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For this Question: Sara's Phone has N apps and each app takes K unit of memory. Now Sara wants to release M units of memory. Your task is to tell the minimum apps Sara needs to delete or say it is not possible.<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>Phone()</b> that takes integers N, K, and M as arguments.
Constraints:-
1 <= N <= 1000
1 <= K <= 100
0 <= M <= 10000Return minimum apps to delete and if it is not possible return -1.Sample Input:-
10 3 10
Sample Output:-
4
Sample Input:-
10 3 40
Sample Output:-
-1, I have written this Solution Code:
int Phone(int n, int k, int m){
if(n*k<m){
return -1;
}
int x = m/k;
if(m%k!=0){x++;}
return x;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara's Phone has N apps and each app takes K unit of memory. Now Sara wants to release M units of memory. Your task is to tell the minimum apps Sara needs to delete or say it is not possible.<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>Phone()</b> that takes integers N, K, and M as arguments.
Constraints:-
1 <= N <= 1000
1 <= K <= 100
0 <= M <= 10000Return minimum apps to delete and if it is not possible return -1.Sample Input:-
10 3 10
Sample Output:-
4
Sample Input:-
10 3 40
Sample Output:-
-1, I have written this Solution Code:
int Phone(int n, int k, int m){
if(n*k<m){
return -1;
}
int x = m/k;
if(m%k!=0){x++;}
return x;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara's Phone has N apps and each app takes K unit of memory. Now Sara wants to release M units of memory. Your task is to tell the minimum apps Sara needs to delete or say it is not possible.<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>Phone()</b> that takes integers N, K, and M as arguments.
Constraints:-
1 <= N <= 1000
1 <= K <= 100
0 <= M <= 10000Return minimum apps to delete and if it is not possible return -1.Sample Input:-
10 3 10
Sample Output:-
4
Sample Input:-
10 3 40
Sample Output:-
-1, I have written this Solution Code: static int Phone(int n, int k, int m){
if(n*k<m){
return -1;
}
int x = m/k;
if(m%k!=0){x++;}
return x;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara's Phone has N apps and each app takes K unit of memory. Now Sara wants to release M units of memory. Your task is to tell the minimum apps Sara needs to delete or say it is not possible.<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>Phone()</b> that takes integers N, K, and M as arguments.
Constraints:-
1 <= N <= 1000
1 <= K <= 100
0 <= M <= 10000Return minimum apps to delete and if it is not possible return -1.Sample Input:-
10 3 10
Sample Output:-
4
Sample Input:-
10 3 40
Sample Output:-
-1, I have written this Solution Code: def Phone(N,K,M):
if N*K < M :
return -1
x = M//K
if M%K!=0:
x=x+1
return x, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: t=int(input())
while t!=0:
m,n=input().split()
m,n=int(m),int(n)
for i in range(m):
arr=input().strip()
if '1' in arr:
arr='1 '*n
else:
arr='0 '*n
print(arr)
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define N 1000
int a[N][N];
// Driver code
int main()
{
int t;
cin>>t;
while(t--){
int n,m;
cin>>n>>m;
bool b[n];
for(int i=0;i<n;i++){
b[i]=false;
}
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
cin>>a[i][j];
if(a[i][j]==1){
b[i]=true;
}
}
}
for(int i=0;i<n;i++){
if(b[i]){
for(int j=0;j<m;j++){
cout<<1<<" ";
}}
else{
for(int j=0;j<m;j++){
cout<<0<<" ";
}
}
cout<<endl;
}
}}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args) throws Exception{
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader bf = new BufferedReader(isr);
int t = Integer.parseInt(bf.readLine());
while (t-- > 0){
String inputs[] = bf.readLine().split(" ");
int m = Integer.parseInt(inputs[0]);
int n = Integer.parseInt(inputs[1]);
String[] matrix = new String[m];
for(int i=0; i<m; i++){
matrix[i] = bf.readLine();
}
StringBuffer ones = new StringBuffer("");
StringBuffer zeros = new StringBuffer("");
for(int i=0; i<n; i++){
ones.append("1 ");
zeros.append("0 ");
}
for(int i=0; i<m; i++){
if(matrix[i].contains("1")){
System.out.println(ones);
}else{
System.out.println(zeros);
}
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a multiset of positive integers S. Initially, a number x is written on the blackboard. Now, you can perform the below operations zero or more times on blackboard numbers.
<ul><li>Choose any number written on the blackboard and then erase it from the blackboard. Suppose the chosen number is a. Now, choose some positive integer b and write two numbers, the quotient and remainder on dividing a by b on the blackboard.</li></ul>
For each x = 1, 2, 3 ... X, determine if it is possible to make the multiset of numbers written on blackboard equal to S after performing some operations, if the initial number is x.The first line contains a single integer T – the number of test cases.
The first line of each test case contains two space-separated integers – N (the size of multiset S) and X.
The second line contains N space-separated integers S<sub>1</sub>, S<sub>2</sub>, ... S<sub>N</sub>.
<b> Constraints: </b>
1 ≤ T ≤ 20
1 ≤ N ≤ 100
1 ≤ X ≤ 5×10<sup>5</sup>
1 ≤ S<sub>i</sub> ≤ 10<sup>9</sup>
Sum of X over all testcases doesn't exceed 5×10<sup>5</sup>.Print a string of length X whose i<sup>th</sup> character is '1' if it is possible to make the blackboard numbers equal to multiset S with i as the initial number on blackboard, and '0' otherwise.Sample Input 1:
1
2 5
1 1
Sample Output 1:
00111
Sample Explanation 1:
If x=4 is the only integer initially written on the blackboard, choose b=3 to have {1,1} as multiset of numbers on blackboard., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define epsi (double)(0.00000000001)
typedef long long int ll;
typedef unsigned long long int ull;
#define vi vector<ll>
#define pii pair<ll,ll>
#define vii vector<pii>
#define vvi vector<vi>
//#define max(a,b) ((a>b)?a:b)
//#define min(a,b) ((a>b)?b:a)
#define min3(a,b,c) min(min(a,b),c)
#define min4(a,b,c,d) min(min(a,b),min(c,d))
#define max3(a,b,c) max(max(a,b),c)
#define max4(a,b,c,d) max(max(a,b),max(c,d))
#define ff(a,b,c) for(int a=b; a<=c; a++)
#define frev(a,b,c) for(int a=c; a>=b; a--)
#define REP(a,b,c) for(int a=b; a<c; a++)
#define pb push_back
#define mp make_pair
#define endl "\n"
#define all(v) v.begin(),v.end()
#define sz(a) (ll)a.size()
#define F first
#define S second
#define ld long double
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define ub upper_bound
#define lb lower_bound
#define setbits(x) __builtin_popcountll(x)
#define trav(a,x) for(auto &a:x)
#define make_unique(v) v.erase(unique(v.begin(), v.end()), v.end())
#define rev(arr) reverse(all(arr))
#define gcd(a,b) __gcd(a,b);
#define ub upper_bound // '>'
#define lb lower_bound // '>='
#define qi queue<ll>
#define fsh cout.flush()
#define si stack<ll>
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define fill(a,b) memset(a, b, sizeof(a))
template<typename T,typename E>bool chmin(T& s,const E& t){bool res=s>t;s=min<T>(s,t);return res;}
const ll INF=1LL<<60;
void solve(){
ll n,x;
cin >> n >> x;
vector<bool> ans(x+1,0);
vi a(n);
rep(i,0,n)cin>>a[i];
ll mn=0;
sort(all(a));
rep(i,0,n){
mn=mn*(1+a[i])+a[i];
if(mn>x){
rep(i,1,x+1) cout << ans[i];
cout << endl;
return;
}
}
ans[mn]=1;
vector<vector<pair<ll,ll>>> v(n,vector<pair<ll,ll>>(x+1,make_pair(INF,0)));
vector<ll> dp(x+1,n);
dp[mn]=0;
rep(k,mn,x+1){
rep(j,0,n){
if(v[j][k%a[j]].first <= k){
chmin(dp[k],v[j][k%a[j]].second);
}
}
if(dp[k]==n)continue;
rep(j,0,n){
//if(i==j)continue;
ll tmp=((mn-a[j])/(a[j]+1)+(k-mn));
tmp=tmp*(a[j]+1)+a[j];
chmin(v[j][tmp%a[j]],make_pair(tmp,dp[k]+1));
}
}
rep(k,mn,x+1){
if(dp[k]<n) ans[k]=1;
}
rep(i,1,x+1)cout<<ans[i];
cout<<endl;
}
int main(){
ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
int t;
cin >> t;
while(t--){
solve();
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, your task is to find the count of repeated elements. Print the repeated elements in ascending order along with their frequency.
Have a look at the example for more understanding.The first line of input contains a single integer N, the next line of input contains N space- separated integers depicting the values of the array.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= 100000For each duplicate element in sorted order in a new line, First, print the duplicate element and then print its number of occurence space- separated.
Note:- It is guaranteed that at least one duplicate element will exist in the given array.Sample Input:-
5
3 2 1 1 2
Sample Output:-
1 2
2 2
Sample Input:-
5
1 1 1 1 5
Sample Output:-
1 4
Explaination:
test 1: Only 1 and 2 are repeated. Both are repeated twice. So, we print:
1 -> frequency of 1
2 -> frequency of 2
1 is printed before 2 as it is smaller than 2, I have written this Solution Code: import numpy as np
from collections import defaultdict
n=int(input())
a=np.array([input().strip().split()],int).flatten()
d=defaultdict(int)
for i in a:
d[i]+=1
d=sorted(d.items())
for i in d:
if(i[1]>1):
print(i[0],end=" ")
print(i[1])
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, your task is to find the count of repeated elements. Print the repeated elements in ascending order along with their frequency.
Have a look at the example for more understanding.The first line of input contains a single integer N, the next line of input contains N space- separated integers depicting the values of the array.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= 100000For each duplicate element in sorted order in a new line, First, print the duplicate element and then print its number of occurence space- separated.
Note:- It is guaranteed that at least one duplicate element will exist in the given array.Sample Input:-
5
3 2 1 1 2
Sample Output:-
1 2
2 2
Sample Input:-
5
1 1 1 1 5
Sample Output:-
1 4
Explaination:
test 1: Only 1 and 2 are repeated. Both are repeated twice. So, we print:
1 -> frequency of 1
2 -> frequency of 2
1 is printed before 2 as it is smaller than 2, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
InputStreamReader read = new InputStreamReader(System.in);
BufferedReader in = new BufferedReader(read);
int noOfElements = Integer.parseInt(in.readLine());
String [] elem = in.readLine().trim().split(" ");
int [] elements = new int [noOfElements];
for(int i=0 ; i< noOfElements; i++){
elements[i] = Integer.parseInt(elem[i]);
}
Arrays.sort(elements);
int count =0;
for(int i = 0 ; i < noOfElements-1 ; i++){
if(elements[i] == elements[i+1]){
count ++;
}
else if(count != 0){
System.out.print(elements[i]+" "+(count+1));
count =0;
System.out.println();
}
}
if(count != 0)
System.out.print(elements[noOfElements-1]+" "+(count+1));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, your task is to find the count of repeated elements. Print the repeated elements in ascending order along with their frequency.
