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For this Question: Implement the function <code>round</code>, which should take a number which can be a float(decimal)
and return its result as an integer rounded of (Use JS In built functions)Function will take a float as input (can be negative or positive)Function will return a rounded off numberconsole. log(round(1.112)) // prints 1
console. log(round(1.9)) // prints 2
console. log(round(-0.66)) // prints -1, I have written this Solution Code: function round(num){
// write code here
// return the output , do not use console.log here
return Math.round(num)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement the function <code>round</code>, which should take a number which can be a float(decimal)
and return its result as an integer rounded of (Use JS In built functions)Function will take a float as input (can be negative or positive)Function will return a rounded off numberconsole. log(round(1.112)) // prints 1
console. log(round(1.9)) // prints 2
console. log(round(-0.66)) // prints -1, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.math.*;
class Main {
public static void main (String[] args) {
Scanner sc=new Scanner(System.in);
double n=sc.nextDouble();
System.out.println(Math.round(n));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import static java.lang.System.out;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader reader = new FastReader();
int n = reader.nextInt();
String S = reader.next();
int ncount = 0;
int tcount = 0;
for (char c : S.toCharArray()) {
if (c == 'N') ncount++;
else tcount++;
}
if (ncount > tcount) {
out.print("Nutan\n");
} else {
out.print("Tusla\n");
}
out.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: n = int(input())
s = input()
a1 = s.count('N')
a2 = s.count('T')
if(a1 > a2):
print("Nutan")
else:
print('Tusla'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: //Author: Xzirium
//Time and Date: 02:18:28 24 March 2022
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//-----------------------------------------------------------------------------------------------------------//
READV(N);
string S;
cin>>S;
ll n=0,t=0;
FORI(i,0,N)
{
if(S[i]=='N')
{
n++;
}
else if(S[i]=='T')
{
t++;
}
}
if(n>t)
{
cout<<"Nutan"<<endl;
}
else
{
cout<<"Tusla"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N. Find the smallest positive number that is evenly divisible by all numbers between 1 to N(inclusive).Only line will contain N.
<b>Constraints</b>
1 ≤ N ≤ 1000A single integer denoting the answer.Input:
6
Output:
60
Explanation:
No other number smaller than 60 is divisible by all {1, 2, 3, 4, 5, 6}., I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
private static long gcd(long a, long b)
{
if(a<b){
long temp=a;
a=b;
b=temp;
}
if (b == 0)
return a;
else
return gcd(b, a % b);
}
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
int n=Integer.parseInt(in.next());
long ans=1;
for(int i=1;i<=n;i++){
ans=(ans*i)/(gcd(i,ans));
}
out.print(ans);
out.close();
}
static class InputReader {
BufferedReader reader;
StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
public void println(int i) {
writer.println(i);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Our five power rangers have powers P1, P2, P3, P4, and P5.
To ensure the proper distribution of power, the power of every power ranger must remain <b>less</b> than the sum of powers of other power rangers.
If the above condition is not met, there's an emergency. Can you let us know if there's an emergency?The first and the only line of input contains 5 integers P1, P2, P3, P4, and P5.
Constraints
0 <= P1, P2, P3, P4, P5 <= 100Output "SPD Emergency" (without quotes) if there's an emergency, else output "Stable".Sample Input
1 2 3 4 5
Sample Output
Stable
Explanation
The power of every power ranger is less than the sum of powers of other power rangers.
Sample Input
1 2 3 4 100
Sample Output
SPD Emergency
Explanation
The power of the 5th power ranger (100) is not less than the sum of powers of other power rangers (1+2+3+4=10)., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
int[] arr=new int[5];
BufferedReader rd=new BufferedReader(new InputStreamReader(System.in));
String[] s=rd.readLine().split(" ");
int sum=0;
for(int i=0;i<5;i++){
arr[i]=Integer.parseInt(s[i]);
sum+=arr[i];
}
int i=0,j=arr.length-1;
boolean isEmergency=false;
while(i<=j)
{
int temp=arr[i];
sum-=arr[i];
if(arr[i]>= sum)
{
isEmergency=true;
break;
}
sum+=temp;
i++;
}
if(isEmergency==false)
{
System.out.println("Stable");
}
else
{
System.out.println("SPD Emergency");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Our five power rangers have powers P1, P2, P3, P4, and P5.
