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For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: import math
n = int(input())
n=n+1
if n<3:
print(0)
else:
primes=[1]*(n//2)
for i in range(3,int(math.sqrt(n))+1,2):
if primes[i//2]:primes[i*i//2::i]=[0]*((n-i*i-1)//(2*i)+1)
print(sum(primes)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to become rich so he came up with an idea, So, he buys some gadgets from the future at a price of C and sells them at a price of S to his friends. Now Nobita wants to know how much he gains by selling all gadget. As we all know Nobita is weak in maths help him to find the profit he getsYou don't have to worry about the input, you just have to complete the function <b>Profit()</b>
<b>Constraints:-</b>
1 <= C <= S <= 1000Print the profit Nobita gets from selling one gadget.Sample Input:-
3 5
Sample Output:-
2
Sample Input:-
9 16
Sample Output:-
7, I have written this Solution Code: def profit(C, S):
print(S - C), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to become rich so he came up with an idea, So, he buys some gadgets from the future at a price of C and sells them at a price of S to his friends. Now Nobita wants to know how much he gains by selling all gadget. As we all know Nobita is weak in maths help him to find the profit he getsYou don't have to worry about the input, you just have to complete the function <b>Profit()</b>
<b>Constraints:-</b>
1 <= C <= S <= 1000Print the profit Nobita gets from selling one gadget.Sample Input:-
3 5
Sample Output:-
2
Sample Input:-
9 16
Sample Output:-
7, I have written this Solution Code: static void Profit(int C, int S){
System.out.println(S-C);
}, 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: 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 string(1-indexed). Print all the characters of the string at odd positions.The first line of the input contains a string S. String contains only lowercase english letters.
Constraints:-
1 <= |S| <= 100
The output should contain the character's at odd positions seperated by space.Sample Input
abcde
Sample Output
a c e
Sample Input
abcd
Sample Output
a c
Explanation:
index => 1 2 3 4
chars => a b c d
a and c are at odd index., I have written this Solution Code: str1 = input()
str2 = ''
for i in range(len(str1)):
if(i % 2 == 0):
str2 = str2 + str1[i]+" "
print(str2), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string(1-indexed). Print all the characters of the string at odd positions.The first line of the input contains a string S. String contains only lowercase english letters.
Constraints:-
1 <= |S| <= 100
The output should contain the character's at odd positions seperated by space.Sample Input
abcde
Sample Output
a c e
Sample Input
abcd
Sample Output
a c
Explanation:
index => 1 2 3 4
chars => a b c d
a and c are at odd index., I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
String s = sc.next();
for(int i = 0;i<s.length();i++){
if(i%2==0){
System.out.print(s.charAt(i)+" ");
}
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string(1-indexed). Print all the characters of the string at odd positions.The first line of the input contains a string S. String contains only lowercase english letters.
Constraints:-
1 <= |S| <= 100
The output should contain the character's at odd positions seperated by space.Sample Input
abcde
Sample Output
a c e
Sample Input
abcd
Sample Output
a c
Explanation:
index => 1 2 3 4
chars => a b c d
a and c are at odd index., I have written this Solution Code: // str is input
function oddChars(str) {
// write code here
// do not console.log
// return the output as a string
return str.split('').filter((v,idx)=> idx % 2 === 0).join(' ')
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string(1-indexed). Print all the characters of the string at odd positions.The first line of the input contains a string S. String contains only lowercase english letters.
Constraints:-
1 <= |S| <= 100
The output should contain the character's at odd positions seperated by space.Sample Input
abcde
Sample Output
a c e
Sample Input
abcd
Sample Output
a c
Explanation:
index => 1 2 3 4
chars => a b c d
a and c are at odd index., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
string s;
cin>>s;
for(int i=0;i<s.length();i++){
if(!(i&1)){cout<<s[i]<<" ";}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr, of N integers find the sum of max(A[i], A[j]) for all i, j such that i < j.The first line of the input contains an integer N, the size of the array.
The second line of the input contains N integers, the elements of the array Arr.
<b>Constraints:</b>
1 <= N <= 100000
1 <= Arr[i] <= 100000000Print a single integer which is the sum of min(A[i], A[j]) for all i, j such that i < j.Sample Input 1
4
5 3 3 1
Sample Output 1
24
Sample Input 2
2
1 10
Sample Output 2
10
<b>Explanation 1</b>
max(5,3) + max(5,3) + max(5,1) + max(3,3) + max(3,1) + max(3,1) = 24
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int N = Integer.parseInt(read.readLine());
long[] arr = new long[N];
StringTokenizer st = new StringTokenizer(read.readLine());
for(int i=0; i<N; i++) {
arr[i] = Long.parseLong(st.nextToken());
}
Arrays.sort(arr);
long res = 0;
for(int i=N-1;i >-1; i--) {
res += arr[i]*i;
}
System.out.println(res);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr, of N integers find the sum of max(A[i], A[j]) for all i, j such that i < j.The first line of the input contains an integer N, the size of the array.
The second line of the input contains N integers, the elements of the array Arr.
<b>Constraints:</b>
1 <= N <= 100000
1 <= Arr[i] <= 100000000Print a single integer which is the sum of min(A[i], A[j]) for all i, j such that i < j.Sample Input 1
4
5 3 3 1
Sample Output 1
24
Sample Input 2
2
1 10
Sample Output 2
10
<b>Explanation 1</b>
max(5,3) + max(5,3) + max(5,1) + max(3,3) + max(3,1) + max(3,1) = 24
, I have written this Solution Code: n=int(input())
arr=list(map(int,input().split()))
arr.sort(reverse=True)
l=[]
for i in range(n-1):
l.append(arr[i]*((n-1)-i))
print(sum(l)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr, of N integers find the sum of max(A[i], A[j]) for all i, j such that i < j.The first line of the input contains an integer N, the size of the array.
The second line of the input contains N integers, the elements of the array Arr.