Have a look at the example for more understanding.The first line of input contains a single integer N, the next line of input contains N space- separated integers depicting the values of the array.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= 100000For each duplicate element in sorted order in a new line, First, print the duplicate element and then print its number of occurence space- separated.
Note:- It is guaranteed that at least one duplicate element will exist in the given array.Sample Input:-
5
3 2 1 1 2
Sample Output:-
1 2
2 2
Sample Input:-
5
1 1 1 1 5
Sample Output:-
1 4
Explaination:
test 1: Only 1 and 2 are repeated. Both are repeated twice. So, we print:
1 -> frequency of 1
2 -> frequency of 2
1 is printed before 2 as it is smaller than 2, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n;
cin>>n;
int a[n];
map<int,int> m;
for(int i=0;i<n;i++){
cin>>a[i];
m[a[i]]++;
}
for(auto it = m.begin();it!=m.end();it++){
if(it->second>1){
cout<<it->first<<" "<<it->second<<'\n';
}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, multiples of 100 which are not multiples of 400 are not leap years.The input contains a single integer N
<b>Constraint:</b>
1 <= n <= 10<sup>4</sup>Print "YES" if the year is a leap year else print "NO".Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code:
function LeapYear(year){
// write code here
// return the output using return keyword
// do not use console.log here
if ((0 != year % 4) || ((0 == year % 100) && (0 != year % 400))) {
return 0;
} else {
return 1
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, multiples of 100 which are not multiples of 400 are not leap years.The input contains a single integer N
<b>Constraint:</b>
1 <= n <= 10<sup>4</sup>Print "YES" if the year is a leap year else print "NO".Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code: year = int(input())
if year % 4 == 0 and not year % 100 == 0 or year % 400 == 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: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, multiples of 100 which are not multiples of 400 are not leap years.The input contains a single integer N
<b>Constraint:</b>
1 <= n <= 10<sup>4</sup>Print "YES" if the year is a leap year else print "NO".Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code: import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int side = scanner.nextInt();
int area = LeapYear(side);
if(area==1){
System.out.println("YES");}
else{
System.out.println("NO");}
}
static int LeapYear(int year){
if(year%400==0){return 1;}
if(year%100 != 0 && year%4==0){return 1;}
else {
return 0;}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Natsu is fighting with a dragon that has A Health and B attack power and Natsu has C health and D attack power. The fight goes in turns first Natsu will attack the Dragon then Dragon will attack Natsu and this goes on. The fight will stop when either the dragon's or Natsu's health drops zero or below. Your task is to check whether Natsu will able to slay the Dragon 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>DragonSlayer()</b> that takes integers A, B, C, and D as arguments.
Constraints:-
1 <= A, B, C, D <= 1000Return 0 if Dragon wins else return 1.Sample Input:-
8 2 5 3
Sample Output:-
1
Explanation:-
Natsu's attack:- A = 5, B = 2, C = 5, D = 3
Dragon's attack:- A = 5, B = 2, C = 3, D =3
Natsu's attack:- A = 2, B =2, C = 3, D=3
Dragon's attack:- A = 2, B =2, C = 1, D=3
Natsu's attack:- A = -1, B =2, C = 1, D=3
Natsu's win, I have written this Solution Code:
int DragonSlayer(int A, int B, int C,int D){
int x = C/B;
if(C%B!=0){x++;}
int y = A/D;
if(A%D!=0){y++;}
if(x<y){return 0;}
return 1;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Natsu is fighting with a dragon that has A Health and B attack power and Natsu has C health and D attack power. The fight goes in turns first Natsu will attack the Dragon then Dragon will attack Natsu and this goes on. The fight will stop when either the dragon's or Natsu's health drops zero or below. Your task is to check whether Natsu will able to slay the Dragon 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>DragonSlayer()</b> that takes integers A, B, C, and D as arguments.
Constraints:-
1 <= A, B, C, D <= 1000Return 0 if Dragon wins else return 1.Sample Input:-
8 2 5 3
Sample Output:-
1
Explanation:-
Natsu's attack:- A = 5, B = 2, C = 5, D = 3
Dragon's attack:- A = 5, B = 2, C = 3, D =3
Natsu's attack:- A = 2, B =2, C = 3, D=3
Dragon's attack:- A = 2, B =2, C = 1, D=3
Natsu's attack:- A = -1, B =2, C = 1, D=3
Natsu's win, I have written this Solution Code: static int DragonSlayer(int A, int B, int C,int D){
int x = C/B;
if(C%B!=0){x++;}
int y = A/D;
if(A%D!=0){y++;}
if(x<y){return 0;}
return 1;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Natsu is fighting with a dragon that has A Health and B attack power and Natsu has C health and D attack power. The fight goes in turns first Natsu will attack the Dragon then Dragon will attack Natsu and this goes on. The fight will stop when either the dragon's or Natsu's health drops zero or below. Your task is to check whether Natsu will able to slay the Dragon 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>DragonSlayer()</b> that takes integers A, B, C, and D as arguments.
Constraints:-
1 <= A, B, C, D <= 1000Return 0 if Dragon wins else return 1.Sample Input:-
8 2 5 3
Sample Output:-
1
Explanation:-
Natsu's attack:- A = 5, B = 2, C = 5, D = 3
Dragon's attack:- A = 5, B = 2, C = 3, D =3
Natsu's attack:- A = 2, B =2, C = 3, D=3
Dragon's attack:- A = 2, B =2, C = 1, D=3
Natsu's attack:- A = -1, B =2, C = 1, D=3
Natsu's win, I have written this Solution Code: int DragonSlayer(int A, int B, int C,int D){
int x = C/B;
if(C%B!=0){x++;}
int y = A/D;
if(A%D!=0){y++;}
if(x<y){return 0;}
return 1;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Natsu is fighting with a dragon that has A Health and B attack power and Natsu has C health and D attack power. The fight goes in turns first Natsu will attack the Dragon then Dragon will attack Natsu and this goes on. The fight will stop when either the dragon's or Natsu's health drops zero or below. Your task is to check whether Natsu will able to slay the Dragon 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>DragonSlayer()</b> that takes integers A, B, C, and D as arguments.
Constraints:-
1 <= A, B, C, D <= 1000Return 0 if Dragon wins else return 1.Sample Input:-
8 2 5 3
Sample Output:-
1
Explanation:-
Natsu's attack:- A = 5, B = 2, C = 5, D = 3
Dragon's attack:- A = 5, B = 2, C = 3, D =3
Natsu's attack:- A = 2, B =2, C = 3, D=3
Dragon's attack:- A = 2, B =2, C = 1, D=3
Natsu's attack:- A = -1, B =2, C = 1, D=3
Natsu's win, I have written this Solution Code:
def DragonSlayer(A,B,C,D):
x = C//B
if(C%B!=0):
x=x+1
y = A//D
if(A%D!=0):
y=y+1
if(x<y):
return 0
return 1
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: n=int(input())
a=map(int,input().split())
b=[]
mx=-200000
cnt=0
for i in a:
if i>mx:
cnt+=1
mx=i
print(cnt), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: function numberOfRoofs(arr)
{
let count=1;
let max = arr[0];
for(let i=1;i<arrSize;i++)
{
if(arr[i] > max)
{
count++;
max = arr[i];
}
}
return count;
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
public static void main(String args[]){
Scanner s=new Scanner(System.in);
int n=s.nextInt();
int []a=new int[n];
for(int i=0;i<n;i++){
a[i]=s.nextInt();
}
int count=1;
int max = a[0];
for(int i=1;i<n;i++)
{
if(a[i] > max)
{
count++;
max = a[i];
}
}
System.out.println(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
Sample Output:
No, I have written this Solution Code: m,n = map(int , input().split())
if (m%n==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: Given an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
Sample Output:
No, 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 n,m;
cin>>n>>m;
if(n%m==0)
cout<<"Yes";
else
cout<<"No";
#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 an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
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();
Long m = sc.nextLong();
if(n%m==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 two integers a and b, your task is to calculate values for each of the following operations:-
a + b
a - b
a * b
a/bSince this will be a functional problem, you don't have to take input. You have to complete the function
<b>operations()</b> that takes the integer a and b as parameters.
<b>Constraints:</b>
1 ≤ b ≤ a ≤1000
<b> It is guaranteed that a will be divisible by b.</b>Print the mentioned operations each in a new line.Sample Input:
15 3
Sample Output:
18
12
45
5, I have written this Solution Code: def operations(x, y):
print(x+y)
print(x-y)
print(x*y)
print(x//y), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){cout<<"FizzBuzz"<<" ";}
else if(i%5==0){cout<<"Buzz ";}
else if(i%3==0){cout<<"Fizz ";}
else{cout<<i<<" ";}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, 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 x= sc.nextInt();
fizzbuzz(x);
}
static void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("FizzBuzz ");}
else if(i%5==0){System.out.print("Buzz ");}
else if(i%3==0){System.out.print("Fizz ");}
else{System.out.print(i+" ");}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: def fizzbuzz(n):
for i in range (1,n+1):
if (i%3==0 and i%5==0):
print("FizzBuzz",end=' ')
elif i%3==0:
print("Fizz",end=' ')
elif i%5==0:
print("Buzz",end=' ')
else:
print(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 for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){printf("FizzBuzz ");}
else if(i%5==0){printf("Buzz ");}
else if(i%3==0){printf("Fizz ");}
else{printf("%d ",i);}
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Aniket and Swapnil are playing a game in which they have two piles of stones with N1 and N2 stones respectively. They take turns picking any even positive number of stones from one of the pile and keep half of them on the other pile and throw away the other half. Whoever can not make a move loses. Given N1 and N2 find who will win. Swapnil makes the first move.The first and only line of input contains two integers N1 and N2.
Constraints
1 <= N1, N2 <= 1000000000000000Print "Swapnil" if Swapnil wins the game and print "Aniket" if Aniket wins the game.Sample Input 1
2 1
Sample Output 1
Aniket
Sample Input 2
4 8
Sample Output 2
Swapnil, 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[] str = br.readLine().trim().split(" ");
long n1 = Long.parseLong(str[0]);
long n2 = Long.parseLong(str[1]);
long diff = Math.abs(n1-n2);
long c =0;
while(diff>1){
if(diff%2!=0)
diff--;
diff/=2;
if(n1>n2){
n1 -= 2*diff;
n2 += diff;
}
else{
n2 -= 2*diff;
n1 += diff;
}
c += diff;
diff = Math.abs(n1-n2);
}
if(c%2==0){
System.out.print("Aniket");
}else{
System.out.print("Swapnil");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Aniket and Swapnil are playing a game in which they have two piles of stones with N1 and N2 stones respectively. They take turns picking any even positive number of stones from one of the pile and keep half of them on the other pile and throw away the other half. Whoever can not make a move loses. Given N1 and N2 find who will win. Swapnil makes the first move.The first and only line of input contains two integers N1 and N2.