To ensure the proper distribution of power, the power of every power ranger must remain <b>less</b> than the sum of powers of other power rangers.
If the above condition is not met, there's an emergency. Can you let us know if there's an emergency?The first and the only line of input contains 5 integers P1, P2, P3, P4, and P5.
Constraints
0 <= P1, P2, P3, P4, P5 <= 100Output "SPD Emergency" (without quotes) if there's an emergency, else output "Stable".Sample Input
1 2 3 4 5
Sample Output
Stable
Explanation
The power of every power ranger is less than the sum of powers of other power rangers.
Sample Input
1 2 3 4 100
Sample Output
SPD Emergency
Explanation
The power of the 5th power ranger (100) is not less than the sum of powers of other power rangers (1+2+3+4=10)., I have written this Solution Code: arr = list(map(int,input().split()))
m = sum(arr)
f=[]
for i in range(len(arr)):
s = sum(arr[:i]+arr[i+1:])
if(arr[i]<s):
f.append(1)
else:
f.append(0)
if(all(f)):
print("Stable")
else:
print("SPD Emergency"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Our five power rangers have powers P1, P2, P3, P4, and P5.
To ensure the proper distribution of power, the power of every power ranger must remain <b>less</b> than the sum of powers of other power rangers.
If the above condition is not met, there's an emergency. Can you let us know if there's an emergency?The first and the only line of input contains 5 integers P1, P2, P3, P4, and P5.
Constraints
0 <= P1, P2, P3, P4, P5 <= 100Output "SPD Emergency" (without quotes) if there's an emergency, else output "Stable".Sample Input
1 2 3 4 5
Sample Output
Stable
Explanation
The power of every power ranger is less than the sum of powers of other power rangers.
Sample Input
1 2 3 4 100
Sample Output
SPD Emergency
Explanation
The power of the 5th power ranger (100) is not less than the sum of powers of other power rangers (1+2+3+4=10)., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
vector<int> vect(5);
int tot = 0;
for(int i=0; i<5; i++){
cin>>vect[i];
tot += vect[i];
}
sort(all(vect));
tot -= vect[4];
if(vect[4] >= tot){
cout<<"SPD Emergency";
}
else{
cout<<"Stable";
}
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a 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: Shinchan and Kazama are standing in a horizontal line, Shinchan is standing at point A and Kazama is standing at point B. Kazama is very intelligent and recently he learned how to calculate the speed if the distance and time are given and now he wants to check if the formula he learned is correct or not So he starts running at a speed of S unit/s towards Shinhan and noted the time he reaches to Shinhan. Since Kazama is disturbed by Shinchan, can you calculate the time for him?The input contains three integers A, B, and S separated by spaces.
Constraints:-
1 <= A, B, V <= 1000
Note:- It is guaranteed that the calculated distance will always be divisible by V.Print the Time taken in seconds by Kazama to reach Shinchan.
Note:- Remember Distance can not be negativeSample Input:-
5 2 3
Sample Output:-
1
Explanation:-
Distance = 5-2 = 3, Speed = 3
Time = Distance/Speed
Sample Input:-
9 1 2
Sample Output:-
4, I have written this Solution Code: a, b, v = map(int, input().strip().split(" "))
c = abs(a-b)
t = c//v
print(t), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Shinchan and Kazama are standing in a horizontal line, Shinchan is standing at point A and Kazama is standing at point B. Kazama is very intelligent and recently he learned how to calculate the speed if the distance and time are given and now he wants to check if the formula he learned is correct or not So he starts running at a speed of S unit/s towards Shinhan and noted the time he reaches to Shinhan. Since Kazama is disturbed by Shinchan, can you calculate the time for him?The input contains three integers A, B, and S separated by spaces.
Constraints:-
1 <= A, B, V <= 1000
Note:- It is guaranteed that the calculated distance will always be divisible by V.Print the Time taken in seconds by Kazama to reach Shinchan.