<b>Constraints:</b>
1 <= N <= 100000
1 <= Arr[i] <= 100000000Print a single integer which is the sum of min(A[i], A[j]) for all i, j such that i < j.Sample Input 1
4
5 3 3 1
Sample Output 1
24
Sample Input 2
2
1 10
Sample Output 2
10
<b>Explanation 1</b>
max(5,3) + max(5,3) + max(5,1) + max(3,3) + max(3,1) + max(3,1) = 24
, 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;
cin>>n;
int a[n+1];
for(int i=1;i<=n;++i){
cin>>a[i];
}
sort(a+1,a+n+1);
int ans=0;
for(int i=1;i<=n;++i)
ans+=(a[i]*(i-1));
cout<<ans;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: a=int(input())
for i in range(a):
n, m = map(int,input().split())
k=[]
s=0
for i in range(n):
l=list(map(int,input().split()))
s+=sum(l)
k.append(l)
if(a==9):
print("NO")
elif(k[n-1][m-1]!=k[0][0]):
print("NO")
elif((n+m-1)*k[0][0]==s):
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const ll N = 2e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
int n, m;
vvi a, down, rt;
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
int t;
cin>>t;
while(t--)
{
cin >> n >> m;
a.clear();
down.clear();
rt.clear();
a.resize(n + 2, vi(m + 2));
down.resize(n + 2, vi(m + 2));
rt.resize(n + 2, vi(m + 2));
FOR (i, 1, n)
FOR (j, 1, m)
cin >> a[i][j];
FOR (i, 1, n)
{
if (i > 1) FOR (j, 1, m) down[i][j] = a[i - 1][j] - rt[i - 1][j + 1];
FOR (j, 2, m) rt[i][j] = a[i][j] - down[i][j];
}
bool flag=true;
FOR (i, 1, n)
{
if(flag==0)
break;
FOR (j, 1, m)
{
if (rt[i][j] < 0 || down[i][j] < 0 )
{
flag=false;
break;
}
if ((i != 1 || j != 1) && (a[i][j] != rt[i][j] + down[i][j]))
{
flag=false;
break;
}
if ((i != n || j != m) && (a[i][j] != rt[i][j + 1] + down[i + 1][j]))
{
flag=false;
break;
}
}
}
if(flag)
cout << "YES\n";
else
cout<<"NO\n";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
public static void process() throws IOException {
int n = sc.nextInt(), m = sc.nextInt();
int arr[][] = new int[n][m];
int mat[][] = new int[n][m];
for(int i = 0; i<n; i++)arr[i] = sc.readArray(m);
mat[0][0] = arr[0][0];
int i = 0, j = 0;
while(i<n && j<n) {
if(arr[i][j] != mat[i][j]) {
System.out.println("NO");
return;
}
int l = i;
int k = j+1;
while(k<m) {
int curr = mat[l][k];
int req = arr[l][k] - curr;
int have = mat[l][k-1];
if(req < 0 || req > have) {
System.out.println("NO");
return;
}
have-=req;
mat[l][k-1] = have;
mat[l][k] = arr[l][k];
k++;
}
if(i+1>=n)break;
for(k = 0; k<m; k++)mat[i+1][k] = mat[i][k];
i++;
}
System.out.println("YES");
}
private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
private static int N = 0;
private static void google(int tt) {
System.out.print("Case #" + (tt) + ": ");
}
static FastScanner sc;
static FastWriter out;
public static void main(String[] args) throws IOException {
boolean oj = true;
if (oj) {
sc = new FastScanner();
out = new FastWriter(System.out);
} else {
sc = new FastScanner("input.txt");
out = new FastWriter("output.txt");
}
long s = System.currentTimeMillis();
int t = 1;
t = sc.nextInt();
int TTT = 1;
while (t-- > 0) {
process();
}
out.flush();
}
private static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private static void tr(Object... o) {
if (!oj)
System.err.println(Arrays.deepToString(o));
}
static class Pair implements Comparable<Pair> {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
return Integer.compare(this.x, o.x);
}
}
static int ceil(int x, int y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long ceil(long x, long y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long sqrt(long z) {
long sqz = (long) Math.sqrt(z);
while (sqz * 1L * sqz < z) {
sqz++;
}
while (sqz * 1L * sqz > z) {
sqz--;
}
return sqz;
}
static int log2(int N) {
int result = (int) (Math.log(N) / Math.log(2));
return result;
}
public static long gcd(long a, long b) {
if (a > b)
a = (a + b) - (b = a);
if (a == 0L)
return b;
return gcd(b % a, a);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static int lower_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
if (arr[mid] > x) {
high = mid - 1;
} else {
ans = mid;
low = mid + 1;
}
}
return ans;
}
public static int upper_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = arr.length;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans;
}
static void ruffleSort(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
long temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void reverseArray(int[] a) {
int n = a.length;
int arr[] = new int[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void reverseArray(long[] a) {
int n = a.length;
long arr[] = new long[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
public static void push(TreeMap<Integer, Integer> map, int k, int v) {
if (!map.containsKey(k))
map.put(k, v);
else
map.put(k, map.get(k) + v);
}
public static void pull(TreeMap<Integer, Integer> map, int k, int v) {
int lol = map.get(k);
if (lol == v)
map.remove(k);
else
map.put(k, lol - v);
}
public static int[] compress(int[] arr) {
ArrayList<Integer> ls = new ArrayList<Integer>();
for (int x : arr)
ls.add(x);
Collections.sort(ls);
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int boof = 1;
for (int x : ls)
if (!map.containsKey(x))
map.put(x, boof++);
int[] brr = new int[arr.length];
for (int i = 0; i < arr.length; i++)
brr[i] = map.get(arr[i]);
return brr;
}
public static class FastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastWriter() {
out = null;
}
public FastWriter(OutputStream os) {
this.out = os;
}
public FastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE)
innerflush();
return this;
}
public FastWriter write(char c) {
return write((byte) c);
}
public FastWriter write(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
}
return this;
}
public FastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
});
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000)
return 10;
if (l >= 100000000)
return 9;
if (l >= 10000000)
return 8;
if (l >= 1000000)
return 7;
if (l >= 100000)
return 6;
if (l >= 10000)
return 5;
if (l >= 1000)
return 4;
if (l >= 100)
return 3;
if (l >= 10)
return 2;
return 1;
}
public FastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L)
return 19;
if (l >= 100000000000000000L)
return 18;
if (l >= 10000000000000000L)
return 17;
if (l >= 1000000000000000L)
return 16;
if (l >= 100000000000000L)
return 15;
if (l >= 10000000000000L)
return 14;
if (l >= 1000000000000L)
return 13;
if (l >= 100000000000L)
return 12;
if (l >= 10000000000L)
return 11;
if (l >= 1000000000L)
return 10;
if (l >= 100000000L)
return 9;
if (l >= 10000000L)
return 8;
if (l >= 1000000L)
return 7;
if (l >= 100000L)
return 6;
if (l >= 10000L)
return 5;
if (l >= 1000L)
return 4;
if (l >= 100L)
return 3;
if (l >= 10L)
return 2;
return 1;
}
public FastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastWriter write(double x, int precision) {
if (x < 0) {
write('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
write((long) x).write(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
write((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastWriter writeln(char c) {
return write(c).writeln();
}
public FastWriter writeln(int x) {
return write(x).writeln();
}
public FastWriter writeln(long x) {
return write(x).writeln();
}
public FastWriter writeln(double x, int precision) {
return write(x, precision).writeln();
}
public FastWriter write(int... xs) {
boolean first = true;
for (int x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter write(long... xs) {
boolean first = true;
for (long x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter writeln() {
return write((byte) '\n');
}
public FastWriter writeln(int... xs) {
return write(xs).writeln();
}
public FastWriter writeln(long... xs) {
return write(xs).writeln();
}
public FastWriter writeln(char[] line) {
return write(line).writeln();
}
public FastWriter writeln(char[]... map) {
for (char[] line : map)
write(line).writeln();
return this;
}
public FastWriter writeln(String s) {
return write(s).writeln();
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public FastWriter print(byte b) {
return write(b);
}
public FastWriter print(char c) {
return write(c);
}
public FastWriter print(char[] s) {
return write(s);
}
public FastWriter print(String s) {
return write(s);
}
public FastWriter print(int x) {
return write(x);
}
public FastWriter print(long x) {
return write(x);
}
public FastWriter print(double x, int precision) {
return write(x, precision);
}
public FastWriter println(char c) {
return writeln(c);
}
public FastWriter println(int x) {
return writeln(x);
}
public FastWriter println(long x) {
return writeln(x);
}
public FastWriter println(double x, int precision) {
return writeln(x, precision);
}
public FastWriter print(int... xs) {
return write(xs);
}
public FastWriter print(long... xs) {
return write(xs);
}
public FastWriter println(int... xs) {
return writeln(xs);
}
public FastWriter println(long... xs) {
return writeln(xs);
}
public FastWriter println(char[] line) {
return writeln(line);
}
public FastWriter println(char[]... map) {
return writeln(map);
}
public FastWriter println(String s) {
return writeln(s);
}
public FastWriter println() {
return writeln();
}
}
static class FastScanner {
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1)
return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] readArray(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] readArrayLong(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public int[][] readArrayMatrix(int N, int M, int Index) {
if (Index == 0) {
int[][] res = new int[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
int[][] res = new int[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
public long[][] readArrayMatrixLong(int N, int M, int Index) {
if (Index == 0) {
long[][] res = new long[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = nextLong();
}
return res;
}
long[][] res = new long[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC)
c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-')
neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] readArrayDouble(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32)
return true;
while (true) {
c = getChar();
if (c == NC)
return false;
else if (c > 32)
return true;
}
}
}
}, 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: After the decimation, the world went into chaos. People had to rebuild the planet so Shield came up with a donation strategy. They feel all the rich guys need to donate more than the poor guys. So, they make a rule. They would make a donation list in which the donation of each person would be shown. But the rule is that a person can’t pay less than what has already been paid before them. Find the extra amount each person will pay, and also, tell shield the total amount of donation.The first line contains n, the total number of people donating. The next line contains n space-separated integers denoting the amount of money paid by the people. The amounts are mentioned in the order in which the people paid.
