and X is an element. Second line contains elements of array space separated. If not present then print "Not found" without quotes
<strong>Constraints</strong>:
1 <= T <= 100
1 <= N, X <= 10000
1 <= arr[i] <= 100000For each test case in new line you need to print all the positions where you find the X separated by space.
Assume 0-indexingInput:
2
5 6
2 3 6 5 6
4 3
2 4 6 5
Output:
2 4
Not found, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int t; cin >> t;
while(t--){
vector<int> v;
int n, x; cin >> n >> x;
for(int i = 1; i <= n; i++){
int p; cin >> p;
if(p == x)
v.push_back(i-1);
}
if(v.size() == 0)
cout << "Not found\n";
else{
for(auto i: v)
cout << i << " ";
cout << endl;
}
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array <strong>arr[]</strong> of size <strong>N</strong> and an element <strong>X</strong>. The task is to find and print the indices of the given element if it is present in array if not then print “<strong>Not found</strong>” without quotes.
<strong>Note</strong>: The elements may be present more than once.The first line of input contains T, denoting the number of test cases.
The first line of each test case contains N and X, N is the size of array and X is an element. Second line contains elements of array space separated. If not present then print "Not found" without quotes
<strong>Constraints</strong>:
1 <= T <= 100
1 <= N, X <= 10000
1 <= arr[i] <= 100000For each test case in new line you need to print all the positions where you find the X separated by space.
Assume 0-indexingInput:
2
5 6
2 3 6 5 6
4 3
2 4 6 5
Output:
2 4
Not found, I have written this Solution Code: def position(n,arr,x):
res = []
cnt = 0
for i in arr:
if(i == x):
res.append(cnt)
cnt += 1
return res
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array <strong>arr[]</strong> of size <strong>N</strong> and an element <strong>X</strong>. The task is to find and print the indices of the given element if it is present in array if not then print “<strong>Not found</strong>” without quotes.
<strong>Note</strong>: The elements may be present more than once.The first line of input contains T, denoting the number of test cases.
The first line of each test case contains N and X, N is the size of array and X is an element. Second line contains elements of array space separated. If not present then print "Not found" without quotes
<strong>Constraints</strong>:
1 <= T <= 100
1 <= N, X <= 10000
1 <= arr[i] <= 100000For each test case in new line you need to print all the positions where you find the X separated by space.
Assume 0-indexingInput:
2
5 6
2 3 6 5 6
4 3
2 4 6 5
Output:
2 4
Not found, I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main
{
public static void main (String[] args)throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t =Integer.parseInt(read.readLine());
while(t-- > 0)
{
String str[] = read.readLine().trim().split(" ");
int n = Integer.parseInt(str[0]);
int x = Integer.parseInt(str[1]);
int arr[] = new int[n];
str = read.readLine().trim().split(" ");
for(int i = 0; i < n; i++)
arr[i] = Integer.parseInt(str[i]);
findPositions(arr, n, x);
}
}
static void findPositions(int arr[], int n, int x)
{
boolean flag = false;
StringBuffer sb = new StringBuffer();
for(int i = 0; i < n; i++)
{
if(arr[i] == x)
{
sb.append(i + " ");
flag = true;
}
}
if(flag ==true)
System.out.println(sb.toString());
else System.out.println("Not found");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Insertion is a basic but frequently used operation. Arrays in most languages cannnot be dynamically shrinked or expanded. Here, we will work with such arrays and try to insert an element at the end of array.
You are given an array arr. The size of the array is N. You need to insert an element at its end and print this newly modified array.The first line of input contains T denoting the number of testcases.
T testcases follow. Each testcase contains two lines of input.
The first line contains size of the array denoted by N and element to be inserted.
The third line contains N elements separated by spaces.
Constraints:
1 <= T <= 20
2 <= N <= 10000
0 <= element, arri <= 10^6For each testcase, in a new line, print the modified array.Input:
2
5 90
1 2 3 4 5
3 50
1 2 3
Output:
1 2 3 4 5 90
1 2 3 50
Explanation:
Testcase 1: After inserting 90 at end, we have array elements as 1 2 3 4 5 90.