Note:- Remember Distance can not be negativeSample Input:-
5 2 3
Sample Output:-
1
Explanation:-
Distance = 5-2 = 3, Speed = 3
Time = Distance/Speed
Sample Input:-
9 1 2
Sample Output:-
4, I have written this Solution Code: /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int m = sc.nextInt();
int k = sc.nextInt();
System.out.print(Time(n,m,k));
}
static int Time(int A, int B, int S){
if(B>A){
return (B-A)/S;
}
return (A-B)/S;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer K, find a positive integer x such that <b>K = x<sup>2</sup> + 3*x</b>. If no such positive integer x exists, print -1.First and the only line of the input contains an integer K.
Constraints:
1 <= K <= 10<sup>18</sup>Print a positive integer x such that the above equation satisfies. If no such integer x exists, print -1.Sample Input:
28
Sample Output:
4
Explaination:
4<sup>2</sup> + 3*4 = 28
There is no other positive integer that will give such result., I have written this Solution Code: def FindIt(n):
sqrt_n = int(K**0.5)
check = -1
for i in range(sqrt_n - 1, sqrt_n - 2, -1):
check = i**2 + (i * 3)
if check == n:
return i
return -1
K = int(input())
print(FindIt(K)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer K, find a positive integer x such that <b>K = x<sup>2</sup> + 3*x</b>. If no such positive integer x exists, print -1.First and the only line of the input contains an integer K.
Constraints:
1 <= K <= 10<sup>18</sup>Print a positive integer x such that the above equation satisfies. If no such integer x exists, print -1.Sample Input:
28
Sample Output:
4
Explaination:
4<sup>2</sup> + 3*4 = 28
There is no other positive integer that will give such result., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL);
#define int long long
#define pb push_back
#define ff first
#define ss second
#define endl '\n'
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
using T = pair<int, int>;
typedef long double ld;
const int mod = 1e9 + 7;
const int INF = 1e9;
void solve(){
int n;
cin >> n;
int l = 1, r = 1e9, ans = -1;
while(l <= r){
int m = (l + r)/2;
int val = m*m + 3*m;
if(val == n){
ans = m;
break;
}
if(val < n){
l = m + 1;
}
else r = m - 1;
}
cout << ans;
}
signed main(){
fast
int t = 1;
// cin >> t;
for(int i = 1; i <= t; i++){
solve();
if(i != t) cout << endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer K, find a positive integer x such that <b>K = x<sup>2</sup> + 3*x</b>. If no such positive integer x exists, print -1.First and the only line of the input contains an integer K.
Constraints:
1 <= K <= 10<sup>18</sup>Print a positive integer x such that the above equation satisfies. If no such integer x exists, print -1.Sample Input:
28
Sample Output:
4
Explaination:
4<sup>2</sup> + 3*4 = 28
There is no other positive integer that will give such result., 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));
long K = Long.parseLong(br.readLine());
long ans = -1;
for(long x =0;((x*x)+(3*x))<=K;x++){
if(K==((x*x)+(3*x))){
ans = x;
break;
}
}
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara's Phone has N apps and each app takes K unit of memory. Now Sara wants to release M units of memory. Your task is to tell the minimum apps Sara needs to delete or say it is not possible.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Phone()</b> that takes integers N, K, and M as arguments.
Constraints:-
1 <= N <= 1000
1 <= K <= 100
0 <= M <= 10000Return minimum apps to delete and if it is not possible return -1.Sample Input:-
10 3 10
Sample Output:-
4
Sample Input:-
10 3 40
Sample Output:-
-1, I have written this Solution Code:
int Phone(int n, int k, int m){
if(n*k<m){
return -1;
}
int x = m/k;
if(m%k!=0){x++;}
return x;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara's Phone has N apps and each app takes K unit of memory. Now Sara wants to release M units of memory. Your task is to tell the minimum apps Sara needs to delete or say it is not possible.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Phone()</b> that takes integers N, K, and M as arguments.