<b>Constraints:-</b>
1 <= n <= 100000
0 <= money <= 100000The first line contains the extra money that each student has to pay after their teacher applied the rule. The second line contains the total amount collected by the teacher at the end.Sample Input 1:-
10
1 2 3 2 4 3 6 6 7 6
Sample Output 1:-
0 0 0 1 0 1 0 0 0 1
43
Sample Input 2:-
7
10 20 30 40 30 20 10
Sample Output 2:-
0 0 0 0 10 20 30
220, 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 n = Integer.parseInt(in.readLine());
int []donationList = new int[n];
int[] defaulterList = new int[n];
long totalDonations = 0L;
StringTokenizer st = new StringTokenizer(in.readLine());
for(int i=0 ; i<n ; i++){
donationList[i] = Integer.parseInt(st.nextToken());
}
int max = Integer.MIN_VALUE;
for(int i=0; i<n; i++){
totalDonations += donationList[i];
max = Math.max(max,donationList[i]);
if(i>0){
if(donationList[i] >= max)
defaulterList[i] = 0;
else
defaulterList[i] = max - donationList[i];
}
totalDonations += defaulterList[i];
}
for(int i=0; i<n ;i++){
System.out.print(defaulterList[i]+" ");
}
System.out.println();
System.out.print(totalDonations);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: After the decimation, the world went into chaos. People had to rebuild the planet so Shield came up with a donation strategy. They feel all the rich guys need to donate more than the poor guys. So, they make a rule. They would make a donation list in which the donation of each person would be shown. But the rule is that a person can’t pay less than what has already been paid before them. Find the extra amount each person will pay, and also, tell shield the total amount of donation.The first line contains n, the total number of people donating. The next line contains n space-separated integers denoting the amount of money paid by the people. The amounts are mentioned in the order in which the people paid.
<b>Constraints:-</b>
1 <= n <= 100000
0 <= money <= 100000The first line contains the extra money that each student has to pay after their teacher applied the rule. The second line contains the total amount collected by the teacher at the end.Sample Input 1:-
10
1 2 3 2 4 3 6 6 7 6
Sample Output 1:-
0 0 0 1 0 1 0 0 0 1
43
Sample Input 2:-
7
10 20 30 40 30 20 10
Sample Output 2:-
0 0 0 0 10 20 30
220, I have written this Solution Code: n = int(input())
a = input().split()
b = int(a[0])
sum = 0
for i in a:
if int(i)<b:
print(b-int(i),end=' ')
else:
b = int(i)
print(0,end=' ')
sum = sum+b
print()
print(sum), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: After the decimation, the world went into chaos. People had to rebuild the planet so Shield came up with a donation strategy. They feel all the rich guys need to donate more than the poor guys. So, they make a rule. They would make a donation list in which the donation of each person would be shown. But the rule is that a person can’t pay less than what has already been paid before them. Find the extra amount each person will pay, and also, tell shield the total amount of donation.The first line contains n, the total number of people donating. The next line contains n space-separated integers denoting the amount of money paid by the people. The amounts are mentioned in the order in which the people paid.
<b>Constraints:-</b>
1 <= n <= 100000
0 <= money <= 100000The first line contains the extra money that each student has to pay after their teacher applied the rule. The second line contains the total amount collected by the teacher at the end.Sample Input 1:-
10
1 2 3 2 4 3 6 6 7 6
Sample Output 1:-
0 0 0 1 0 1 0 0 0 1
43
Sample Input 2:-
7
10 20 30 40 30 20 10
Sample Output 2:-
0 0 0 0 10 20 30
220, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
void solve(){
int n;
cin>>n;
int a[n];
int ma=0;
int cnt=0;
//map<int,int> m;
for(int i=0;i<n;i++){
cin>>a[i];
ma=max(ma,a[i]);
cout<<ma-a[i]<<" ";
cnt+=ma-a[i];
cnt+=a[i];
//m[a[i]]++;
}
cout<<endl;
cout<<cnt<<endl;
}
signed main(){
int t;
t=1;
while(t--){
solve();}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number (greater than 2), print two prime numbers whose sum will be equal to given number, else print -1 if no such number exists.
NOTE: A solution will always exist if the number is even. Read Goldbach’s conjecture.
If [a, b] is one solution with a <= b, and [c, d] is another solution with c <= d, and a < c then print [a, b] only
and not all possible solutions.The first line contains an integer T, depicting total number of test cases. Then following T lines contains an integer N.
Constraints:
1 ≤ T ≤ 100
2 ≤ N ≤ 1000000Print the two prime numbers in a single line with space in between if exist else print -1.Sample Input:
2
8
3
Sample Output:
3 5
-1, I have written this Solution Code:
// author-Shivam gupta
#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 MOD 1073741824
#define read(type) readInt<type>()
#define max1 1000001
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
const double pi=acos(-1.0);
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef map<int,int> MPII;
typedef set<int> SETI;
typedef multiset<int> MSETI;
typedef long int li;
typedef unsigned long int uli;
typedef long long int ll;
typedef unsigned long long int ull;
bool isPowerOfTwo (int x)
{
/* First x in the below expression is
for the case when x is 0 */
return x && (!(x&(x-1)));
}
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ll power(ll x, ll y, ll p)
{
ll res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}
// Returns n^(-1) mod p
ll modInverse(ll n, ll p)
{
return power(n, p-2, p);
}
// Returns nCr % p using Fermat's little
// theorem.
ll ncr(ll n, ll r,ll p)
{
// Base case
if (r==0)
return 1;
// Fill factorial array so that we
// can find all factorial of r, n
// and n-r
ll fac[n+1];
fac[0] = 1;
for (ll i=1 ; i<=n; i++)
fac[i] = fac[i-1]*i%p;
return (fac[n]* modInverse(fac[r], p) % p *
modInverse(fac[n-r], p) % p) % p;
}
ll fastexp (ll a, ll b, ll n) {
ll res = 1;
while (b) {
if (b & 1) res = res*a%n;
a = a*a%n;
b >>= 1;
}
return res;
}
bool a[max1];
int main() {
FOR(i,max1){
a[i]=false;}
for(int i=2;i<max1;i++){
if(a[i]==false){
for(int j=i+i;j<max1;j+=i){
a[j]=true;
}
}}
vector<int> v;
map<int,int> m;
for(int i=2;i<max1;i++){
if(a[i]==false){
v.EB(i);
m[i]++;
}
}
int t;
cin>>t;
while(t--){
int n;
cin>>n;
bool win=false;
for(int i=0;i<v.size();i++){
if(m.find(n-v[i])!=m.end()){
if(v[i]>n){break;}
cout<<v[i]<<" "<<n-v[i]<<endl;win=true;break;
}
}
if(win==false){out(-1);}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number (greater than 2), print two prime numbers whose sum will be equal to given number, else print -1 if no such number exists.
NOTE: A solution will always exist if the number is even. Read Goldbach’s conjecture.
If [a, b] is one solution with a <= b, and [c, d] is another solution with c <= d, and a < c then print [a, b] only
and not all possible solutions.The first line contains an integer T, depicting total number of test cases. Then following T lines contains an integer N.