Testcase 2: After inserting 50 at end, we have array elements as 1 2 3 50., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main() {
int t;
cin>>t;
while(t--) {
long int n, el;
cin>>n>>el;
long int arr[n+1];
for(long int i=0; i<n; i++) {
cin>>arr[i];
}
arr[n] = el;
for(long int i=0; i<=n; i++) {
cout<<arr[i];
if (i != n) {
cout<<" ";
}
else if(t != 0) {
cout<<endl;
}
}
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Insertion is a basic but frequently used operation. Arrays in most languages cannnot be dynamically shrinked or expanded. Here, we will work with such arrays and try to insert an element at the end of array.
You are given an array arr. The size of the array is N. You need to insert an element at its end and print this newly modified array.The first line of input contains T denoting the number of testcases.
T testcases follow. Each testcase contains two lines of input.
The first line contains size of the array denoted by N and element to be inserted.
The third line contains N elements separated by spaces.
Constraints:
1 <= T <= 20
2 <= N <= 10000
0 <= element, arri <= 10^6For each testcase, in a new line, print the modified array.Input:
2
5 90
1 2 3 4 5
3 50
1 2 3
Output:
1 2 3 4 5 90
1 2 3 50
Explanation:
Testcase 1: After inserting 90 at end, we have array elements as 1 2 3 4 5 90.
Testcase 2: After inserting 50 at end, we have array elements as 1 2 3 50., I have written this Solution Code: t=int(input())
while t>0:
t-=1
li = list(map(int,input().strip().split()))
n=li[0]
num=li[1]
a= list(map(int,input().strip().split()))
a.insert(len(a),num)
for i in a:
print(i,end=" ")
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Insertion is a basic but frequently used operation. Arrays in most languages cannnot be dynamically shrinked or expanded. Here, we will work with such arrays and try to insert an element at the end of array.
You are given an array arr. The size of the array is N. You need to insert an element at its end and print this newly modified array.The first line of input contains T denoting the number of testcases.
T testcases follow. Each testcase contains two lines of input.
The first line contains size of the array denoted by N and element to be inserted.
The third line contains N elements separated by spaces.
Constraints:
1 <= T <= 20
2 <= N <= 10000
0 <= element, arri <= 10^6For each testcase, in a new line, print the modified array.Input:
2
5 90
1 2 3 4 5
3 50
1 2 3
Output:
1 2 3 4 5 90
1 2 3 50
Explanation:
Testcase 1: After inserting 90 at end, we have array elements as 1 2 3 4 5 90.
Testcase 2: After inserting 50 at end, we have array elements as 1 2 3 50., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n =Integer.parseInt(br.readLine().trim());
while(n-->0){
String str[]=br.readLine().trim().split(" ");
String newel =br.readLine().trim()+" "+str[1];
System.out.println(newel);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For an integer N, your task is to calculate sum of first N natural numbers.<b>User Task:</b>
Since this will be a functional problem, you don't have to worry about input. You just have to complete the function <b>sum()</b> which takes the integer N as a parameter.
Constraints:
1 <= N < = 100000000Print the sum of first N natural numbers.Sample Input:-
5
Sample Output:-
15
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code:
static void sum(int N){
long x=N;
x=x*(x+1);
x=x/2;
System.out.print(x);
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The flower shop near his house sells flowers of N types. It is found that the store has Ai flowers of the i-th type. We like odd numbers. Therefore, we have decided that there should be an odd number of flowers of each type in the bouquet, and the total number of flowers in the bouquet should also be odd.