Constraints:-
1 <= N <= 1000
1 <= K <= 100
0 <= M <= 10000Return minimum apps to delete and if it is not possible return -1.Sample Input:-
10 3 10
Sample Output:-
4
Sample Input:-
10 3 40
Sample Output:-
-1, I have written this Solution Code:
int Phone(int n, int k, int m){
if(n*k<m){
return -1;
}
int x = m/k;
if(m%k!=0){x++;}
return x;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara's Phone has N apps and each app takes K unit of memory. Now Sara wants to release M units of memory. Your task is to tell the minimum apps Sara needs to delete or say it is not possible.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Phone()</b> that takes integers N, K, and M as arguments.
Constraints:-
1 <= N <= 1000
1 <= K <= 100
0 <= M <= 10000Return minimum apps to delete and if it is not possible return -1.Sample Input:-
10 3 10
Sample Output:-
4
Sample Input:-
10 3 40
Sample Output:-
-1, I have written this Solution Code: static int Phone(int n, int k, int m){
if(n*k<m){
return -1;
}
int x = m/k;
if(m%k!=0){x++;}
return x;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara's Phone has N apps and each app takes K unit of memory. Now Sara wants to release M units of memory. Your task is to tell the minimum apps Sara needs to delete or say it is not possible.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Phone()</b> that takes integers N, K, and M as arguments.
Constraints:-
1 <= N <= 1000
1 <= K <= 100
0 <= M <= 10000Return minimum apps to delete and if it is not possible return -1.Sample Input:-
10 3 10
Sample Output:-
4
Sample Input:-
10 3 40
Sample Output:-
-1, I have written this Solution Code: def Phone(N,K,M):
if N*K < M :
return -1
x = M//K
if M%K!=0:
x=x+1
return x, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
Sample Output:
No, I have written this Solution Code: m,n = map(int , input().split())
if (m%n==0):
print("Yes")
else:
print("No");, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
Sample Output:
No, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n,m;
cin>>n>>m;
if(n%m==0)
cout<<"Yes";
else
cout<<"No";
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
Sample Output:
No, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
long n = sc.nextLong();
Long m = sc.nextLong();
if(n%m==0){
System.out.print("Yes");
}
else{
System.out.print("No");
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have N buckets of apples, i<sup>th</sup> bucket has bucket[i] apples.
You plan to eat apples for the next M days. On j<sup>th</sup> day you take the bucket with maximum apples and eat day[j] apples from the bucket. if there is not enough apple in the bucket then you eat all the apples from that bucket.
After M days, what is the maximum number of apples in any bucket?First line contains a single integer N.
Next line contains n space separated integers.
Third line contains M.
Next line contains M space separated integers.
<b>Constraints</b>
1 <= N, M<= 10<sup>5</sup>
1 <= bucket[i], day[i] <= 10<sup>9</sup>A single integer denoting required answer.Input:
3
12 17 10
4
10 8 6 9
Output:
4
Explanation:
day1 =10 => buckets = {12, 17, 10 } => {12, 7, 10}
day2 = 8 => buckets = {12, 7, 10 } => {4, 7, 10}
day3 = 6 => buckets = {4, 7, 10} => { 4, 7, 4 }
day4 = 9 => buckets = {4, 7, 4} => { 4, 0, 4 }
on day4 we needed 9 apples, but the maximum in a bucket was only 7., I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
int n=Integer.parseInt(in.next());
int bucket[] = new int [n];
for(int i=0;i<n;i++){
bucket[i] = Integer.parseInt(in.next());
}
int m=Integer.parseInt(in.next());
int day[] = new int [m];
for(int i=0;i<m;i++){
day[i] = Integer.parseInt(in.next());
}
PriorityQueue<Integer> pq=new PriorityQueue<>(Collections.reverseOrder());
for(int i=0;i<n;i++){
pq.add(bucket[i]);
}
for(int i=0;i<m && !pq.isEmpty();i++){
int x=pq.poll();
x -= day[i];
if(x>0)pq.add(x);
}
int ans=0;
if(!pq.isEmpty())ans=pq.peek();
out.print(ans);
out.close();
}
static class InputReader {
BufferedReader reader;
StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
public void println(int i) {
writer.println(i);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: 4 friends are playing Make Believe Agents. You have 4 chairs fixed to the ground.