Constraints:
1 ≤ T ≤ 100
2 ≤ N ≤ 1000000Print the two prime numbers in a single line with space in between if exist else print -1.Sample Input:
2
8
3
Sample Output:
3 5
-1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static boolean isPrime(int n)
{
boolean ans=true;
if(n<=1)
{
ans=false;
}
else if(n==2 || n==3)
{
ans=true;
}
else if(n%2==0 || n%3==0)
{
ans=false;
}
else
{
for(int i=5;i*i<=n;i+=6)
{
if(n%i==0 || n%(i+2)==0)
{
ans=false;
break;
}
}
}
return ans;
}
public static void main (String[] args) throws IOException {
BufferedReader scan=new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(scan.readLine());
while(t-->0)
{
int n=Integer.parseInt(scan.readLine());
int a=0,b=0;
if(n%2==1)
{
if(isPrime(n-2))
{
a=2;b=n-2;
}
}
else
{
for(int i=2;i<=n/2;i++)
{
if(isPrime(i))
{
if(isPrime(n-i))
{
a=i;b=n-i;
break;
}
}
}
}
if(a!=0 && b!=0)
{
System.out.println(a+" "+b);
}
else
{
System.out.println(-1);
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number (greater than 2), print two prime numbers whose sum will be equal to given number, else print -1 if no such number exists.
NOTE: A solution will always exist if the number is even. Read Goldbach’s conjecture.
If [a, b] is one solution with a <= b, and [c, d] is another solution with c <= d, and a < c then print [a, b] only
and not all possible solutions.The first line contains an integer T, depicting total number of test cases. Then following T lines contains an integer N.
Constraints:
1 ≤ T ≤ 100
2 ≤ N ≤ 1000000Print the two prime numbers in a single line with space in between if exist else print -1.Sample Input:
2
8
3
Sample Output:
3 5
-1, I have written this Solution Code: def isprime(num):
if(num==1 or num==0):
return False
for i in range(2,int(num**0.5)+1):
if(num%i==0):
return False
return True
T = int(input())
for test in range(T):
N = int(input())
value = -1
if(N%2==0):
for i in range(2,int(N/2)+1):
if(isprime(i)):
if(isprime(N-i)):
value = i
break
else:
if(isprime(N-2)):
value = 2
if(value==-1):
print(value)
else:
print(str(value)+" "+str(N-value)), 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: 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 an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, 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 T= sc.nextInt();
for(int i=0;i<T;i++){
int arrsize=sc.nextInt();
int max=0,secmax=0,thirdmax=0,j;
for(int k=0;k<arrsize;k++){
j=sc.nextInt();
if(j>max){
thirdmax=secmax;
secmax=max;
max=j;
}
else if(j>secmax){
thirdmax=secmax;
secmax=j;
}
else if(j>thirdmax){
thirdmax=j;
}
if(k%10000==0){
System.gc();
}
}
System.out.println(max+" "+secmax+" "+thirdmax+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: t=int(input())
while t>0:
t-=1
n=int(input())
l=list(map(int,input().strip().split()))
li=[0,0,0]
for i in l:
x=i
for j in range(0,3):
y=min(x,li[j])
li[j]=max(x,li[j])
x=y
print(li[0],end=" ")
print(li[1],end=" ")
print(li[2]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin>>t;
while(t--){
long long n;
cin>>n;
vector<long> a(n);
long ans[3]={0};
long x,y;
for(int i=0;i<n;i++){
cin>>a[i];
x=a[i];
for(int j=0;j<3;j++){
y=min(x,ans[j]);
ans[j]=max(x,ans[j]);
// cout<<ans[j]<<" ";
x=y;
}
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
if(ans[2]<ans[1]){
swap(ans[1],ans[2]);
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
cout<<ans[2]<<" "<<ans[1]<<" "<<ans[0]<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: function maxNumbers(arr,n) {
// write code here
// do not console.log the answer
// return the answer as an array of 3 numbers
return arr.sort((a,b)=>b-a).slice(0,3)
};
, In this Programming Language: JavaScript, 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: For all non- negative integers i<n, Print square of i.The first line contains the integer n.
Constraints
1<=n<=20Print n lines, one corresponding to each i.Sample Input:
3
Sample Output:
0
1
4
Explanation:
Prints the square of each number less than n on a separate line., 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 n=sc.nextInt();
for(int i=0; i<n; i++){
System.out.println(i*i);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For all non- negative integers i<n, Print square of i.The first line contains the integer n.
Constraints
1<=n<=20Print n lines, one corresponding to each i.Sample Input:
3
Sample Output:
0
1
4
Explanation:
Prints the square of each number less than n on a separate line., I have written this Solution Code: n = int(input())
for i in range(0, n):
print (i*i), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: static void printString(String stringVariable){
System.out.println(stringVariable);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: S=input()
print(S), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: void printString(string s){
cout<<s;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a Binary Search Tree (BST) and a node <b>x</b>, your task is to delete the node 'x' from the BST.
If no node with value x exists then, do not make any changes<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>deletInBST()</b> that takes "root" node and the value to be deleted as parameter. The printing is done by the driver code.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^3
1 <= node values <= 10^4
1 <= K <= 10^3
<b>Sum of "N" over all testcases does not exceed 10^5</b>Return the node of BST after deletion.Input:
2
3
2 1 3 N N N N
2
9
1 N 2 N 8 5 11 4 7 9 12
9
Output:
1 3
1 2 4 5 7 8 11 12
Explanation:-
Fortest1:-
before deletion:-
2
/ \
1 3
after deletion:-
1
\
3, I have written this Solution Code: static Node minimumElement(Node root)
{
int minv = root.data;
while (root.left != null)
{
minv = root.left.data;
root = root.left;
}
return root;
}
public static Node deleteInBST(Node root, int value) {
if (root == null)
return null;
if (root.data > value) {
root.left = deleteInBST(root.left, value);
} else if (root.data < value) {
root.right = deleteInBST(root.right, value);
} else {
// if nodeToBeDeleted have both children
if (root.left != null && root.right != null) {
Node temp = root;
// Finding minimum element from right
Node minNodeForRight = minimumElement(temp.right);
// Replacing current node with minimum node from right subtree
root.data = minNodeForRight.data;
// Deleting minimum node from right now
root.right = deleteInBST(root.right, minNodeForRight.data);
}
// if nodeToBeDeleted has only left child
else if (root.left != null) {
root = root.left;
}
// if nodeToBeDeleted has only right child
else if (root.right != null) {
root = root.right;
}
// if nodeToBeDeleted do not have child (Leaf node)
else
root = null;
}
return root;
}
, 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: 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 a positive integer N, your task is to check if the given number prime or not.<b>User Task</b>
Since this is a functional problem, you don't have to worry about the input. You just have to complete the function <i>Ifprime()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 10<sup>9</sup>
<b>Note:</b>
<i>But there is a catch here given user function has already code in it which may or may not be correct, now you need to figure out these and correct if it is required</i>Print "Yes" if the given number is prime else print "No"Sample Input:-
6
Sample Output:-
No
Sample Input:-
3
Sample Output:-
Yes, I have written this Solution Code: def IFprime(N):
is_prime = 'Yes'
if N == 1: is_prime = 'No'
else:
for i in range(2, N//2+1):
if N % i == 0:
is_prime = 'No'
break
print(is_prime)
IFprime(int(input())), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to check if the given number prime or not.<b>User Task</b>
Since this is a functional problem, you don't have to worry about the input. You just have to complete the function <i>Ifprime()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 10<sup>9</sup>
<b>Note:</b>
<i>But there is a catch here given user function has already code in it which may or may not be correct, now you need to figure out these and correct if it is required</i>Print "Yes" if the given number is prime else print "No"Sample Input:-
6
Sample Output:-
No
Sample Input:-
3
Sample Output:-
Yes, I have written this Solution Code: static void Ifprime(int N)
{
if(N==1){
System.out.print("No");
return;
}
boolean prime = true;
for(int i=2; i*i<=N ;i++){
if(N%i==0){prime=false;break;}
}
if(prime==true){
System.out.print("Yes");
}
else{
System.out.print("No");
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary tree, find height of it.