Constraints
1 <= N1, N2 <= 1000000000000000Print "Swapnil" if Swapnil wins the game and print "Aniket" if Aniket wins the game.Sample Input 1
2 1
Sample Output 1
Aniket
Sample Input 2
4 8
Sample Output 2
Swapnil, I have written this Solution Code: n1,n2 = map(int,input().split())
print('Aniket' if abs(n1-n2)==1 or n1==n2 else 'Swapnil'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Aniket and Swapnil are playing a game in which they have two piles of stones with N1 and N2 stones respectively. They take turns picking any even positive number of stones from one of the pile and keep half of them on the other pile and throw away the other half. Whoever can not make a move loses. Given N1 and N2 find who will win. Swapnil makes the first move.The first and only line of input contains two integers N1 and N2.
Constraints
1 <= N1, N2 <= 1000000000000000Print "Swapnil" if Swapnil wins the game and print "Aniket" if Aniket wins the game.Sample Input 1
2 1
Sample Output 1
Aniket
Sample Input 2
4 8
Sample Output 2
Swapnil, 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 inf 1e8+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();
//////////////
#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,y;
cin>>x>>y;
if(abs(x-y)<=1)
{
cout<<"Aniket";
}
else
cout<<"Swapnil";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, the task is to find the number of divisors of N which are divisible by 2.The input line contains T, denoting the number of testcases. First line of each testcase contains integer N
Constraints:
1 <= T <= 50
1 <= N <= 10^9For each testcase in new line, you need to print the number of divisors of N which are exactly divisble by 2Input:
2
9
8
Output
0
3, I have written this Solution Code: import math
n = int(input())
for i in range(n):
x = int(input())
count = 0
for i in range(1, int(math.sqrt(x))+1):
if x % i == 0:
if (i%2 == 0):
count+=1
if ((x/i) %2 == 0):
count+=1
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, the task is to find the number of divisors of N which are divisible by 2.The input line contains T, denoting the number of testcases. First line of each testcase contains integer N
Constraints:
1 <= T <= 50
1 <= N <= 10^9For each testcase in new line, you need to print the number of divisors of N which are exactly divisble by 2Input:
2
9
8
Output
0
3, 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());
int count=0;
for(int i=1;i<=Math.sqrt(n);i++){
if(n%i == 0)
{
if(i%2==0) {
count++;
}
if(i*i != n && (n/i)%2==0) {
count++;
}
}
}
System.out.println(count);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, the task is to find the number of divisors of N which are divisible by 2.The input line contains T, denoting the number of testcases. First line of each testcase contains integer N
Constraints:
1 <= T <= 50
1 <= N <= 10^9For each testcase in new line, you need to print the number of divisors of N which are exactly divisble by 2Input:
2
9
8
Output
0
3, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
long long n;
cin>>n;
if(n&1){cout<<0<<endl;continue;}
long x=sqrt(n);
int cnt=0;
for(long long i=1;i<=x;i++){
if(!(n%i)){
if(!(i%2)){cnt++;}
if(i*i!=n){
if(!((n/i)%2)){cnt++;}
}
}
}
cout<<cnt<<endl;}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: static void simpleSum(int a, int b, int c){
System.out.println(a+b+c);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: void simpleSum(int a, int b, int c){
cout<<a+b+c;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: x = input()
a, b, c = x.split()
a = int(a)
b = int(b)
c = int(c)
print(a+b+c), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of opening and closing parenthesis '(' and ')'. Find length of the longest valid parenthesis substring.Each test case have one line string S of character '(' and ')' of length N.
Constraints:
1 <= N <= 10^5Print the length of the longest valid parenthesis substring.Input
((()
Output
2
Input
)()())
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));
String s = br.readLine();
int n = s.length();
int left = 0, right = 0;
int maxlength = 0;
for(int i = 0; i < n; i++)
{
if (s.charAt(i) == '(')
left++;
else
right++;
if (left == right)
maxlength = Math.max(maxlength, 2 * right);
else if (right > left)
left = right = 0;
}
left = right = 0;
for(int i = n - 1; i >= 0; i--)
{
if (s.charAt(i) == '(')
left++;
else
right++;
if (left == right)
maxlength = Math.max(maxlength, 2 * left);
else if (left > right)
left = right = 0;
}
System.out.println(maxlength);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of opening and closing parenthesis '(' and ')'. Find length of the longest valid parenthesis substring.Each test case have one line string S of character '(' and ')' of length N.
Constraints:
1 <= N <= 10^5Print the length of the longest valid parenthesis substring.Input
((()
Output
2
Input
)()())
Output
4, I have written this Solution Code: def longestValidParentheses( S):
stack, ans = [-1], 0
for i in range(len(S)):
if S[i] == '(': stack.append(i)
elif len(stack) == 1: stack[0] = i
else:
stack.pop()
ans = max(ans, i - stack[-1])
return ans
n=input()
print(longestValidParentheses(n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of opening and closing parenthesis '(' and ')'. Find length of the longest valid parenthesis substring.Each test case have one line string S of character '(' and ')' of length N.
Constraints:
1 <= N <= 10^5Print the length of the longest valid parenthesis substring.Input
((()
Output
2
Input
)()())
Output
4, 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;
int a[N];
signed main() {
IOS;
string s;
cin >> s;
int n = s.length();
s = "#" + s;
stack<int> st;
st.push(0);
int ans = 0;
for(int i = 1; i <= n; i++){
if(s[i] == '(')
st.push(i);
else{
if(s[st.top()] == '('){
int x = st.top();
a[i] = i-x+1 + a[x-1];
ans = max(ans, a[i]);
st.pop();
}
else
st.push(i);
}
}
cout << ans;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a Singly linked list and an integer K. Your task is to insert the integer K at the head of the given linked list<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>addElement()</b> that takes head node and the element K as a parameter.
Constraints:
1 <=N<= 1000
1 <=K, value<= 1000Return the head of the modified linked listSample Input:-
5 2
1 2 3 4 5
Sample Output:
2 1 2 3 4 5
, I have written this Solution Code: public static Node addElement(Node head,int k) {
Node temp =new Node(k);
temp.next=head;
return temp;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array A of N integers, find the pair of integers in the array which have minimum XOR value. Report the minimum XOR value.First line denotes N, the size of the array.
Next line denotes N space-separated array elements.
Constraints:
2 <= N <= 100000
0 <= A[i] <= 10^7Print a single integer denoting minimum xor valueSample Input
4
0 2 5 7
Sample Output
2
Explanation:
0 xor 2 = 2
Sample Input
4
0 4 7 9
Sample Output
3, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
int ans = 0;
int mini = Integer.MAX_VALUE;
Scanner sc = new Scanner(System.in);
int array_size = sc.nextInt();
int N[] = new int[array_size];
for (int i = 0; i < array_size; i++) {
if(mini == 0){
break;
}
N[i] = sc.nextInt();
for (int j = i - 1; j >= 0; j--) {
ans = N[i] ^ N[j];
if (mini > ans) {
mini = ans;
}
}
}
System.out.println(mini);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array A of N integers, find the pair of integers in the array which have minimum XOR value. Report the minimum XOR value.First line denotes N, the size of the array.
Next line denotes N space-separated array elements.
Constraints:
2 <= N <= 100000
0 <= A[i] <= 10^7Print a single integer denoting minimum xor valueSample Input
4
0 2 5 7
Sample Output
2
Explanation:
0 xor 2 = 2
Sample Input
4
0 4 7 9
Sample Output
3, I have written this Solution Code: a=int(input())
lis = list(map(int,input().split()))
val=666666789
for i in range(a+1):
for j in range(i+1,a):
temp = lis[i]^lis[j]
if(temp<val):
val = temp
if(val==0):
break
print(val), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array A of N integers, find the pair of integers in the array which have minimum XOR value. Report the minimum XOR value.First line denotes N, the size of the array.
Next line denotes N space-separated array elements.
Constraints:
2 <= N <= 100000
0 <= A[i] <= 10^7Print a single integer denoting minimum xor valueSample Input
4
0 2 5 7
Sample Output
2
Explanation:
0 xor 2 = 2
Sample Input
4
0 4 7 9
Sample Output
3, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
// Function to find minimum XOR pair
int minXOR(int arr[], int n)
{
// Sort given array
sort(arr, arr + n);
int minXor = INT_MAX;
int val = 0;
// calculate min xor of consecutive pairs
for (int i = 0; i < n - 1; i++) {
val = arr[i] ^ arr[i + 1];
minXor = min(minXor, val);
}
return minXor;
}
// Driver program
int main()
{
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++)
{
cin>>arr[i];
}
cout << minXOR(arr, n) << endl;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are fighting against a monster.
The monster has health M while you have initial health Y.
To defeat him, you need health strictly greater than the monster's health.
You have access to two types of pills, red and green. Eating a red pill adds R to your health while eating a green pill multiplies your health by G.
Determine if it is possible to defeat the monster by eating at most one pill of any kind.Input contains four integers M (0 <= M <= 10<sup>4</sup>), Y (0 <= Y <= 10<sup>4</sup>), R (0 <= R <= 10<sup>4</sup>) and G (1 <= G <= 10<sup>4</sup>).Print 1 if it is possible to defeat the monster and 0 if it is impossible to defeat it.Sample Input 1:
10 2 2 2
Sample Output 1:
0
Sample Input 2:
8 7 2 1
Sample Output 2:
1
Explanation for Sample 2:
You can eat a red pill to make your health : 7 + 2 = 9 > 8., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException
{
InputStreamReader pk = new InputStreamReader(System.in);
BufferedReader in = new BufferedReader(pk);
String[] strNums;
int num[] = new int[4];
strNums = in.readLine().split(" ");
for (int i = 0; i < strNums.length; i++) {
num[i] = Integer.parseInt(strNums[i]);
}
if((num[0]<(num[1]+num[2])) || (num[0]<(num[1]*num[3])))
System.out.println("1");
else
System.out.println("0");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are fighting against a monster.
The monster has health M while you have initial health Y.
To defeat him, you need health strictly greater than the monster's health.
You have access to two types of pills, red and green. Eating a red pill adds R to your health while eating a green pill multiplies your health by G.