Now, you want to draw a square outline such that all the chairs are inside the square. This square will be the secret base. Now each of the "Agent" wants a favourite side of the square. A side is favourite if it is exactly 13 units aways from the Agent's chair and it is the nearest side among the 4 sides of square from the Agent.
To avoid internal fights no two agents should have same favourite side. You have to find the side length of the biggest room that satisfies all the above conditions or report -1 if it is not possible.First line of input contains T, denoting number of test cases.
Next T lines contain 8 integers denoting x1, y1, x2, y2, x3, y3, x4, and y4.
Constraints:
1 <= T <= 100
-100 <= xi, yi <= 100
All the four (x, y) coordinates are distinct.For each test case, print the side length of the biggest room that satisfies all the above conditions or report -1 if it is not possible.
Output <b>must</b> be rounded to two digits after the decimal.
Make sure to print the same format as in the sample.Sample Input
2
0 1 2 5 5 4 3 0
10 10 30 30 20 20 0 10
Sample Output
31.00
-1, 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 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>
#define rep(i,n) for (int i=0; i<(n); i++)
/////////////
#define eps 1e-11
#define bs 1000000007
#define bsize 256
#define MAGIC 113
using namespace std;
int tests;
long double sx[10],sy[10],qx[10],qy[10];
long double eval(long double ang)
{
for (int i=0;i<4;i++)
{
qx[i]=sx[i]*cos(ang)-sy[i]*sin(ang);
qy[i]=sx[i]*sin(ang)+sy[i]*cos(ang);
}
long double minx,maxx,miny,maxy;
minx=miny=1e9;
maxx=maxy=-1e9;
for (int i=0;i<4;i++)
{
minx=min(minx,qx[i]);
maxx=max(maxx,qx[i]);
miny=min(miny,qy[i]);
maxy=max(maxy,qy[i]);
}
return fabs(maxx+-maxy-minx+miny);
}
long double get(long double ang)
{
for (int i=0;i<4;i++)
{
qx[i]=sx[i]*cos(ang)-sy[i]*sin(ang);
qy[i]=sx[i]*sin(ang)+sy[i]*cos(ang);
}
long double minx,maxx,miny,maxy;
minx=miny=1e9;
maxx=maxy=-1e9;
for (int i=0;i<4;i++)
{
minx=min(minx,qx[i]);
maxx=max(maxx,qx[i]);
miny=min(miny,qy[i]);
maxy=max(maxy,qy[i]);
}
for (int i=0;i<4;i++)
{
int cnt=0;
if (fabs(qx[i]-minx)<eps)cnt++;
if (fabs(qx[i]-maxx)<eps)++cnt;
if (fabs(qy[i]-miny)<eps)++cnt;
if (fabs(qy[i]-maxy)<eps)++cnt;
if (cnt!=1)
return -1;
}
return maxx-minx+10;
}
long double solve()
{
long double ans=-1e9;
long double one=M_PI/2/MAGIC;
for (int P=0;P<MAGIC;P++)
{
long double l,r;
l=P*one;
r=(P+1)*one;
for (int iter=1;iter<=100;iter++)
{
double mid1,mid2;
mid1=l*2+r;
mid2=l+r*2;
mid1/=3;
mid2/=3;
long double val1=eval(mid1);
long double val2=eval(mid2);
if (val1<=val2)r=mid2;
else l=mid1;
}
// cout<<eval(l)<<endl;
long double val=eval(l);
if (val<eps)
{
ans=max(ans,get(l));
}
}
for (int P=0;P<MAGIC;P++)
{
long double l,r;
l=P*one;
r=(P+1)*one;
for (int iter=1;iter<=100;iter++)
{
long double mid1,mid2;
mid1=l*2+r;
mid2=l+r*2;
mid1/=3;
mid2/=3;
long double val1=eval(mid1);
long double val2=eval(mid2);
if (val1>=val2)r=mid2;
else l=mid1;
}
//cout<<eval(l)<<endl;
//cin.get();
double val=eval(l);
if (val<eps)
{
ans=max(ans,get(l));
}
}
return ans;
}
int main(){
ios_base::sync_with_stdio(0);
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
cin>>tests;
for (;tests;--tests)
{
for (int i=0;i<4;i++)
{
cin>>sx[i]>>sy[i];
}
long double ans=solve();
cout.precision(2);
if (ans<0)cout<<"-1"<<endl;
else cout<<fixed<<ans+16<<endl;
}
#ifdef ANIKET_GOYAL
cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N coordinates on a two dimensional plane. Find the area of the smallest rectangle such that all the points can lie inside or on the rectangle boundary.