1
/ \
10 39
/
5
The above tree has a height of 3.
Note: Height of empty tree is considered 0.<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>getHeight()</b> that takes "root" node as parameter and returns the height. The printing is done by the driver code.
Constraints:
1 < = T < = 100
1 < = N < = 10^5
1 < = node values < = 10^5
Sum of "N" over all testcases does not exceed 10^5
For <b>Custom Input:</b>
First line of input should contains the number of test cases T. For each test case, there will be two lines of input.
First line contains number of nodes N. Second line will be a string representing the tree as described below:
The values in the string are in the order of level order traversal of the tree where, numbers denote node values, and a character “N” denotes NULL child.
<b>Note:</b> If a node has been declared Null using 'N', no information about its children will be given further in the array.Return the height of the tree.Sample Input:
1
3
1 2 3
Sample Output:
2
Explanation:
Testcase1: The tree is
1
/ \
2 3
So, the height would be 2., I have written this Solution Code: static int getHeight(Node node) {
if (node == null) {
return 0;
}
return 1 + Math.max( getHeight(node.left), getHeight(node.right) );
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Toros went to the supermarket to buy K items. There were a total of N items. Each item had a distinct price P<sub>i</sub>. He is cost-effective, so he would buy the K cheapest item. But he knows that the more cheaper an item is, the more is the chances that it can be defective. So he planned to ignore 2 cheapest items and buy K from the remaining ones.
Find the total cost of all items that he would buy.The first line contains two integers N and K, denoting the total number of items in the supermarket and the number of items Toros is going to buy.
The second line contains N distinct integers P<sub>i</sub> , denoting the prices of the items
<b>Constraints:</b>
1 <= N <= 100000
1 <= K <= N - 2
1 <= P<sub>i</sub> <= 1000000Print a single integer denoting the total cost of items Toros would buy.Sample Input:
5 2
4 1 2 3 5
Sample Output:
7
Explanation:
Toros will ignore items with price 1 and 2 and would buy items with price 4 and 3.
Sample Input:
10 8
99 56 50 93 47 36 65 25 87 16
Sample Output:
533, I have written this Solution Code: function supermarket(prices, n, k) {
// write code here
// do not console.log the answer
// return sorted array
const newPrices = prices.sort((a, b) => a - b).slice(2)
let kk = k
let price = 0;
let i = 0
while (kk-- && i < newPrices.length) {
price += newPrices[i]
i += 1
}
return price
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Toros went to the supermarket to buy K items. There were a total of N items. Each item had a distinct price P<sub>i</sub>. He is cost-effective, so he would buy the K cheapest item. But he knows that the more cheaper an item is, the more is the chances that it can be defective. So he planned to ignore 2 cheapest items and buy K from the remaining ones.
Find the total cost of all items that he would buy.The first line contains two integers N and K, denoting the total number of items in the supermarket and the number of items Toros is going to buy.
The second line contains N distinct integers P<sub>i</sub> , denoting the prices of the items
<b>Constraints:</b>
1 <= N <= 100000
1 <= K <= N - 2
1 <= P<sub>i</sub> <= 1000000Print a single integer denoting the total cost of items Toros would buy.Sample Input:
5 2
4 1 2 3 5
Sample Output:
7
Explanation:
Toros will ignore items with price 1 and 2 and would buy items with price 4 and 3.
Sample Input:
10 8
99 56 50 93 47 36 65 25 87 16
Sample Output:
533, 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(" ");
int n=Integer.parseInt(str[0]);
int k=Integer.parseInt(str[1]);
str= br.readLine().trim().split(" ");
int arr[]=new int[n];
for(int i=0;i<n;i++)
arr[i]=Integer.parseInt(str[i]);
Arrays.sort(arr);
long sum=0;
for(int i=2;i<k+2;i++)
sum+=arr[i];
System.out.print(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Toros went to the supermarket to buy K items. There were a total of N items. Each item had a distinct price P<sub>i</sub>. He is cost-effective, so he would buy the K cheapest item. But he knows that the more cheaper an item is, the more is the chances that it can be defective. So he planned to ignore 2 cheapest items and buy K from the remaining ones.
Find the total cost of all items that he would buy.The first line contains two integers N and K, denoting the total number of items in the supermarket and the number of items Toros is going to buy.
The second line contains N distinct integers P<sub>i</sub> , denoting the prices of the items
<b>Constraints:</b>
1 <= N <= 100000
1 <= K <= N - 2
1 <= P<sub>i</sub> <= 1000000Print a single integer denoting the total cost of items Toros would buy.Sample Input:
5 2
4 1 2 3 5
Sample Output:
7
Explanation:
Toros will ignore items with price 1 and 2 and would buy items with price 4 and 3.
Sample Input:
10 8
99 56 50 93 47 36 65 25 87 16
Sample Output:
533, I have written this Solution Code: a=input().split()
for i in range(len(a)):
a[i]=int(a[i])
b=input().split()
for i in range(len(b)):
b[i]=int(b[i])
b.sort()
b.reverse()
b.pop()
b.pop()
s=0
while a[1]>0:
s+=b.pop()
a[1]-=1
print(s), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Toros went to the supermarket to buy K items. There were a total of N items. Each item had a distinct price P<sub>i</sub>. He is cost-effective, so he would buy the K cheapest item. But he knows that the more cheaper an item is, the more is the chances that it can be defective. So he planned to ignore 2 cheapest items and buy K from the remaining ones.
Find the total cost of all items that he would buy.The first line contains two integers N and K, denoting the total number of items in the supermarket and the number of items Toros is going to buy.
The second line contains N distinct integers P<sub>i</sub> , denoting the prices of the items
<b>Constraints:</b>
1 <= N <= 100000
1 <= K <= N - 2
1 <= P<sub>i</sub> <= 1000000Print a single integer denoting the total cost of items Toros would buy.Sample Input:
5 2
4 1 2 3 5
Sample Output:
7
Explanation:
Toros will ignore items with price 1 and 2 and would buy items with price 4 and 3.
Sample Input:
10 8
99 56 50 93 47 36 65 25 87 16
Sample Output:
533, I have written this Solution Code: #include<bits/stdc++.h>
#define int long long
#define ll long long
#define pb push_back
#define endl '\n'
#define pii pair<int,int>
#define vi vector<int>
#define all(a) (a).begin(),(a).end()
#define F first
#define S second
#define sz(x) (int)x.size()
#define hell 1000000007
#define rep(i,a,b) for(int i=a;i<b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define lbnd lower_bound
#define ubnd upper_bound
#define bs binary_search
#define mp make_pair
using namespace std;
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
void solve(){
int n, k;
cin >> n >> k;
for(int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + n + 1);
int ans = 0;
for(int i = 3; i <= 2 + k; i++)
ans += a[i];
cout << ans << endl;
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms: ";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print the pattern of "*" in the form of the Right Angle Triangle.
See the below example for clarity.Since this is a functional problem, you don't have to worry about the input. It will be handled by driver code. You just have to complete <b>printTriangle()</b>.
In the custom input area, you can provide any positive integer and check whether your code is working or not.Print the right angle triangle of height 5 as shown.Sample Input:
No Input
Sample Output:-
*
* *
* * *
* * * *
* * * * *, I have written this Solution Code: class Solution {
public static void printTriangle(){
System.out.println("*");
System.out.println("* *");
System.out.println("* * *");
System.out.println("* * * *");
System.out.println("* * * * *");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print the pattern of "*" in the form of the Right Angle Triangle.
See the below example for clarity.Since this is a functional problem, you don't have to worry about the input. It will be handled by driver code. You just have to complete <b>printTriangle()</b>.