Determine if it is possible to defeat the monster by eating at most one pill of any kind.Input contains four integers M (0 <= M <= 10<sup>4</sup>), Y (0 <= Y <= 10<sup>4</sup>), R (0 <= R <= 10<sup>4</sup>) and G (1 <= G <= 10<sup>4</sup>).Print 1 if it is possible to defeat the monster and 0 if it is impossible to defeat it.Sample Input 1:
10 2 2 2
Sample Output 1:
0
Sample Input 2:
8 7 2 1
Sample Output 2:
1
Explanation for Sample 2:
You can eat a red pill to make your health : 7 + 2 = 9 > 8., I have written this Solution Code: a = []
a = list(map(int, input().split()))
if a[1]+a[2] > a[0] or a[1]*a[3] > a[0]:
defeat = 1
else:
defeat = 0
print(defeat), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are fighting against a monster.
The monster has health M while you have initial health Y.
To defeat him, you need health strictly greater than the monster's health.
You have access to two types of pills, red and green. Eating a red pill adds R to your health while eating a green pill multiplies your health by G.
Determine if it is possible to defeat the monster by eating at most one pill of any kind.Input contains four integers M (0 <= M <= 10<sup>4</sup>), Y (0 <= Y <= 10<sup>4</sup>), R (0 <= R <= 10<sup>4</sup>) and G (1 <= G <= 10<sup>4</sup>).Print 1 if it is possible to defeat the monster and 0 if it is impossible to defeat it.Sample Input 1:
10 2 2 2
Sample Output 1:
0
Sample Input 2:
8 7 2 1
Sample Output 2:
1
Explanation for Sample 2:
You can eat a red pill to make your health : 7 + 2 = 9 > 8., I have written this Solution Code: //HEADER FILES AND NAMESPACES
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
// #include <sys/resource.h>
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
// DEFINE STATEMENTS
const long long infty = 1e18;
#define num1 1000000007
#define num2 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define REPd(i,a,n) for(ll i=a; i>=n; i--)
#define pb push_back
#define pob pop_back
#define f first
#define s second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define all(x) x.begin(), x.end()
#define M_PI 3.14159265358979323846
#define epsilon (double)(0.000000001)
#define popcount __builtin_popcountll
#define fileio(x) freopen("input.txt", "r", stdin); freopen(x, "w", stdout);
#define out(x) cout << ((x) ? "Yes\n" : "No\n")
#define sz(x) x.size()
#define start_clock() auto start_time = std::chrono::high_resolution_clock::now();
#define measure() auto end_time = std::chrono::high_resolution_clock::now(); cerr << (end_time - start_time)/std::chrono::milliseconds(1) << "ms" << endl;
typedef long long ll;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
// DEBUG FUNCTIONS
#ifdef LOCALY
template<typename T>
void __p(T a) {
cout<<a;
}
template<typename T, typename F>
void __p(pair<T, F> a) {
cout<<"{";
__p(a.first);
cout<<",";
__p(a.second);
cout<<"}";
}
template<typename T>
void __p(std::vector<T> a) {
cout<<"{";
for(auto it=a.begin(); it<a.end(); it++)
__p(*it),cout<<",}"[it+1==a.end()];
}
template<typename T>
void __p(std::set<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T>
void __p(std::multiset<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T, typename F>
void __p(std::map<T,F> a) {
cout<<"{\n";
for(auto it=a.begin(); it!=a.end();++it)
{
__p(it->first);
cout << ": ";
__p(it->second);
cout<<"\n";
}
cout << "}\n";
}
template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) {
__p(a1);
__p(a...);
}
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
cout<<name<<" : ";
__p(arg1);
cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
int bracket=0,i=0;
for(;; i++)
if(names[i]==','&&bracket==0)
break;
else if(names[i]=='(')
bracket++;
else if(names[i]==')')
bracket--;
const char *comma=names+i;
cout.write(names,comma-names)<<" : ";
__p(arg1);
cout<<" | ";
__f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<" ", __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...)
#define error(...)
#endif
// DEBUG FUNCTIONS END
// CUSTOM HASH TO SPEED UP UNORDERED MAP AND TO AVOID FORCED CLASHES
struct custom_hash {
static uint64_t splitmix64(uint64_t x) {
x += 0x9e3779b97f4a7c15;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
return x ^ (x >> 31);
}
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
return splitmix64(x + FIXED_RANDOM);
}
};
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count()); // FOR RANDOM NUMBER GENERATION
ll mod_exp(ll a, ll b, ll c)
{
ll res=1; a=a%c;
while(b>0)
{
if(b%2==1)
res=(res*a)%c;
b/=2;
a=(a*a)%c;
}
return res;
}
ll mymod(ll a,ll b)
{
return (((a = a%b) < 0) ? a + b : a);
}
ll gcdExtended(ll,ll,ll *,ll *);
ll modInverse(ll a, ll m)
{
ll x, y;
ll g = gcdExtended(a, m, &x, &y);
g++; //this line was added just to remove compiler warning
ll res = (x%m + m) % m;
return res;
}
ll gcdExtended(ll a, ll b, ll *x, ll *y)
{
if (a == 0)
{
*x = 0, *y = 1;
return b;
}
ll x1, y1;
ll gcd = gcdExtended(b%a, a, &x1, &y1);
*x = y1 - (b/a) * x1;
*y = x1;
return gcd;
}
struct Graph
{
vector<vector<int>> adj;
Graph(int n)
{
adj.resize(n+1);
}
void add_edge(int a, int b, bool directed = false)
{
adj[a].pb(b);
if(!directed) adj[b].pb(a);
}
};
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll M, Y, R, G;
cin >> M >> Y >> R >> G;
ll maxm = Y;
maxm = max(maxm, Y+R);
maxm = max(maxm, Y*G);
cout << (M < maxm) << "\n";
return 0;
}
/*
1. Check borderline constraints. Can a variable you are dividing by be 0?
2. Use ll while using bitshifts
3. Do not erase from set while iterating it
4. Initialise everything
5. Read the task carefully, is something unique, sorted, adjacent, guaranteed??
6. DO NOT use if(!mp[x]) if you want to iterate the map later
7. Are you using i in all loops? Are the i's conflicting?
*/
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two integers l and r, l ≤ r. Find the largest possible value of a mod b over all pairs (a, b) of integers for which r ≥ a ≥ b ≥ l. As a reminder, a mod b is the remainder we get when dividing a by b. For example, 26 mod 8 = 2.The input consists of two space- separated integers l and r.
<b>Constraints</b>
1 ≤ l ≤ r ≤ 10<sup>9</sup>Print the largest possible value of a mod b over all pairs (a, b) of integers for which r ≥ a ≥ b ≥ l.<b>Sample Input 1</b>
1 1
<b>Sample Output 1</b>
0
<b>Sample Input 2</b>
8 26
<b>Sample Output 2</b>
12, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
int t;
signed main(){
t=1;
// cin>>t;
while(t--){
int l,r;
cin>>l>>r;
cout<<r%max(r/2+1,l)<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the scores of a cricket player in n innings and a target score k, you can choose an inning score x from the list, erase x from the list, and subtract the value of x from all the remaining innings scores. Find if there is a sequence of n-1 operations such that, after applying the operations, the only remaining inning score is equal to a target score k.The first line contains two integers n and k, the number of integers in the list, and the target value, respectively.
The second line contains the n innings a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>...a<sub>n</sub>.
<b>Constraints </b>
2 ≤ n ≤ 2*10<sup>5</sup>
1 ≤ k ≤ 10<sup>9</sup>
-10<sup>9</sup> ≤ a[i] ≤ 10<sup>9</sup>Print "YES" if there is some sequence of n−1 operations such that, after applying the operations, the only remaining inning score is equal to k. Else print "NO".Input:
4 5
4 2 2 7
Output:
YES
Explanation :
We have the list {4, 2, 2, 7}, and we have the target k=5. One way to achieve it is the following: first, we choose the third element, obtaining the list {2, 0, 5}. Next, we choose the first element, obtaining the list {−2, 3}. Finally, we choose the first element, obtaining the list {5}., I have written this Solution Code: #include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main() {
int t;
t=1;
while(t--) {
int n, a;
cin >> n >> a;
vector<int> v(n);
for(int& x : v) cin >> x;
bool ans = false;
if(n == 1) ans = (v[0] == a);
else {
sort(v.begin(), v.end());
int i = 0;
int j = 1;
while(j < n and i < n) {
if(v[i] + abs(a) == v[j]) {
ans = true;
break;
}
else if(v[i] + abs(a) < v[j]) ++i;
else ++j;
}
}
cout << (ans? "YES" : "NO") << '\n';
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has 2*N chocolate boxes with uneven chocolates in them. She wants to distribute the boxes to N of her students. For each student, Sara will pick one box from the start and one box from the end. Since the chocolates are uneven Sara wants to know the maximum number of chocolates a student received.
The boxes are represented by a singly linked list in which each node represents the number of chocolates in the current box.<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>maxChocolates()</b> that takes the head node of the linked list as a parameter.
<b>Constraints:</b>
1 <=N <= 5000
1 <=node.data<= 10000Print the maximum number of chocolates a student received.Sample Input:-
5
1 2 3 4 5 3 2 1 4 5
Sample Output:-
8
Explanation:-
Given list:- 1- >2- >3- >4- >5- >3- >2- >1- >4- >5
Student 1 received:- 1 + 5 = 6
Student 2 received:- 2 + 4 = 6
Student 3 received:- 3 + 1 = 4
Student 4 received:- 4 + 2 = 6
Student 5 received:- 5 + 3 = 8
Sample Input:-
2
1 4 2 3
Sample Output:-
6, I have written this Solution Code: public static int maxChocolates(Node head) {
Node temp = head;
int n=0;
while(temp!=null){
temp=temp.next;
n++;
}
int x = n/2;
temp=head;
while(x-->0){
temp=temp.next;
}
Node p = temp.next;
temp.next = null;
Node h;
while(p!=null){
h=p.next;
p.next=temp;
temp=p;
p=h;
}
x=n/2;
int ans=0;
while(x-->0){
ans=Math.max(ans,head.val+temp.val);
temp=temp.next;
head=head.next;
}
return ans;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.Math.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
String[] st = bf.readLine().split(" ");
if(Integer.parseInt(st[1])==0)
System.out.print(-1);
else {
int f = (Integer.parseInt(st[0])/Integer.parseInt(st[1]));
System.out.print(f);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: D,Q = input().split()
D = int(D)
Q = int(Q)
if(0<=D and Q<=100 and Q >0):
print(int(D/Q))
else:
print('-1'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: #include <iostream>
using namespace std;
int main(){
int n,m;
cin>>n>>m;
if(m==0){cout<<-1;return 0;}
cout<<n/m;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are playing a game in which they are given a binary string S. The rules of the game are given as:-
Players play in turns
The game ends when all the characters in the string are same, and the player who has the current turn wins.