Note - the sides of rectangle should be parallel to x and y axis.First line of input contains N.
Next N lines contains two integers x[i] and y[i].
Constraints:
2 <= N <= 100000
0 <= x[i], y[i] <= 1000000000
Note the required rectangle will never have 0 area.Print the area of the smallest rectangle such that all the points can lie inside or on the rectangle the boundary.Sample Input
2
0 0
1 1
Sample Output
1
Explanation: required rectangle has corners at (0, 0) (0, 1) (1, 1) (1, 0), I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws Exception {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
int N = Integer.parseInt(reader.readLine());
int minX = Integer.MAX_VALUE, maxX = Integer.MIN_VALUE;
int minY = Integer.MAX_VALUE, maxY = Integer.MIN_VALUE;
for(int i = 0; i < N; i++) {
StringTokenizer tokens = new StringTokenizer(reader.readLine());
int x = Integer.parseInt(tokens.nextToken());
int y = Integer.parseInt(tokens.nextToken());
minX = Math.min(minX, x);
maxX = Math.max(maxX, x);
minY = Math.min(minY, y);
maxY = Math.max(maxY, y);
}
long X = maxX - minX;
long Y = maxY - minY;
long area = X * Y;
System.out.println(X * Y);
reader.close();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N coordinates on a two dimensional plane. Find the area of the smallest rectangle such that all the points can lie inside or on the rectangle boundary.
Note - the sides of rectangle should be parallel to x and y axis.First line of input contains N.
Next N lines contains two integers x[i] and y[i].
Constraints:
2 <= N <= 100000
0 <= x[i], y[i] <= 1000000000
Note the required rectangle will never have 0 area.Print the area of the smallest rectangle such that all the points can lie inside or on the rectangle the boundary.Sample Input
2
0 0
1 1
Sample Output
1
Explanation: required rectangle has corners at (0, 0) (0, 1) (1, 1) (1, 0), I have written this Solution Code:
minx = 1000000000
maxx= 0
miny=1000000000
maxy=0
n = int(input())
while(n):
a,b = map(int,input().split())
minx = min(minx,a);
maxx = max(maxx,a);
miny = min(miny,b);
maxy = max(maxy,b);
n -=1
l = (maxx-minx);
b = (maxy-miny);
print(l*b), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N coordinates on a two dimensional plane. Find the area of the smallest rectangle such that all the points can lie inside or on the rectangle boundary.
Note - the sides of rectangle should be parallel to x and y axis.First line of input contains N.
Next N lines contains two integers x[i] and y[i].
Constraints:
2 <= N <= 100000
0 <= x[i], y[i] <= 1000000000
Note the required rectangle will never have 0 area.Print the area of the smallest rectangle such that all the points can lie inside or on the rectangle the boundary.Sample Input
2
0 0
1 1
Sample Output
1
Explanation: required rectangle has corners at (0, 0) (0, 1) (1, 1) (1, 0), I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define 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 mix=infi,maax=0;
int miy=infi,may=0;
for(int i=0;i<n;++i){
int x,y;
cin>>x>>y;
mix=min(mix,x);
miy=min(miy,y);
maax=max(maax,x);
may=max(may,y);
}
cout<<(maax-mix)*(may-miy);
#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: Implement the function <code>doesItContain</code>, which takes 2 strings as argument,
return true or false whether the first string contains second string or not(Use JS In built functions)Function will take 2 args,
1) string in which to check
2) the string which is to be checked is it in 1st stringFunction will return boolean whether it contains given substring or notconsole. log(doesItContain("Hi World world", "world")) // prints true
console. log(doesItContain("hi hi hi", "hello")) // prints false, I have written this Solution Code: function doesItContain(line, charToBeReplaced) {
// write code here
// return the output , do not use console.log here
return line.includes(charToBeReplaced)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A (0 indexed) of n integers from 0 to n-1 (each integer exactly once), output an array such that A[i] becomes A[A[i]].