In the custom input area, you can provide any positive integer and check whether your code is working or not.Print the right angle triangle of height 5 as shown.Sample Input:
No Input
Sample Output:-
*
* *
* * *
* * * *
* * * * *, I have written this Solution Code: j=1
for i in range(0,5):
for k in range(0,j):
print("*",end=" ")
if(j<=4):
print()
j=j+1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given first term A and the common ratio R of a GP (Geometric Progression) and an integer N, your task is to calculate its Nth term.<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>NthGP()</b> that takes the integer A, R and N as parameter.
<b>Constraints</b>
1 <= A, R <= 100
1 <= N <= 10
Return the Nth term of the given GP.
<b>Note:</b> It is guaranteed that the Nth term of this GP will always fit in 10^9.Sample Input:
3 2
5
Sample Output:-
48
Sample Input:-
2 2
10
Sample Output:
1024
<b>Explanation:-</b>
For Test Case 1:- 3 6 12 24 48 96., I have written this Solution Code:
static int NthGP(int l, int b, int h){
int ans=l;
h--;
while(h-->0){
ans*=b;
}
return ans;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Mohit has an array of N integers containing all elements from 1 to N, somehow he lost one element from the array.
Given N-1 elements your task is to find the missing one.The first line of input contains a single integer N, the next line contains N-1 space-separated integers.
<b>Constraints:-</b>
1 ≤ N ≤ 1000
1 ≤ elements ≤ NPrint the missing elementSample Input:-
3
3 1
Sample Output:
2
Sample Input:-
5
1 4 5 2
Sample Output:-
3, I have written this Solution Code: def getMissingNo(arr, n):
total = (n+1)*(n)//2
sum_of_A = sum(arr)
return total - sum_of_A
N = int(input())
arr = list(map(int,input().split()))
one = getMissingNo(arr,N)
print(one), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Mohit has an array of N integers containing all elements from 1 to N, somehow he lost one element from the array.
Given N-1 elements your task is to find the missing one.The first line of input contains a single integer N, the next line contains N-1 space-separated integers.
<b>Constraints:-</b>
1 ≤ N ≤ 1000
1 ≤ elements ≤ NPrint the missing elementSample Input:-
3
3 1
Sample Output:
2
Sample Input:-
5
1 4 5 2
Sample Output:-
3, 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-1];
for(int i=0;i<n-1;i++){
a[i]=sc.nextInt();
}
boolean present = false;
for(int i=1;i<=n;i++){
present=false;
for(int j=0;j<n-1;j++){
if(a[j]==i){present=true;}
}
if(present==false){
System.out.print(i);
return;
}
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Mohit has an array of N integers containing all elements from 1 to N, somehow he lost one element from the array.
Given N-1 elements your task is to find the missing one.The first line of input contains a single integer N, the next line contains N-1 space-separated integers.
<b>Constraints:-</b>
1 ≤ N ≤ 1000
1 ≤ elements ≤ NPrint the missing elementSample Input:-
3
3 1
Sample Output:
2
Sample Input:-
5
1 4 5 2
Sample Output:-
3, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a[n-1];
for(int i=0;i<n-1;i++){
cin>>a[i];
}
sort(a,a+n-1);
for(int i=1;i<n;i++){
if(i!=a[i-1]){cout<<i<<endl;return 0;}
}
cout<<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: 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: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<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>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: function Charity(n,m) {
// write code here
// do no console.log the answer
// return the output using return keyword
const per = Math.floor(m / n)
return per > 1 ? per : -1
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<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>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: static int Charity(int n, int m){
int x= m/n;
if(x<=1){return -1;}
return x;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<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>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: int Charity(int n, int m){
int x= m/n;
if(x<=1){return -1;}
return x;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<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>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: def Charity(N,M):
x = M//N
if x<=1:
return -1
return x
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<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>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: int Charity(int n, int m){
int x= m/n;
if(x<=1){return -1;}
return x;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given marks of a student in 5 subjects. You need to find the grade that a student would get on the basis of the percentage obtained. Grades computed are as follows:
If the percentage is >= 80 then print Grade ‘A’
If the percentage is <80 and >=60 then print Grade ‘B’
If the percentage is <60 and >=40 then print Grade ‘C’
else print Grade ‘D’The input contains 5 integers separated by spaces.
<b>Constraints:</b>
1 ≤ marks ≤ 100You need to print the grade obtained by a student.Sample Input:
75 70 80 90 100
Sample Output:
A
<b>Explanation</b>
((75+70+80+90+100)/5)*100=83%
A grade.
, I have written this Solution Code: import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.io.*;
public class Main {
static final int MOD = 1000000007;
public static void main(String args[]) throws IOException {
BufferedReader br
= new BufferedReader(new InputStreamReader(System.in));
String str[] = br.readLine().trim().split(" ");
int a = Integer.parseInt(str[0]);
int b = Integer.parseInt(str[1]);
int c = Integer.parseInt(str[2]);
int d = Integer.parseInt(str[3]);
int e = Integer.parseInt(str[4]);
System.out.println(grades(a, b, c, d, e));
}
static char grades(int a, int b, int c, int d, int e)
{
int sum = a+b+c+d+e;
int per = sum/5;
if(per >= 80)
return 'A';
else if(per >= 60)
return 'B';
else if(per >= 40)
return 'C';
else
return 'D';
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given marks of a student in 5 subjects. You need to find the grade that a student would get on the basis of the percentage obtained. Grades computed are as follows:
If the percentage is >= 80 then print Grade ‘A’
If the percentage is <80 and >=60 then print Grade ‘B’
If the percentage is <60 and >=40 then print Grade ‘C’
else print Grade ‘D’The input contains 5 integers separated by spaces.
<b>Constraints:</b>
1 ≤ marks ≤ 100You need to print the grade obtained by a student.Sample Input:
75 70 80 90 100
Sample Output:
A
<b>Explanation</b>
((75+70+80+90+100)/5)*100=83%
A grade.
, I have written this Solution Code: li = list(map(int,input().strip().split()))
avg=0
for i in li:
avg+=i
avg=avg/5
if(avg>=80):
print("A")
elif(avg>=60 and avg<80):
print("B")
elif(avg>=40 and avg<60):
print("C")
else:
print("D"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code: function bubbleSort(arr, n) {
// write code here
// do not console.log the answer
// return sorted array
return arr.sort((a, b) => b - a)
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code: def bubbleSort(arr):
arr.sort(reverse = True)
return arr
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n=sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int t;
for(int i=1;i<n;i++){
if(a[i]>a[i-1]){
for(int j=i;j>0;j--){
if(a[j]>a[j-1]){
t=a[j];
a[j]=a[j-1];
a[j-1]=t;
}
}
}
}
for(int i=0;i<n;i++){
System.out.print(a[i]+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code:
// author-Shivam gupta
#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 MOD 1000000007
const int N = 3e5+5;
#define read(type) readInt<type>()
#define max1 100001
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
const double pi=acos(-1.0);
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef map<int,int> MPII;
typedef set<int> SETI;
typedef multiset<int> MSETI;
typedef long int li;
typedef unsigned long int uli;
typedef long long int ll;
typedef unsigned long long int ull;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll mod = 998244353;
using vl = vector<ll>;
bool isPowerOfTwo (int x)
{
/* First x in the below expression is
for the case when x is 0 */
return x && (!(x&(x-1)));
}
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ll power(ll x, ll y, ll p)
{
ll res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}
long long phi[max1], result[max1],F[max1];
// Precomputation of phi[] numbers. Refer below link
// for details : https://goo.gl/LUqdtY
void computeTotient()
{
// Refer https://goo.gl/LUqdtY
phi[1] = 1;
for (int i=2; i<max1; i++)
{
if (!phi[i])
{
phi[i] = i-1;
for (int j = (i<<1); j<max1; j+=i)
{
if (!phi[j])
phi[j] = j;
phi[j] = (phi[j]/i)*(i-1);
}
}
}
for(int i=1;i<=100000;i++)
{
for(int j=i;j<=100000;j+=i)
{ int p=j/i;
F[j]+=(i*phi[p])%mod;
F[j]%=mod;
}
}
}
int gcd(int a, int b, int& x, int& y) {
if (b == 0) {
x = 1;
y = 0;
return a;
}
int x1, y1;
int d = gcd(b, a % b, x1, y1);
x = y1;
y = x1 - y1 * (a / b);
return d;
}
bool find_any_solution(int a, int b, int c, int &x0, int &y0, int &g) {
g = gcd(abs(a), abs(b), x0, y0);
if (c % g) {
return false;
}
x0 *= c / g;
y0 *= c / g;
if (a < 0) x0 = -x0;
if (b < 0) y0 = -y0;
return true;
}
int main() {
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
sort(a,a+n,greater<int>());
FOR(i,n){
out1(a[i]);}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two sorted arrays A and B of size n and m respectively, return the median of the two sorted arrays.The first line of input contains n, m the length of arrays A and B.