The current player has to delete the i<sup>th</sup> character from the string. (1 <= i <= |S|)
If both the players play optimally and Naruto goes first, find who will win the game.The Input contains only string S.
Constraints:-
1 <= |S| <= 100000Print "Naruto" if Naruto wins else print "Sasuke".Sample Input:-
0101
Sample Output:-
Sasuke
Explanation:-
First Naruto will delete the character 0 from index 1. string:- 101
Then Sasuke will delete the character 1 from index 1. string 01
It doesn't matter which character Naruto removes now, Sasuke will definitely win.
Sample Input:-
1111
Sample Output:-
Naruto
Explanation:-
All the characters in the string are already same hence Naruto wins, 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 s1=br.readLine();
int N=s1.length();
int index=0;
int count=0;
for(int i=N-2;i>=0;i--)
{
if(s1.charAt(i)!=s1.charAt(i+1))
{
index=i+1;
break;
}
}
if(index==N-1)
{
if(N%2==0)
{
System.out.print("Sasuke");
}
else{
System.out.print("Naruto");
}
}
else if(index==0)
{
System.out.print("Naruto");
}
else{
for(int i=0;i<index;i++)
{
count++;
}
if(count%2==0)
{
System.out.print("Naruto");
}
else{
System.out.print("Sasuke");
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are playing a game in which they are given a binary string S. The rules of the game are given as:-
Players play in turns
The game ends when all the characters in the string are same, and the player who has the current turn wins.
The current player has to delete the i<sup>th</sup> character from the string. (1 <= i <= |S|)
If both the players play optimally and Naruto goes first, find who will win the game.The Input contains only string S.
Constraints:-
1 <= |S| <= 100000Print "Naruto" if Naruto wins else print "Sasuke".Sample Input:-
0101
Sample Output:-
Sasuke
Explanation:-
First Naruto will delete the character 0 from index 1. string:- 101
Then Sasuke will delete the character 1 from index 1. string 01
It doesn't matter which character Naruto removes now, Sasuke will definitely win.
Sample Input:-
1111
Sample Output:-
Naruto
Explanation:-
All the characters in the string are already same hence Naruto wins, I have written this Solution Code: binary_string = input()
n = len(binary_string)-1
index = n
for i in range(n-1, -1, -1):
if binary_string[i] == binary_string[n]:
index = i
else:
break
if index % 2 == 0:
print("Naruto")
else:
print("Sasuke"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are playing a game in which they are given a binary string S. The rules of the game are given as:-
Players play in turns
The game ends when all the characters in the string are same, and the player who has the current turn wins.
The current player has to delete the i<sup>th</sup> character from the string. (1 <= i <= |S|)
If both the players play optimally and Naruto goes first, find who will win the game.The Input contains only string S.
Constraints:-
1 <= |S| <= 100000Print "Naruto" if Naruto wins else print "Sasuke".Sample Input:-
0101
Sample Output:-
Sasuke
Explanation:-
First Naruto will delete the character 0 from index 1. string:- 101
Then Sasuke will delete the character 1 from index 1. string 01
It doesn't matter which character Naruto removes now, Sasuke will definitely win.
Sample Input:-
1111
Sample Output:-
Naruto
Explanation:-
All the characters in the string are already same hence Naruto wins, 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 1000000000000007
#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 cnt[max1];
signed main(){
string s;
cin>>s;
int cnt=0;
FOR(i,s.length()){
if(s[i]=='0'){cnt++;}
}
if(cnt==s.length() || cnt==0){out("Naruto");return 0;}
if(s.length()%2==0){
out("Sasuke");
}
else{
out("Naruto");
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, if the year is a multiple of 100 and not a multiple of 400, then it is not a leap year.<b>User Task:</b>
Complete the function <b>LeapYear()</b> that takes integer n as a parameter.
<b>Constraint:</b>
1 <= n <= 5000If it is a leap year then print <b>YES</b> and if it is not a leap year, then print <b>NO</b>Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code: n = int(input())
if (n%4==0 and n%100!=0 or n%400==0):
print("YES")
elif n==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: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, if the year is a multiple of 100 and not a multiple of 400, then it is not a leap year.<b>User Task:</b>
Complete the function <b>LeapYear()</b> that takes integer n as a parameter.
<b>Constraint:</b>
1 <= n <= 5000If it is a leap year then print <b>YES</b> and if it is not a leap year, then print <b>NO</b>Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code:
import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int n = scanner.nextInt();
LeapYear(n);
}
static void LeapYear(int year){
if(year%400==0 || (year%100 != 0 && year%4==0)){System.out.println("YES");}
else {
System.out.println("NO");}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){cout<<"FizzBuzz"<<" ";}
else if(i%5==0){cout<<"Buzz ";}
else if(i%3==0){cout<<"Fizz ";}
else{cout<<i<<" ";}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, 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 x= sc.nextInt();
fizzbuzz(x);
}
static void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("FizzBuzz ");}
else if(i%5==0){System.out.print("Buzz ");}
else if(i%3==0){System.out.print("Fizz ");}
else{System.out.print(i+" ");}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: def fizzbuzz(n):
for i in range (1,n+1):
if (i%3==0 and i%5==0):
print("FizzBuzz",end=' ')
elif i%3==0:
print("Fizz",end=' ')
elif i%5==0:
print("Buzz",end=' ')
else:
print(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 for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){printf("FizzBuzz ");}
else if(i%5==0){printf("Buzz ");}
else if(i%3==0){printf("Fizz ");}
else{printf("%d ",i);}
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
Sample Output:
No, I have written this Solution Code: m,n = map(int , input().split())
if (m%n==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: Given an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
Sample Output:
No, 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 n,m;
cin>>n>>m;
if(n%m==0)
cout<<"Yes";
else
cout<<"No";
#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 an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
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();
Long m = sc.nextLong();
if(n%m==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: You are given a matrix <b>A</b> of dimensions <b>n x m</b>. The task is to perform <b>boundary traversal</b> on the matrix in clockwise manner.The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase has two lines of input. The first line contains dimensions of the matrix A, n and m. The second line contains n*m elements separated by spaces.
Constraints:
1 <= T <= 100
1 <= n, m <= 30
0 <= A[i][j] <= 100For each testcase, in a new line, print the boundary traversal of the matrix A.Input:
4
4 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
3 4
12 11 10 9 8 7 6 5 4 3 2 1
1 4
1 2 3 4
4 1
1 2 3 4
Output:
1 2 3 4 8 12 16 15 14 13 9 5
12 11 10 9 5 1 2 3 4 8
1 2 3 4
1 2 3 4
Explanation:
Testcase1: The matrix is:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
The boundary traversal is 1 2 3 4 8 12 16 15 14 13 9 5
Testcase 2: Boundary Traversal will be 12 11 10 9 5 1 2 3 4 8.
Testcase 3: Boundary Traversal will be 1 2 3 4.
Testcase 4: Boundary Traversal will be 1 2 3 4., I have written this Solution Code:
def boundaryTraversal(matrix, N, M): #N = 3, M = 4
start_row = 0
start_col = 0
count = 0
if N != 1 and M != 1:
total = 2 * (N - 1) + 2 * (M - 1)
else:
total = max(M, N)
#First row
for i in range(start_col, M, 1): #0,1, 2, 3
count += 1
print(matrix[start_row][i], end = " ")
if count == total:
return
start_row += 1
#Last column
for i in range(start_row, N, 1): # 1, 2
count += 1
print(matrix[i][M - 1], end = " ")
if count == total:
return
M -= 1
#Last Row
for i in range(M - 1, start_col - 1, -1): # 2, 1, 0
count += 1
print(matrix[N - 1][i], end = " ")
if count == total:
return
#First Column
N -= 1 # 2
for i in range(N - 1, start_row - 1, -1):
count += 1
print(matrix[i][start_col], end = " ")
start_col += 1
testcase = int(input().strip())
for test in range(testcase):
dim = input().strip().split(" ")
N = int(dim[0])
M = int(dim[1])
matrix = []
elements = input().strip().split(" ") #N * M
start = 0
for i in range(N):
matrix.append(elements[start : start + M]) # (0, M - 1) | (M, 2*m-1)
start = start + M
boundaryTraversal(matrix, N, M)
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix <b>A</b> of dimensions <b>n x m</b>. The task is to perform <b>boundary traversal</b> on the matrix in clockwise manner.The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase has two lines of input. The first line contains dimensions of the matrix A, n and m. The second line contains n*m elements separated by spaces.
Constraints:
1 <= T <= 100
1 <= n, m <= 30
0 <= A[i][j] <= 100For each testcase, in a new line, print the boundary traversal of the matrix A.Input:
4
4 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
3 4
12 11 10 9 8 7 6 5 4 3 2 1
1 4
1 2 3 4
4 1
1 2 3 4
Output:
1 2 3 4 8 12 16 15 14 13 9 5
12 11 10 9 5 1 2 3 4 8
1 2 3 4
1 2 3 4
Explanation:
Testcase1: The matrix is:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
The boundary traversal is 1 2 3 4 8 12 16 15 14 13 9 5
Testcase 2: Boundary Traversal will be 12 11 10 9 5 1 2 3 4 8.
Testcase 3: Boundary Traversal will be 1 2 3 4.
Testcase 4: Boundary Traversal will be 1 2 3 4., I have written this Solution Code: import java.io.*;
import java.lang.*;
import java.util.*;
class Main
{
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-- > 0)
{
int n1 = sc.nextInt();
int m1 = sc.nextInt();
int arr1[][] = new int[n1][m1];
for(int i = 0; i < n1; i++)
{
for(int j = 0; j < m1; j++)
arr1[i][j] = sc.nextInt();
}
boundaryTraversal(n1, m1,arr1);
System.out.println();
}
}
static void boundaryTraversal( int n1, int m1, int arr1[][])
{
// base cases
if(n1 == 1)
{
int i = 0;
while(i < m1)
System.out.print(arr1[0][i++] + " ");
}
else if(m1 == 1)
{
int i = 0;
while(i < n1)
System.out.print(arr1[i++][0]+" ");
}
else
{
// traversing the first row
for(int j=0;j<m1;j++)
{
System.out.print(arr1[0][j]+" ");
}
// traversing the last column
for(int j=1;j<n1;j++)
{
System.out.print(arr1[j][m1-1]+ " ");
}
// traversing the last row
for(int j=m1-2;j>=0;j--)
{
System.out.print(arr1[n1-1][j]+" ");
}
// traversing the first column
for(int j=n1-2;j>=1;j--)
{
System.out.print(arr1[j][0]+" ");
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are fighting against a monster.
The monster has health M while you have initial health Y.
To defeat him, you need health strictly greater than the monster's health.
You have access to two types of pills, red and green. Eating a red pill adds R to your health while eating a green pill multiplies your health by G.