Determine the maximum number of flowers the bouquet can consist of.The first line contains an integer N — the number of types of flowers that are sold in the store
The second line contains N integers— the number of flowers of each type
1 <= N <= 100000
1 <= Ai <= 1000Print one number — the maximum number of flowers the bouquet can consist of.Sample input
3
3 5 8
Sample output
15
Sample input
3
1 1 1
Sample output
3, I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in)) ;
int n = Integer.parseInt(read.readLine()) ;
String[] str = read.readLine().trim().split(" ") ;
int[] arr = new int[n] ;
for(int i=0; i<n; i++) {
arr[i] = Integer.parseInt(str[i]) ;
}
long ans = 0L;
if((n & 1) == 1) {
for(int i=0; i<n; i++) {
ans += ((arr[i] & 1) == 1) ? arr[i] : arr[i] - 1 ;
}
}
else {
Arrays.sort(arr);
for(int i=1; i<n; i++) {
ans += ((arr[i] & 1) == 1) ? arr[i] : arr[i] - 1 ;
}
}
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The flower shop near his house sells flowers of N types. It is found that the store has Ai flowers of the i-th type. We like odd numbers. Therefore, we have decided that there should be an odd number of flowers of each type in the bouquet, and the total number of flowers in the bouquet should also be odd.
Determine the maximum number of flowers the bouquet can consist of.The first line contains an integer N — the number of types of flowers that are sold in the store
The second line contains N integers— the number of flowers of each type
1 <= N <= 100000
1 <= Ai <= 1000Print one number — the maximum number of flowers the bouquet can consist of.Sample input
3
3 5 8
Sample output
15
Sample input
3
1 1 1
Sample output
3, I have written this Solution Code: n=int(input())
arr=map(int,input().split())
result=[]
for item in arr:
if item%2==0:
result.append(item-1)
else:
result.append(item)
if sum(result)%2==0:
result.sort()
result.pop(0)
print(sum(result))
else:
print(sum(result)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The flower shop near his house sells flowers of N types. It is found that the store has Ai flowers of the i-th type. We like odd numbers. Therefore, we have decided that there should be an odd number of flowers of each type in the bouquet, and the total number of flowers in the bouquet should also be odd.
Determine the maximum number of flowers the bouquet can consist of.The first line contains an integer N — the number of types of flowers that are sold in the store
The second line contains N integers— the number of flowers of each type
1 <= N <= 100000
1 <= Ai <= 1000Print one number — the maximum number of flowers the bouquet can consist of.Sample input
3
3 5 8
Sample output
15
Sample input
3
1 1 1
Sample output
3, I have written this Solution Code: #include <bits/stdc++.h>
#define ll long long int
using namespace std;
int main(){
int n; cin >> n ;
long long int ans = 0;
// int arr[n];
int odd = 0,minn=INT_MAX;
for(int i=0;i<n;i++){
ll temp; cin >> temp;
// cin >> arr[i];
ans += (temp&1) ? temp : temp-1;
if(minn>temp)
minn = temp;
}
if(n&1)
cout << ans ;
else{
if(minn&1)
cout <<ans-minn;
else
cout << ans - minn + 1;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code:
int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code:
int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code: static int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code: def mythOrFact(N):
prime=1
cnt=N+1
for i in range (2,N):
if N%i==0:
prime=0
cnt=cnt+i
x = 3*N
x=x//2
if(cnt <= x and prime==1) or (cnt>x and prime==0):
return 1
return 0
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a circular linked list consisting of N nodes and an integer K, your task is to add the integer K at the end of the list.
<b>Note:
Sample Input and Output just show how a linked list will look depending on the questions. Do not copy-paste as it is in custom input</b><b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Insertion()</b> that takes head node of circular linked list and the integer K as parameter.