Make sure you don't use extra memory than the array itself.The first line of the input contains a number n, the number of integers in the array.
The second line of the input contains n numbers, the elements of A.
<b>Constraints:</b>
1 <= n <= 100000
0 <= A[i] <= n-1Output the required array of n integers.Sample Input 1:
5
4 2 3 0 1
Sample Output 1:
1 3 0 4 2
Sample Input 2:
10
9 5 1 4 7 8 0 6 3 2
Sample Output 2:
2 8 5 7 6 3 9 0 4 1
<b>Explanation 1:</b>
A[0] will be A[A[0]]=A[4]=1
A[1] will be A[A[1]]=A[2]=3
A[2] will be A[A[2]]=A[3]=0
A[3] will be A[A[3]]=A[0]=4
A[4] will be A[A[4]]=A[1]=2, I have written this Solution Code: function simpleArrangement(n, arr) {
// write code here
// do not console.log
// return the output as an array
const newArr = []
// for (let i = 0; i < n; i++) {
// arr[i] += (arr[arr[i]] % n) * n;
// }
// Second Step: Divide all values by n
for (let i = 0; i < n; i++) {
newArr.push(arr[arr[i]])
}
return newArr
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A (0 indexed) of n integers from 0 to n-1 (each integer exactly once), output an array such that A[i] becomes A[A[i]].
Make sure you don't use extra memory than the array itself.The first line of the input contains a number n, the number of integers in the array.
The second line of the input contains n numbers, the elements of A.
<b>Constraints:</b>
1 <= n <= 100000
0 <= A[i] <= n-1Output the required array of n integers.Sample Input 1:
5
4 2 3 0 1
Sample Output 1:
1 3 0 4 2
Sample Input 2:
10
9 5 1 4 7 8 0 6 3 2
Sample Output 2:
2 8 5 7 6 3 9 0 4 1
<b>Explanation 1:</b>
A[0] will be A[A[0]]=A[4]=1
A[1] will be A[A[1]]=A[2]=3
A[2] will be A[A[2]]=A[3]=0
A[3] will be A[A[3]]=A[0]=4
A[4] will be A[A[4]]=A[1]=2, I have written this Solution Code: n = int(input())
a = input().split()
print(" ".join(a[int(x)] for x in a)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A (0 indexed) of n integers from 0 to n-1 (each integer exactly once), output an array such that A[i] becomes A[A[i]].
Make sure you don't use extra memory than the array itself.The first line of the input contains a number n, the number of integers in the array.
The second line of the input contains n numbers, the elements of A.
<b>Constraints:</b>
1 <= n <= 100000
0 <= A[i] <= n-1Output the required array of n integers.Sample Input 1:
5
4 2 3 0 1
Sample Output 1:
1 3 0 4 2
Sample Input 2:
10
9 5 1 4 7 8 0 6 3 2
Sample Output 2:
2 8 5 7 6 3 9 0 4 1
<b>Explanation 1:</b>
A[0] will be A[A[0]]=A[4]=1
A[1] will be A[A[1]]=A[2]=3
A[2] will be A[A[2]]=A[3]=0
A[3] will be A[A[3]]=A[0]=4
A[4] will be A[A[4]]=A[1]=2, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
String s = br.readLine();
String[] str = s.split(" ");
int arr[] = new int[n];
for(int i=0;i<n;i++){
arr[i]=Integer.parseInt(str[i]);
}
for(int i =0;i<n;i++){
System.out.print(arr[arr[i]]+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A (0 indexed) of n integers from 0 to n-1 (each integer exactly once), output an array such that A[i] becomes A[A[i]].
Make sure you don't use extra memory than the array itself.The first line of the input contains a number n, the number of integers in the array.