Determine if it is possible to defeat the monster by eating at most one pill of any kind.Input contains four integers M (0 <= M <= 10<sup>4</sup>), Y (0 <= Y <= 10<sup>4</sup>), R (0 <= R <= 10<sup>4</sup>) and G (1 <= G <= 10<sup>4</sup>).Print 1 if it is possible to defeat the monster and 0 if it is impossible to defeat it.Sample Input 1:
10 2 2 2
Sample Output 1:
0
Sample Input 2:
8 7 2 1
Sample Output 2:
1
Explanation for Sample 2:
You can eat a red pill to make your health : 7 + 2 = 9 > 8., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException
{
InputStreamReader pk = new InputStreamReader(System.in);
BufferedReader in = new BufferedReader(pk);
String[] strNums;
int num[] = new int[4];
strNums = in.readLine().split(" ");
for (int i = 0; i < strNums.length; i++) {
num[i] = Integer.parseInt(strNums[i]);
}
if((num[0]<(num[1]+num[2])) || (num[0]<(num[1]*num[3])))
System.out.println("1");
else
System.out.println("0");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are fighting against a monster.
The monster has health M while you have initial health Y.
To defeat him, you need health strictly greater than the monster's health.
You have access to two types of pills, red and green. Eating a red pill adds R to your health while eating a green pill multiplies your health by G.
Determine if it is possible to defeat the monster by eating at most one pill of any kind.Input contains four integers M (0 <= M <= 10<sup>4</sup>), Y (0 <= Y <= 10<sup>4</sup>), R (0 <= R <= 10<sup>4</sup>) and G (1 <= G <= 10<sup>4</sup>).Print 1 if it is possible to defeat the monster and 0 if it is impossible to defeat it.Sample Input 1:
10 2 2 2
Sample Output 1:
0
Sample Input 2:
8 7 2 1
Sample Output 2:
1
Explanation for Sample 2:
You can eat a red pill to make your health : 7 + 2 = 9 > 8., I have written this Solution Code: a = []
a = list(map(int, input().split()))
if a[1]+a[2] > a[0] or a[1]*a[3] > a[0]:
defeat = 1
else:
defeat = 0
print(defeat), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are fighting against a monster.
The monster has health M while you have initial health Y.
To defeat him, you need health strictly greater than the monster's health.
You have access to two types of pills, red and green. Eating a red pill adds R to your health while eating a green pill multiplies your health by G.
Determine if it is possible to defeat the monster by eating at most one pill of any kind.Input contains four integers M (0 <= M <= 10<sup>4</sup>), Y (0 <= Y <= 10<sup>4</sup>), R (0 <= R <= 10<sup>4</sup>) and G (1 <= G <= 10<sup>4</sup>).Print 1 if it is possible to defeat the monster and 0 if it is impossible to defeat it.Sample Input 1:
10 2 2 2
Sample Output 1:
0
Sample Input 2:
8 7 2 1
Sample Output 2:
1
Explanation for Sample 2:
You can eat a red pill to make your health : 7 + 2 = 9 > 8., I have written this Solution Code: //HEADER FILES AND NAMESPACES
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
// #include <sys/resource.h>
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
// DEFINE STATEMENTS
const long long infty = 1e18;
#define num1 1000000007
#define num2 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define REPd(i,a,n) for(ll i=a; i>=n; i--)
#define pb push_back
#define pob pop_back
#define f first
#define s second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define all(x) x.begin(), x.end()
#define M_PI 3.14159265358979323846
#define epsilon (double)(0.000000001)
#define popcount __builtin_popcountll
#define fileio(x) freopen("input.txt", "r", stdin); freopen(x, "w", stdout);
#define out(x) cout << ((x) ? "Yes\n" : "No\n")
#define sz(x) x.size()
#define start_clock() auto start_time = std::chrono::high_resolution_clock::now();
#define measure() auto end_time = std::chrono::high_resolution_clock::now(); cerr << (end_time - start_time)/std::chrono::milliseconds(1) << "ms" << endl;
typedef long long ll;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
// DEBUG FUNCTIONS
#ifdef LOCALY
template<typename T>
void __p(T a) {
cout<<a;
}
template<typename T, typename F>
void __p(pair<T, F> a) {
cout<<"{";
__p(a.first);
cout<<",";
__p(a.second);
cout<<"}";
}
template<typename T>
void __p(std::vector<T> a) {
cout<<"{";
for(auto it=a.begin(); it<a.end(); it++)
__p(*it),cout<<",}"[it+1==a.end()];
}
template<typename T>
void __p(std::set<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T>
void __p(std::multiset<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T, typename F>
void __p(std::map<T,F> a) {
cout<<"{\n";
for(auto it=a.begin(); it!=a.end();++it)
{
__p(it->first);
cout << ": ";
__p(it->second);
cout<<"\n";
}
cout << "}\n";
}
template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) {
__p(a1);
__p(a...);
}
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
cout<<name<<" : ";
__p(arg1);
cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
int bracket=0,i=0;
for(;; i++)
if(names[i]==','&&bracket==0)
break;
else if(names[i]=='(')
bracket++;
else if(names[i]==')')
bracket--;
const char *comma=names+i;
cout.write(names,comma-names)<<" : ";
__p(arg1);
cout<<" | ";
__f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<" ", __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...)
#define error(...)
#endif
// DEBUG FUNCTIONS END
// CUSTOM HASH TO SPEED UP UNORDERED MAP AND TO AVOID FORCED CLASHES
struct custom_hash {
static uint64_t splitmix64(uint64_t x) {
x += 0x9e3779b97f4a7c15;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
return x ^ (x >> 31);
}
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
return splitmix64(x + FIXED_RANDOM);
}
};
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count()); // FOR RANDOM NUMBER GENERATION
ll mod_exp(ll a, ll b, ll c)
{
ll res=1; a=a%c;
while(b>0)
{
if(b%2==1)
res=(res*a)%c;
b/=2;
a=(a*a)%c;
}
return res;
}
ll mymod(ll a,ll b)
{
return (((a = a%b) < 0) ? a + b : a);
}
ll gcdExtended(ll,ll,ll *,ll *);
ll modInverse(ll a, ll m)
{
ll x, y;
ll g = gcdExtended(a, m, &x, &y);
g++; //this line was added just to remove compiler warning
ll res = (x%m + m) % m;
return res;
}
ll gcdExtended(ll a, ll b, ll *x, ll *y)
{
if (a == 0)
{
*x = 0, *y = 1;
return b;
}
ll x1, y1;
ll gcd = gcdExtended(b%a, a, &x1, &y1);
*x = y1 - (b/a) * x1;
*y = x1;
return gcd;
}
struct Graph
{
vector<vector<int>> adj;
Graph(int n)
{
adj.resize(n+1);
}
void add_edge(int a, int b, bool directed = false)
{
adj[a].pb(b);
if(!directed) adj[b].pb(a);
}
};
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll M, Y, R, G;
cin >> M >> Y >> R >> G;
ll maxm = Y;
maxm = max(maxm, Y+R);
maxm = max(maxm, Y*G);
cout << (M < maxm) << "\n";
return 0;
}
/*
1. Check borderline constraints. Can a variable you are dividing by be 0?
2. Use ll while using bitshifts
3. Do not erase from set while iterating it
4. Initialise everything
5. Read the task carefully, is something unique, sorted, adjacent, guaranteed??
6. DO NOT use if(!mp[x]) if you want to iterate the map later
7. Are you using i in all loops? Are the i's conflicting?
*/
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: t=int(input())
while t!=0:
m,n=input().split()
m,n=int(m),int(n)
for i in range(m):
arr=input().strip()
if '1' in arr:
arr='1 '*n
else:
arr='0 '*n
print(arr)
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define N 1000
int a[N][N];
// Driver code
int main()
{
int t;
cin>>t;
while(t--){
int n,m;
cin>>n>>m;
bool b[n];
for(int i=0;i<n;i++){
b[i]=false;
}
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
cin>>a[i][j];
if(a[i][j]==1){
b[i]=true;
}
}
}
for(int i=0;i<n;i++){
if(b[i]){
for(int j=0;j<m;j++){
cout<<1<<" ";
}}
else{
for(int j=0;j<m;j++){
cout<<0<<" ";
}
}
cout<<endl;
}
}}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args) throws Exception{
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader bf = new BufferedReader(isr);
int t = Integer.parseInt(bf.readLine());
while (t-- > 0){
String inputs[] = bf.readLine().split(" ");
int m = Integer.parseInt(inputs[0]);
int n = Integer.parseInt(inputs[1]);
String[] matrix = new String[m];
for(int i=0; i<m; i++){
matrix[i] = bf.readLine();
}
StringBuffer ones = new StringBuffer("");
StringBuffer zeros = new StringBuffer("");
for(int i=0; i<n; i++){
ones.append("1 ");
zeros.append("0 ");
}
for(int i=0; i<m; i++){
if(matrix[i].contains("1")){
System.out.println(ones);
}else{
System.out.println(zeros);
}
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array of N integers a[], and Q queries. For each query, you will be given a positive integer K and your task is to print the number of elements in array a[] that are smaller than or equal to K.<b>In case of Java only</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>smallerElements()</b> that takes the array a[], integer N and integer k as arguments.
<b>Custom Input</b>
The first line of input contains a single integer N.
The second line of input contains N space- separated integers depicting the values of the array.
The third line of input contains a single integer Q, the number of queries.
Each of the next Q lines of input contain a single integer, the value of K.
<b>Constraints:-</b>
1 <= N <= 10<sup>5</sup>
1 <= K, Arr[i] <= 10<sup>12</sup>
1 <= Q <= 10<sup>4</sup>Return the count of elements smaller than or equal to K.Sample Input:-
5
2 5 6 11 15
5
2
4
8
1
16
Sample Output:-
1
1
3
0
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 1001
#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);
}
signed main(){
int n;
cin>>n;
vector<int> v(n);
FOR(i,n){
cin>>v[i];}
int q;
cin>>q;
int x;
while(q--){
cin>>x;
auto it = upper_bound(v.begin(),v.end(),x);
out(it-v.begin());
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array of N integers a[], and Q queries. For each query, you will be given a positive integer K and your task is to print the number of elements in array a[] that are smaller than or equal to K.<b>In case of Java only</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>smallerElements()</b> that takes the array a[], integer N and integer k as arguments.
<b>Custom Input</b>
The first line of input contains a single integer N.
The second line of input contains N space- separated integers depicting the values of the array.
The third line of input contains a single integer Q, the number of queries.
Each of the next Q lines of input contain a single integer, the value of K.