Constraints:
1 <=N <= 1000
1 <= Node.data, K<= 1000Return the head node of the modified circular linked list.Sample Input 1:-
3
1- >2- >3
4
Sample Output 1:-
1- >2- >3- >4
Sample Input 2:-
3
1- >3- >2
1
Sample Output 2:-
1- >3- >2- >1, I have written this Solution Code: public static Node Insertion(Node head, int K){
Node node=head;
while ( node.next != head)
{node = node.next; }
Node temp = new Node(K);
node.next=temp;
temp.next=head;
return head;}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Complete the function isArray which takes an input which can be any data type and returns true
if it's an array else false.Could be any datatype string number object or an arraytrue or falseSample Input:-
1
[2, 3]
Sample Output
false
true, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner s = new Scanner(System.in);
String str= s.next();
if(str.charAt(0)== '['){
System.out.println("true");
}
else{
System.out.println("false");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Complete the function isArray which takes an input which can be any data type and returns true
if it's an array else false.Could be any datatype string number object or an arraytrue or falseSample Input:-
1
[2, 3]
Sample Output
false
true, I have written this Solution Code: function isArray(input){
if(Array.isArray(input)) {
console.log(true)
}else{
console.log(false)
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Create a new class VipCustomer
it should have 3 fields name, creditLimit(double), and email address, there default value is as {name:"XYZ", creditLimit:"10", email:"xyz@abc. com"} respectively means ordering of parameter should be same.
E.g constructor(name,creditLimit,email);
create 3 constructor
1st constructor empty should call the constructor with 3 parameters with default values
2nd constructor should pass on the 2 values it receives as name and creditLimit respectively and, add a default value for the 3rd
3rd constructor should save all fields.
create getters only for this name getName, getCreditLimit and getEmail and confirm it works.
Note: each methods and variables should of public typeYou don't have to take any input, You only have to write class <b>VipCustomer</b>.Output will be printed by tester, "Correct" if your code is perfectly fine otherwise "Wrong".Sample Input:
class VipCustomer{
// if your code works fine for the tester
}
Sample Output:
Correct, I have written this Solution Code:
class VipCustomer{
public double creditLimit;
public String name;
public String email;
VipCustomer(){
this("XYZ",10.0,"[email protected]");
}
VipCustomer(String name,double creditLimit){
this(name,creditLimit,"[email protected]");
}
VipCustomer(String _name,double _creditLimit,String _email){
email=_email;
name=_name;
creditLimit=_creditLimit;
}
public String getName(){
return this.name;
}
public String getEmail(){
return this.email;
}
public double getCreditLimit(){
return this.creditLimit;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number (n) is represented in Linked List such that each digit corresponds to a node in linked list. Add 1 to it.
<b>Note:-</b> Linked list representation of a number is from left to right i.e if the number is 123 than in linked list it is represented as 3->2->1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>addOne()</b> that takes head node of the linked list as parameter.
Constraints:
1 <=length of n<= 1000Return the head of the modified linked list.Input 1:
456
Output 1:
457
Input 2:
999
Output 2:
1000, I have written this Solution Code:
static Node addOne(Node head)
{
Node res = head;
Node temp = null, prev = null;
int carry = 1, sum;
while (head != null) //while both lists exist
{
sum = carry + head.data;
carry = (sum >= 10)? 1 : 0;
sum = sum % 10;
head.data = sum;
temp = head;
head = head.next;
}
if (carry > 0) {
Node x=new Node(carry);
temp.next=x;}
return res;
} , In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: An array of numbers is given. It has all unique elements except one. Find the only duplicate element in that array having all other unique elements.The first line contains an integer N, the number of elements in the array.
The second line contains N integers.
Constraints:
1 <= N <= 10^5
1 <= Elements of Array <= 10^5Print the single duplicate number in the arrayInput
6
1 2 3 4 4 5
Output
4
Explanation: 4 is repeated in this array
Input:
3
2 1 2
Output:
2, I have written this Solution Code: n = int(input())
arr = list(map(int, input().split()))
l = [0] * (max(arr) + 1)
for i in arr:
l[i] += 1
for i in range(len(l)):
if l[i] > 1:
print(i), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: An array of numbers is given. It has all unique elements except one. Find the only duplicate element in that array having all other unique elements.The first line contains an integer N, the number of elements in the array.
The second line contains N integers.