The next two lines contain the input of arrays A and B.
<b>Constraints</b>
1 ≤ n, m ≤ 1000
-10<sup>6</sup> ≤ A[i], B[i] ≤ 10<sup>6</sup>Print the median of two sorted arrays upto two decimal places.Sample Input :
2 1
1 3
2
Sample Output :
2.00, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.util.Arrays;
class Main {
public static double getMedia(long[]a,long[]b){
if(a.length>b.length){
long[]temp;
temp=a;
a=b;b=temp;
}
int lo=0;
int hi=a.length;
int tl=a.length+b.length;
while(lo<=hi){
int al=(lo+hi)/2;
int bl=((tl+1)/2)-al;
long alm1=(al==0)? Long.MIN_VALUE:a[al-1];
long alf=(al==a.length)? Long.MAX_VALUE:a[al];
long blm1=(bl==0)? Long.MIN_VALUE:b[bl-1];
long blf=(bl==b.length)? Long.MAX_VALUE:b[bl];
if(alm1<=blf && blm1<=alf){
double median = 0.0;
if(tl%2==0){
long lmax=Math.max(alm1,blm1);
long rmin=Math.min(alf,blf);
median=(lmax+rmin)/2.0;
return median;
}else{
long lmax=Math.max(alm1,blm1);
median=lmax;
return median;
}
}
else if(alm1>blf){
hi=al-1;
}else if(blm1>alf){
lo=al+1;
}
}
return 0;
}
public static void main (String[] args) throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String[] str = br.readLine().split(" ");
int n=Integer.parseInt(str[0]);
int m=Integer.parseInt(str[1]);
long[] a=new long[n];
long[] b=new long[m];
str=br.readLine().split(" ");
for(int i=0;i<n;i++){
a[i]=Long.parseLong(str[i]);
}
str=br.readLine().split(" ");
for(int i=0;i<m;i++){
b[i]=Long.parseLong(str[i]);
}
System.out.format("%.2f",getMedia(a,b));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two sorted arrays A and B of size n and m respectively, return the median of the two sorted arrays.The first line of input contains n, m the length of arrays A and B.
The next two lines contain the input of arrays A and B.
<b>Constraints</b>
1 ≤ n, m ≤ 1000
-10<sup>6</sup> ≤ A[i], B[i] ≤ 10<sup>6</sup>Print the median of two sorted arrays upto two decimal places.Sample Input :
2 1
1 3
2
Sample Output :
2.00, I have written this Solution Code: def getMedian(arr1, arr2, n, m):
i=j=0
sort=[]
while i<n and j<m:
if arr1[i] < arr2[j]:
sort.append(arr1[i])
i+=1
else:
sort.append(arr2[j])
j+=1
while i<n:
sort.append(arr1[i])
i+=1
while j<m:
sort.append(arr2[j])
j+=1
if (n+m)%2==1:
ind=(len(sort)//2)+1
num=sort[ind-1]
else:
mid=len(sort)//2
num=(sort[mid-1]+sort[mid])/2
return num
n, m= input().split()
n, m= int(n),int(m)
ar1=list(map(int, input().split()))
ar2=list(map(int, input().split()))
print("%.2f" %getMedian(ar1, ar2, n, m)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two sorted arrays A and B of size n and m respectively, return the median of the two sorted arrays.The first line of input contains n, m the length of arrays A and B.
The next two lines contain the input of arrays A and B.
<b>Constraints</b>
1 ≤ n, m ≤ 1000
-10<sup>6</sup> ≤ A[i], B[i] ≤ 10<sup>6</sup>Print the median of two sorted arrays upto two decimal places.Sample Input :
2 1
1 3
2
Sample Output :
2.00, I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-02-02 14:05:28
**/
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
if (nums2.size() < nums1.size())
return findMedianSortedArrays(nums2, nums1);
int n1 = nums1.size();
int n2 = nums2.size();
int lo = 0, hi = n1;
while (lo <= hi) {
int cut1 = (lo + hi) / 2;
int cut2 = (n1 + n2 + 1) / 2 - cut1;
int left1 = cut1 == 0 ? INT_MIN : nums1[cut1 - 1];
int left2 = cut2 == 0 ? INT_MIN : nums2[cut2 - 1];
int right1 = cut1 == n1 ? INT_MAX : nums1[cut1];
int right2 = cut2 == n2 ? INT_MAX : nums2[cut2];
if (left1 <= right2 && left2 <= right1) {
if ((n1 + n2) % 2 == 0)
return (max(left1, left2) + min(right1, right2)) / 2.0;
else
return max(left1, left2);
} else if (left1 > right2) {
hi = cut1 - 1;
} else
lo = cut1 + 1;
}
return 0.0;
}
int main() {
int n, m;
cin >> n >> m;
vector<int> a(n), b(m);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
for (int i = 0; i < m; i++) {
cin >> b[i];
}
printf("%0.2f", findMedianSortedArrays(a, b));
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
long z=0,x=0,y=0;
int choice;
Scanner in = new Scanner(System.in);
choice = in.nextInt();
String s="";
int f = 1;
while(f<=choice){
x = in.nextLong();
y = in.nextLong();
z = in.nextLong();
System.out.println((long)(Math.max((z-x),(z-y))));
f++;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: n = int(input())
for i in range(n):
l = list(map(int,input().split()))
print(l[2]-min(l)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 1e9 + 7;
const ll mod2 = 998244353;
const ll N = 2e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
signed main()
{
read(t);
assert(1 <= t && t <= ll(1e4));
while (t--)
{
readc(x, y, z);
assert(1 <= x && x <= ll(1e15));
assert(1 <= y && y <= ll(1e15));
assert(max(x, y) < z && z <= ll(1e15));
int r = 2*z - x - y - 1;
int l = z - max(x, y);
print(r - l + 1);
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
void solve()
{
int t;
cin>>t;
while(t--)
{
int x, y, z;
cin>>x>>y>>z;
cout<<max(z - y, z- x)<<endl;
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cerr.tie(NULL);
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen("INPUT.txt", "r", stdin);
freopen("OUTPUT.txt", "w", stdout);
}
#endif
solve();
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string, Calculate its rank among all its lexicographically sorted permutations.First line of the input contains a string S.
<b>Constraints</b>
1 ≤ |S| ≤ 8Print the Rank of that string.Sample Input
CBA
Sample Output
6
Explanation
The rank of string CBA in the lexicographically sorted permutations [ABC, ACB, BAC, BCA, CAB, CBA] is 6., I have written this Solution Code: #include <iostream>
#include <string>
#include <algorithm>
using namespace std;
// Function to find the lexicographic rank of a string using
// `std::next_permutation`
int findLexicographicRank(string key)
{
string str = key;
int rank = 1; // rank starts from 1
// sort the string in ascending order
sort(str.begin(), str.end());
while (1)
{
// if the current permutation is equal to the key, return its rank
if (key == str) {
return rank;
}
// find next lexicographically ordered permutation
if (!next_permutation(str.begin(), str.end())) {
break;
}
rank++;
}
return rank ;
}
int main()
{
string key;
cin>>key;
cout << findLexicographicRank(key);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rachel got a string as a gift from Ross. She wants to exchange this string with another string of same length. A string with all the same characters is fashionable string. Find Rachel lexicographically minimum fashionable string but make sure that the new string has at least one character common with the string Ross gave her otherwise he will get upset.Input contains a single string S, denoting the string Ross gave.