<b>Constraints:-</b>
1 <= N <= 10<sup>5</sup>
1 <= K, Arr[i] <= 10<sup>12</sup>
1 <= Q <= 10<sup>4</sup>Return the count of elements smaller than or equal to K.Sample Input:-
5
2 5 6 11 15
5
2
4
8
1
16
Sample Output:-
1
1
3
0
5, I have written this Solution Code: static int smallerElements(int a[], int n, int k){
int l=0;
int h=n-1;
int m;
int ans=n;
while(l<=h){
m=l+h;
m/=2;
if(a[m]<=k){
l=m+1;
}
else{
h=m-1;
ans=m;
}
}
return ans;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
1) S is a palindrome.
2) Let N be the length of S. The string formed by the 1- st through ((N−1)/2)<sub>th</sub> characters of S is a palindrome.
3) The string consisting of the (N+3)/2- st through N<sub>th</sub> characters of S is a palindrome.
Determine whether S is a strong palindrome.The first and the only line of the input contains a single string S.
<b>Constraints:</b>
3 ≤ |S| ≤ 99If S is a strong palindrome, print "Yes"; otherwise, print "No".<b>Sample Input 1:</b>
akasaka
<b>Sample Output 1:</b>
Yes
<b>Sample Explanation 1:</b>
1) S is Akasaka.
2) The string formed by the 1<sub>st</sub> through the 3<sub>rd</sub> characters is aka.
3) The string formed by the 5<sub>th</sub> through the 7<sub>th</sub> characters is aka. All of these are palindromes, so S is a strong palindrome., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define pi acos(-1.0)
#define mod 1000000007
#define endl '\n'
#define ha cout<<"YES"<<endl
#define na cout<<"NO"<<endl
#define mod 1000000007
#define ll long long
int main()
{
string s;
cin >> s;
int f=0,fl=0,flag=0;
int l=s.length();//7
for(int i=0,j=l-1;i<l/2;i++,j--)
{
if(s[i]!=s[j])
{
f=1;
break;
}
}
for(int i=0,j=l/2-1;i<l/2;i++,j--)
{
if(s[i]!=s[j])
{
fl=1;
break;
}
}
for(int i=l/2+1,j=l-1;i<j;i++,j--)
{
if(s[i]!=s[j])
{
flag=1;
break;
}
}
//cout<<f<<fl<<flag<<endl;
if (f == 0 && fl==0 && flag==0) cout<<"Yes";
else cout<<"No";
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<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>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code: static int focal_length(int R, char Mirror)
{
int f=R/2;
if((R%2==1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<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>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code:
int focal_length(int R, char Mirror)
{
int f=R/2;
if((R&1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<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>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code:
int focal_length(int R, char Mirror)
{
int f=R/2;
if((R&1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<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>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code: def focal_length(R,Mirror):
f=R/2;
if(Mirror == ')'):
f=-f
if R%2==1:
f=f-1
return int(f)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers your task is to calculate the maximum integer among the given integers.The input contains three integers a, b, and c
<b>Constraint:</b>
1<=integers<=10000Print the maximum integer among the given integers.Sample Input:-
2 6 3
Sample Output:-
6
Sample Input:-
48 100 100
Sample Output:
100, I have written this Solution Code: a,b,c=[int(a) for a in input().split()]
print(max(a,b,c)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers your task is to calculate the maximum integer among the given integers.The input contains three integers a, b, and c
<b>Constraint:</b>
1<=integers<=10000Print the maximum integer among the given integers.Sample Input:-
2 6 3
Sample Output:-
6
Sample Input:-
48 100 100
Sample Output:
100, I have written this Solution Code: import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int p = scanner.nextInt();
int tm = scanner.nextInt();
int r = scanner.nextInt();
int intrst = MaxInteger(p,tm,r);
System.out.println(intrst);
}
static int MaxInteger(int a ,int b, int c){
if(a>=b && a>=c){return a;}
if(b>=a && b>=c){return b;}
return c;}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of N integers, find whether there exists three consecutive same integers in the array.The first line of the input contains an integer N.
The second line contains N space separated integers of the array A.
Constraints
3 <= N <= 1000
1 <= A[i] <= 100Output "Yes" if there exists three consecutive equal integers in the array, else output "No" (without quotes).Sample Input
5
1 2 2 2 4
Sample Output
Yes
Explanation: The segment [2, 2, 2] follows the criterion.
Sample Input
5
1 2 2 3 4
Sample Output
No, I have written this Solution Code: n=int(input())
li = list(map(int,input().strip().split()))
for i in range(0,n-2):
if li[i]==li[i+1] and li[i+1]==li[i+2]:
print("Yes",end="")
exit()
print("No",end=""), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of N integers, find whether there exists three consecutive same integers in the array.The first line of the input contains an integer N.
The second line contains N space separated integers of the array A.
Constraints
3 <= N <= 1000
1 <= A[i] <= 100Output "Yes" if there exists three consecutive equal integers in the array, else output "No" (without quotes).Sample Input
5
1 2 2 2 4
Sample Output
Yes
Explanation: The segment [2, 2, 2] follows the criterion.
Sample Input
5
1 2 2 3 4
Sample Output
No, 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(n);
bool fl = false;
For(i, 0, n){
cin>>a[i];
if(i>=2){
if(a[i]==a[i-1] && a[i]==a[i-2]){
fl = true;
}
}
}
if(fl){
cout<<"Yes";
}
else
cout<<"No";
}
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 an array A of N integers, find whether there exists three consecutive same integers in the array.The first line of the input contains an integer N.
The second line contains N space separated integers of the array A.
Constraints
3 <= N <= 1000
1 <= A[i] <= 100Output "Yes" if there exists three consecutive equal integers in the array, else output "No" (without quotes).Sample Input
5
1 2 2 2 4
Sample Output
Yes
Explanation: The segment [2, 2, 2] follows the criterion.
Sample Input
5
1 2 2 3 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);
int n = sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
for(int i=0;i<n-2;i++){
if(a[i]==a[i+1] && a[i+1]==a[i+2]){
System.out.print("Yes");
return;
}
}
System.out.print("No");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import static java.lang.System.out;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
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());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader reader = new FastReader();
int n = reader.nextInt();
String S = reader.next();
int ncount = 0;
int tcount = 0;
for (char c : S.toCharArray()) {
if (c == 'N') ncount++;
else tcount++;
}
if (ncount > tcount) {
out.print("Nutan\n");
} else {
out.print("Tusla\n");
}
out.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: n = int(input())
s = input()
a1 = s.count('N')
a2 = s.count('T')
if(a1 > a2):
print("Nutan")
else:
print('Tusla'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: //Author: Xzirium
//Time and Date: 02:18:28 24 March 2022
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//-----------------------------------------------------------------------------------------------------------//
READV(N);
string S;
cin>>S;
ll n=0,t=0;
FORI(i,0,N)
{
if(S[i]=='N')
{
n++;
}
else if(S[i]=='T')
{
t++;
}
}
if(n>t)
{
cout<<"Nutan"<<endl;
}
else
{
cout<<"Tusla"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Take an integer as input and print it.The first line contains integer as input.
<b>Constraints</b>
1 <= N <= 10Print the input integer in a single lineSample Input:-
2
Sample Output:-
2
Sample Input:-
4
Sample Output:-
4, I have written this Solution Code: /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Main
{
public static void printVariable(int variable){
System.out.println(variable);
}
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int num = sc.nextInt();
printVariable(num);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given length l, width b and height h of a cuboid. The task is to find the total surface area and volume of cuboid.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions:-
<b>Surface_Area()</b> that takes the integer l, b and h as parameter.
<b>Volume()</b> that takes the integer l, b and h as parameter.
<b>Constraints:</b>
1 <= l, b, h <= 10^2Return the surface area of cuboid in function Surface_Area() and return Volume of cuboid in function Volume().Sample Input:
1 2 3
Sample Output:
22 6
Explanation:
Test Case 1: The total surface area for the given cuboid is 22 and its volume 6., I have written this Solution Code: static int Surface_Area(int l, int b, int h){
return 2*(l*b + b*h + h*l);
}
static int Volume(int l, int b, int h){
return l*b*h;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers, find two numbers such that they add up to a specific target number k.
Output the indices of the elements that add up to the sum (array is 1 indexed).
If multiple solutions exist, output the one where index2 is minimum. If there are multiple solutions with the minimum index2, choose the one with minimum index1 out of them.
For example:
array A[] = 1 1 3 2 2
k = 3
Output: 1 4
Explanation: pair of indices which gives target k are: (1, 4), (1, 5), (2, 4), (2, 5). Among all such pairs (1, 4) satisfies above condition.The first line contain integers N and k, the number of elements in the array and the target sum.
The second line of the input contains N singly spaces integers.
Constraints:-
2 <= N <= 100000
1 <= A[i] <= 1000000000
1 <= k <= 2000000000Output two integers the indices of the two elements. It is guaranteed that the ans will always existSample Input
5 3
1 1 3 2 2
Sample Output
1 4
Explanation: The satisfying pairs are (1, 4), (2, 4), (1, 5), (2, 5).
Sample Input:
2 2
1 1
Sample Output:
1 2, I have written this Solution Code: import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.HashMap;
import java.util.InputMismatchException;
import static java.lang.Math.pow;
class InputReader {
private boolean finished = false;
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public static InputReader getInputReader(boolean readFromTextFile) throws FileNotFoundException {
return ((readFromTextFile) ? new InputReader(new FileInputStream("src/input.txt"))
: new InputReader(System.in));
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int peek() {
if (numChars == -1) {
return -1;
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
return -1;
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String nextString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private String readLine0() {
StringBuilder buf = new StringBuilder();
int c = read();
while (c != '\n' && c != -1) {
if (c != '\r') {
buf.appendCodePoint(c);
}
c = read();
}
return buf.toString();
}
public String readLine() {
String s = readLine0();
while (s.trim().length() == 0) {
s = readLine0();
}
return s;
}
public String readLine(boolean ignoreEmptyLines) {
if (ignoreEmptyLines) {
return readLine();
} else {
return readLine0();
}
}
public BigInteger readBigInteger() {
try {
return new BigInteger(nextString());
} catch (NumberFormatException e) {
throw new InputMismatchException();
}
}
public char nextCharacter() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
return (char) c;
}
public double nextDouble() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E') {
return res * pow(10, nextInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c)) {
if (c == 'e' || c == 'E') {
return res * pow(10, nextInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public boolean isExhausted() {
int value;
while (isSpaceChar(value = peek()) && value != -1) {
read();
}
return value == -1;
}
public String next() {
return nextString();
}
public SpaceCharFilter getFilter() {
return filter;
}
public void setFilter(SpaceCharFilter filter) {
this.filter = filter;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
public int[] nextIntArray(int n) {
int[] array = new int[n];
for (int i = 0; i < n; ++i) array[i] = nextInt();
return array;
}
public int[] nextSortedIntArray(int n) {
int array[] = nextIntArray(n);
Arrays.sort(array);
return array;
}
public int[] nextSumIntArray(int n) {
int[] array = new int[n];
array[0] = nextInt();
for (int i = 1; i < n; ++i) array[i] = array[i - 1] + nextInt();
return array;
}
public long[] nextLongArray(int n) {
long[] array = new long[n];
for (int i = 0; i < n; ++i) array[i] = nextLong();
return array;
}
public long[] nextSumLongArray(int n) {
long[] array = new long[n];
array[0] = nextInt();
for (int i = 1; i < n; ++i) array[i] = array[i - 1] + nextInt();
return array;
}
public long[] nextSortedLongArray(int n) {
long array[] = nextLongArray(n);
Arrays.sort(array);
return array;
}
}
class Main {
public static void main (String[] args) throws FileNotFoundException {
InputReader sc = InputReader.getInputReader(false);
int n = sc.nextInt();
long target = sc.nextLong();
long[] arr = new long[n+1];
HashMap<Long ,Integer> mp = new HashMap<>();
for(int i=1;i<=n;i++){
arr[i] = sc.nextLong();
}
for(int i=1;i<=n;i++){
if (mp.containsKey(target-arr[i])){
System.out.println(mp.get(target-arr[i])+ " "+ i);
break;
}
if(!mp.containsKey(arr[i])){
mp.put(arr[i] , i);
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers, find two numbers such that they add up to a specific target number k.