Constraints:
1 <= N <= 10^5
1 <= Elements of Array <= 10^5Print the single duplicate number in the arrayInput
6
1 2 3 4 4 5
Output
4
Explanation: 4 is repeated in this array
Input:
3
2 1 2
Output:
2, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int m = 100001;
int main(){
int n,k;
cin>>n;
long long a[n],sum=0;
for(int i=0;i<n;i++){
cin>>a[i];
if(a[i]<0){
a[i]=-a[i];
}
}
sort(a,a+n);
for(int i=0;i<n-1;i++){
if(a[i]==a[i+1]){cout<<a[i];return 0;}
}
cout<<sum;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string s, find the first non-repeating character in the string and return its index. If it does not exist, return -1.First line of the input contains the string s.
<b>Constraints</b>
1<= s. length <= 100000Print the index of the first non- repeating character in a stringInput
s = "newtonschool"
Output
1
Explanation
"e" is the first non- repeating character in a string, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s = br.readLine();
System.out.print(nonRepeatChar(s));
}
static int nonRepeatChar(String s){
char count[] = new char[256];
for(int i=0; i< s.length(); i++){
count[s.charAt(i)]++;
}
for (int i=0; i<s.length(); i++) {
if (count[s.charAt(i)]==1){
return i;
}
}
return -1;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string s, find the first non-repeating character in the string and return its index. If it does not exist, return -1.First line of the input contains the string s.
<b>Constraints</b>
1<= s. length <= 100000Print the index of the first non- repeating character in a stringInput
s = "newtonschool"
Output
1
Explanation
"e" is the first non- repeating character in a string, I have written this Solution Code: from collections import defaultdict
s=input()
d=defaultdict(int)
for i in s:
d[i]+=1
ans=-1
for i in range(len(s)):
if(d[s[i]]==1):
ans=i
break
print(ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string s, find the first non-repeating character in the string and return its index. If it does not exist, return -1.First line of the input contains the string s.
<b>Constraints</b>
1<= s. length <= 100000Print the index of the first non- repeating character in a stringInput
s = "newtonschool"
Output
1
Explanation
"e" is the first non- repeating character in a string, I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-01-10 12:51:16
**/
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
int firstUniqChar(string s) {
map<char, int> charCount;
int len = s.length();
for (int i = 0; i < len; i++) {
charCount[s[i]]++;
}
for (int i = 0; i < len; i++) {
if (charCount[s[i]] == 1)
return i;
}
return -1;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
string str;
cin>>str;
cout<<firstUniqChar(str)<<"\n";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ram has a military dog who has the special ability of hearing, He can hear frequencies starting from 70Hz to 44000Hz (both inclusive). Now Ram wants to detect some unusual sounds across him. Tell "Yes" or "No" whether his dog can detect the sound or not.First line of the input contains the value of Test cases T.
Each test case consists of a single integer X - the frequency of unusual sound in Hertz.
<b>Constraints</b>
1 ≤ T ≤ 10<sup>4</sup>
1 ≤ X ≤ 10<sup>5</sup>Print Yes or No wether his dog can detect them or not for each test case.Sample Input
3
21
453
45000
Sample Output
No
Yes
No, I have written this Solution Code: /**
* author: tourist1256
* created: 2022-09-20 14:02:39
**/
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <class key, class value, class cmp = std::less<key>>
using ordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order(k) returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
#define int long long
#define yn(ans) printf("%s\n", (ans) ? "Yes" : "No");
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
inline int64_t random_long(int l = LLONG_MIN, int r = LLONG_MAX) {
uniform_int_distribution<int64_t> generator(l, r);
return generator(rng);
}
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
auto start = std::chrono::high_resolution_clock::now();
int tt;
cin >> tt;
while (tt--) {
int N;
cin >> N;
yn(N >= 70 && N <= 44000);
}
auto stop = std::chrono::high_resolution_clock::now();
auto duration = std::chrono::duration_cast<std::chrono::nanoseconds>(stop - start);
cerr << "Time taken : " << ((long double)duration.count()) / ((long double)1e9) << "s " << endl;
return 0;
};
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
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