Constraints:
1 <= |S| <= 10000
S contains only lowercase english letters.Print the new string Rachel gets.Sample Input
bccde
Sample Output
bbbbb
Explanation: "bbbbb" is lexicographically minimum fashionable string which has atleast one character common with the initial string.
Sample Input
a
Sample Output
a, 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().trim();
char n[] = s.toCharArray();
int posMin = 0;
for(int i=0;i<n.length;i++){
if(n[posMin]>n[i])
posMin = i;
}
for(int i=0;i<n.length;i++)
n[i] = n[posMin];
System.out.print(String.valueOf(n));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rachel got a string as a gift from Ross. She wants to exchange this string with another string of same length. A string with all the same characters is fashionable string. Find Rachel lexicographically minimum fashionable string but make sure that the new string has at least one character common with the string Ross gave her otherwise he will get upset.Input contains a single string S, denoting the string Ross gave.
Constraints:
1 <= |S| <= 10000
S contains only lowercase english letters.Print the new string Rachel gets.Sample Input
bccde
Sample Output
bbbbb
Explanation: "bbbbb" is lexicographically minimum fashionable string which has atleast one character common with the initial string.
Sample Input
a
Sample Output
a, 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();
//////////////
template<class C> void mini(C&a4, C b4){a4=min(a4,b4);}
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define mod 1000000007ll
#define pii pair<int,int>
/////////////
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
string s;
cin>>s;
char c='z';
for(auto r:s)
c=min(c,r);
for(int i=0;i<s.length();++i)
cout<<c;
#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: Rachel got a string as a gift from Ross. She wants to exchange this string with another string of same length. A string with all the same characters is fashionable string. Find Rachel lexicographically minimum fashionable string but make sure that the new string has at least one character common with the string Ross gave her otherwise he will get upset.Input contains a single string S, denoting the string Ross gave.
Constraints:
1 <= |S| <= 10000
S contains only lowercase english letters.Print the new string Rachel gets.Sample Input
bccde
Sample Output
bbbbb
Explanation: "bbbbb" is lexicographically minimum fashionable string which has atleast one character common with the initial string.
Sample Input
a
Sample Output
a, I have written this Solution Code: s=input()
l=len(s)
k=min(s)
m=k*l
print(m), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, For each i (1<=i<=n) if i is even print "<b>even</b>" else print "<b>odd</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 functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 ≤ n ≤ 100Print even or odd for each i, separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: public static void For_Loop(int n){
for(int i=1;i<=n;i++){
if(i%2==1){System.out.print("odd ");}
else{
System.out.print("even ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, For each i (1<=i<=n) if i is even print "<b>even</b>" else print "<b>odd</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 functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 ≤ n ≤ 100Print even or odd for each i, separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: n = int(input())
for i in range(1, n+1):
if(i%2)==0:
print("even ",end="")
else:
print("odd ",end=""), 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: 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: To seek revenge on Midgard, Loki devises a plan to torture the humans by making them take part in one of his silly games. He makes N people sit in a circle. He says he will kill every kth person sitting in the circle, starting from 1st person.
Loki performs his revenge prank until and unless 1 survivor remains.
What is the initial position of the survivor, if the indexing is clockwise?The first line of input contains a single integer T. The next T line of input contains Two space separated integers each containing value of N and k.
<b>Constraints:</b>
1 <= T <= 100
1 <= k, N <= 20Print the initial position of the survivor.Sample Input:
2
3 2
5 3
Sample Output
3
4
Explanation:
Test case 1: There are 3 people so skipping 1 person i.e 1st person 2nd person will be killed in next step 3rd person will be skipped and 1st person will be killed. Thus the safe position is 3.
Test case 2: 2 people i.e 1and 2 are skipped and person 3 will be killed in next step 4 and 5 will be skipped and 1st person will be killed next step 2 and 4 will be skipped and 5th person will be killed next step first 2 will be skipped then 4 will be skipped and so coming back to 2 therefore person 2 will be killed. Thus the safe position is 4., I have written this Solution Code: t=int(input())
while t>0:
n,k=map(int,input().split())
s=set()
remove=0
for i in range(1,n+1):
s.add(i)
i=1
while True:
if len(s)==1:
break
if i in s:
remove+=1
if remove==k:
if i in s:
s.remove(i)
remove=0
i+=1
if i>n:
i=1
print(*s)
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: To seek revenge on Midgard, Loki devises a plan to torture the humans by making them take part in one of his silly games. He makes N people sit in a circle. He says he will kill every kth person sitting in the circle, starting from 1st person.
Loki performs his revenge prank until and unless 1 survivor remains.
What is the initial position of the survivor, if the indexing is clockwise?The first line of input contains a single integer T. The next T line of input contains Two space separated integers each containing value of N and k.
<b>Constraints:</b>
1 <= T <= 100
1 <= k, N <= 20Print the initial position of the survivor.Sample Input:
2
3 2
5 3
Sample Output
3
4
Explanation:
Test case 1: There are 3 people so skipping 1 person i.e 1st person 2nd person will be killed in next step 3rd person will be skipped and 1st person will be killed. Thus the safe position is 3.
Test case 2: 2 people i.e 1and 2 are skipped and person 3 will be killed in next step 4 and 5 will be skipped and 1st person will be killed next step 2 and 4 will be skipped and 5th person will be killed next step first 2 will be skipped then 4 will be skipped and so coming back to 2 therefore person 2 will be killed. Thus the safe position is 4., I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.*;
class Main{
public static void main (String[] args)throws IOException
{
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine().trim());
while(t-->0)
{
String str[] = read.readLine().trim().split(" ");
int n = Integer.parseInt(str[0]);
int k = Integer.parseInt(str[1]);
System.out.println(safe_Position(n,k));
}
}
public static int safe_Position(int n, int k)
{
if (n == 1) //base case
return 1;
else
/* The position returned by safe_Position(n - 1, k) is adjusted because the
recursive call safe_Position(n - 1, k) considers the original position
k%n + 1 as position 1 */
return (safe_Position(n - 1, k) + k-1) % n + 1; //recursion
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given 2 non-negative integers m and n, find gcd(m, n)
GCD of 2 integers m and n is defined as the greatest integer g such that g is a divisor of both m and n. Both m and n fit in a 32 bit signed integer.
NOTE: DO NOT USE LIBRARY FUNCTIONSInput contains two space separated integers, m and n
Constraints:-
1 < = m, n < = 10^18Output the single integer denoting the gcd of the above integers.Sample Input:
6 9
Sample Output:
3
Sample Input:-
5 6
Sample Output:-
1, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int n, m;
cin >> n >> m;
cout << __gcd(n, m);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given 2 non-negative integers m and n, find gcd(m, n)
GCD of 2 integers m and n is defined as the greatest integer g such that g is a divisor of both m and n. Both m and n fit in a 32 bit signed integer.
NOTE: DO NOT USE LIBRARY FUNCTIONSInput contains two space separated integers, m and n
Constraints:-
1 < = m, n < = 10^18Output the single integer denoting the gcd of the above integers.Sample Input:
6 9
Sample Output:
3
Sample Input:-
5 6
Sample Output:-
1, I have written this Solution Code: def hcf(a, b):
if(b == 0):
return a
else:
return hcf(b, a % b)
li= list(map(int,input().strip().split()))
print(hcf(li[0], li[1])), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
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