Output the indices of the elements that add up to the sum (array is 1 indexed).
If multiple solutions exist, output the one where index2 is minimum. If there are multiple solutions with the minimum index2, choose the one with minimum index1 out of them.
For example:
array A[] = 1 1 3 2 2
k = 3
Output: 1 4
Explanation: pair of indices which gives target k are: (1, 4), (1, 5), (2, 4), (2, 5). Among all such pairs (1, 4) satisfies above condition.The first line contain integers N and k, the number of elements in the array and the target sum.
The second line of the input contains N singly spaces integers.
Constraints:-
2 <= N <= 100000
1 <= A[i] <= 1000000000
1 <= k <= 2000000000Output two integers the indices of the two elements. It is guaranteed that the ans will always existSample Input
5 3
1 1 3 2 2
Sample Output
1 4
Explanation: The satisfying pairs are (1, 4), (2, 4), (1, 5), (2, 5).
Sample Input:
2 2
1 1
Sample Output:
1 2, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
long long n,k;
cin>>n>>k;
long long a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
unordered_map<long long ,int > m;
long long x;
for(int i=0;i<n;i++){
x=k-a[i];
if(m.find(x)!=m.end()){
cout<<m[x]+1<<" "<<i+1;
return 0;
}
if(m.find(a[i])==m.end()){m[a[i]]=i;}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given 2 parameters: a low and high number. Your goal is to print all numbers between low and high,
and for each of these numbers print whether or not the number is divisible by 3. If the number is divisible by 3,
print the word "div3" directly after the number.2 numbers, one will be low and other high.
0<=low<=high<=10000If the number is divisible by 3, print the word "div3" directly after the number.Sample input:-
1 6
Sample output:-
1 2 3 div3 4 5 6 div3, 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 low = sc.nextInt();
int high = sc.nextInt();
for(int i = low; i <= high; i++){
if(i%3 == 0){
System.out.print(i);
System.out.print(" ");
System.out.print("div"+3);
System.out.print(" ");
}
else{
System.out.print(i);
System.out.print(" ");
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given 2 parameters: a low and high number. Your goal is to print all numbers between low and high,
and for each of these numbers print whether or not the number is divisible by 3. If the number is divisible by 3,
print the word "div3" directly after the number.2 numbers, one will be low and other high.
0<=low<=high<=10000If the number is divisible by 3, print the word "div3" directly after the number.Sample input:-
1 6
Sample output:-
1 2 3 div3 4 5 6 div3, I have written this Solution Code: inp = input("").split(" ")
init = []
for i in range(int(inp[0]),int(inp[1])+1):
if(i%3 == 0):
init.append(str(i)+" div3")
else:
init.append(str(i))
print(" ".join(init)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given 2 parameters: a low and high number. Your goal is to print all numbers between low and high,
and for each of these numbers print whether or not the number is divisible by 3. If the number is divisible by 3,
print the word "div3" directly after the number.2 numbers, one will be low and other high.
0<=low<=high<=10000If the number is divisible by 3, print the word "div3" directly after the number.Sample input:-
1 6
Sample output:-
1 2 3 div3 4 5 6 div3, I have written this Solution Code: function test_divisors(low, high) {
// we'll store all numbers and strings within an array
// instead of printing directly to the console
const output = [];
for (let i = low; i <= high; i++) {
// simply store the current number in the output array
output.push(i);
// check if the current number is evenly divisible by 3
if (i % 3 === 0) { output.push('div3'); }
}
// return all numbers and strings
console.log(output.join(" "));
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a 2D matrix of size [M, N], Q number of queries. In each query, you will be given a number X to check whether it is present in the matrix or not.The first line contains three integers M(number of rows), N(Number of columns), and Q(number of queries)
Next M lines contain N integers which are the elements of the matrix.
Next, Q lines will contain a single integer X.
Constraints:-
1<=M,N<=1000
1<=Q<=10000
1<=X, Arr[i]<=1000000000For each query, in a new line print "Yes" if the element is present in matrix or print "No" if the element is absent.Input:-
3 3 2
1 2 3
5 6 7
8 9 10
7
11
Output:-
Yes
No
Input:-
3 4 4
4 8 11 14
15 54 45 47
1 2 3 4
5
15
45
26
Output:-
No
Yes
Yes
No, I have written this Solution Code: import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
// don't change the name of this class
// you can add inner classes if needed
class Main {
public static void main (String[] args) {
// Your code here
Scanner sc = new Scanner(System.in);
int m = sc.nextInt();
int n = sc.nextInt();
int q = sc.nextInt();
int mat[] = new int[m*n];
int matSize = m*n;
for(int i = 0; i < m*n; i++)
{
int ele = sc.nextInt();
mat[i] = ele;
}
Arrays.sort(mat);
for(int i = 1; i <= q; i++)
{
int qs = sc.nextInt();
System.out.println(isPresent(mat, matSize, qs));
}
}
static String isPresent(int mat[], int size, int ele)
{
int l = 0, h = size-1;
while(l <= h)
{
int mid = l + (h-l)/2;
if(mat[mid] == ele)
return "Yes";
else if(mat[mid] > ele)
h = mid - 1;
else l = mid+1;
}
return "No";
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a 2D matrix of size [M, N], Q number of queries. In each query, you will be given a number X to check whether it is present in the matrix or not.The first line contains three integers M(number of rows), N(Number of columns), and Q(number of queries)
Next M lines contain N integers which are the elements of the matrix.
Next, Q lines will contain a single integer X.
Constraints:-
1<=M,N<=1000
1<=Q<=10000
1<=X, Arr[i]<=1000000000For each query, in a new line print "Yes" if the element is present in matrix or print "No" if the element is absent.Input:-
3 3 2
1 2 3
5 6 7
8 9 10
7
11
Output:-
Yes
No
Input:-
3 4 4
4 8 11 14
15 54 45 47
1 2 3 4
5
15
45
26
Output:-
No
Yes
Yes
No, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define N 1000000
long a[N];
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n,m,q;
cin>>n>>m>>q;
n=n*m;
long long sum=0,sum1=0;
for(int i=0;i<n;i++){
cin>>a[i];
}
sort(a,a+n);
while(q--){
long x;
cin>>x;
int l=0;
int r=n-1;
while (r >= l) {
int mid = l + (r - l) / 2;
if (a[mid] == x) {
cout<<"Yes"<<endl;goto f;}
if (a[mid] > x)
{
r=mid-1;
}
else {l=mid+1;
}
}
cout<<"No"<<endl;
f:;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a 2D matrix of size [M, N], Q number of queries. In each query, you will be given a number X to check whether it is present in the matrix or not.The first line contains three integers M(number of rows), N(Number of columns), and Q(number of queries)
Next M lines contain N integers which are the elements of the matrix.
Next, Q lines will contain a single integer X.
Constraints:-
1<=M,N<=1000
1<=Q<=10000
1<=X, Arr[i]<=1000000000For each query, in a new line print "Yes" if the element is present in matrix or print "No" if the element is absent.Input:-
3 3 2
1 2 3
5 6 7
8 9 10
7
11
Output:-
Yes
No
Input:-
3 4 4
4 8 11 14
15 54 45 47
1 2 3 4
5
15
45
26
Output:-
No
Yes
Yes
No, I have written this Solution Code: import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
// don't change the name of this class
// you can add inner classes if needed
class Main {
public static void main (String[] args) {
// Your code here
Scanner sc = new Scanner(System.in);
int m = sc.nextInt();
int n = sc.nextInt();
int q = sc.nextInt();
int mat[] = new int[m*n];
int matSize = m*n;
for(int i = 0; i < m*n; i++)
{
int ele = sc.nextInt();
mat[i] = ele;
}
Arrays.sort(mat);
for(int i = 1; i <= q; i++)
{
int qs = sc.nextInt();
System.out.println(isPresent(mat, matSize, qs));
}
}
static String isPresent(int mat[], int size, int ele)
{
int l = 0, h = size-1;
while(l <= h)
{
int mid = l + (h-l)/2;
if(mat[mid] == ele)
return "Yes";
else if(mat[mid] > ele)
h = mid - 1;
else l = mid+1;
}
return "No";
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a 2D matrix of size [M, N], Q number of queries. In each query, you will be given a number X to check whether it is present in the matrix or not.The first line contains three integers M(number of rows), N(Number of columns), and Q(number of queries)
Next M lines contain N integers which are the elements of the matrix.
Next, Q lines will contain a single integer X.
Constraints:-
1<=M,N<=1000
1<=Q<=10000
1<=X, Arr[i]<=1000000000For each query, in a new line print "Yes" if the element is present in matrix or print "No" if the element is absent.Input:-
3 3 2
1 2 3
5 6 7
8 9 10
7
11
Output:-
Yes
No
Input:-
3 4 4
4 8 11 14
15 54 45 47
1 2 3 4
5
15
45
26
Output:-
No
Yes
Yes
No, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define N 1000000
long a[N];
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n,m,q;
cin>>n>>m>>q;
n=n*m;
long long sum=0,sum1=0;
for(int i=0;i<n;i++){
cin>>a[i];
}
sort(a,a+n);
while(q--){
long x;
cin>>x;
int l=0;
int r=n-1;
while (r >= l) {
int mid = l + (r - l) / 2;
if (a[mid] == x) {
cout<<"Yes"<<endl;goto f;}
if (a[mid] > x)
{
r=mid-1;
}
else {l=mid+1;
}
}
cout<<"No"<<endl;
f:;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
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