The second line of the input contains n numbers, the elements of A.
<b>Constraints:</b>
1 <= n <= 100000
0 <= A[i] <= n-1Output the required array of n integers.Sample Input 1:
5
4 2 3 0 1
Sample Output 1:
1 3 0 4 2
Sample Input 2:
10
9 5 1 4 7 8 0 6 3 2
Sample Output 2:
2 8 5 7 6 3 9 0 4 1
<b>Explanation 1:</b>
A[0] will be A[A[0]]=A[4]=1
A[1] will be A[A[1]]=A[2]=3
A[2] will be A[A[2]]=A[3]=0
A[3] will be A[A[3]]=A[0]=4
A[4] will be A[A[4]]=A[1]=2, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
long long n,k;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
for(int i=0;i<n;i++){
cout<<a[a[i]]<<" ";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, check if it contains any negative integer.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
-10000 < = Arr[i] < = 10000Print "Yes" if the array contains any negative integer else print "No".Sample Input:-
4
1 2 3 -3
Sample Output:-
Yes
Sample Input:-
3
1 2 3
Sample Output:-
No, I have written this Solution Code: N = int(input())
Nums = list(map(int,input().split()))
f = False
for n in Nums:
if n < 0:
f = True
break
if (f):
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, check if it contains any negative integer.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
-10000 < = Arr[i] < = 10000Print "Yes" if the array contains any negative integer else print "No".Sample Input:-
4
1 2 3 -3
Sample Output:-
Yes
Sample Input:-
3
1 2 3
Sample Output:-
No, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a;
bool win=false;
for(int i=0;i<n;i++){
cin>>a;
if(a<0){win=true;}}
if(win){
cout<<"Yes";
}
else{
cout<<"No";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, check if it contains any negative integer.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
-10000 < = Arr[i] < = 10000Print "Yes" if the array contains any negative integer else print "No".Sample Input:-
4
1 2 3 -3
Sample Output:-
Yes
Sample Input:-
3
1 2 3
Sample Output:-
No, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i] = sc.nextInt();
}
for(int i=0;i<n;i++){
if(a[i]<0){System.out.print("Yes");return;}
}
System.out.print("No");
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Harry is trying to find the chamber of secrets but the chamber of secrets is well protected by dark spells. To find the chamber Harry needs to find the initial position of the chamber. Harry knows the final position of the chamber is X. He also knows that in total, the chamber has been repositioned N times. During the i<sup>th</sup> repositioning, the position of the chamber is changed to (previous position * Arr[i]) % 1000000007.
Given X and Arr, help Harry find the initial position of the chamber.
Note:- The initial position of the chamber is less than 1000000007.The first line of the input contains two integers N and X.
The second line of the input contains N integers denoting Arr.
Constraints
1 <= N <= 100000
1 <= Arr[i] < 1000000007Print a single integer denoting the initial position of the chamber.Sample Input
5 480
1 4 4 3 1
Sample Output
10
Explanation: Initial position = 10
After first repositioning position = (10*1)%1000000007 = 10
After second repositioning position = (10*4)%1000000007 = 40
After third repositioning position = (40*4)%1000000007 = 160
After fourth repositioning position = (160*3)%1000000007 = 480
After fifth repositioning position = (480*1)%1000000007 = 480, I have written this Solution Code: def inv(a,b,m):
res = 1
while(b):
if b&1:
res = (res*a)%m
a = (a*a)%m
b >>= 1
return res
n,x = map(int,input().split())
a = list(map(int,input().split()))
m = 1000000007
for i in a:
x = (x*inv(i,m-2,m))%m
print(x), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Harry is trying to find the chamber of secrets but the chamber of secrets is well protected by dark spells. To find the chamber Harry needs to find the initial position of the chamber. Harry knows the final position of the chamber is X. He also knows that in total, the chamber has been repositioned N times. During the i<sup>th</sup> repositioning, the position of the chamber is changed to (previous position * Arr[i]) % 1000000007.
Given X and Arr, help Harry find the initial position of the chamber.
Note:- The initial position of the chamber is less than 1000000007.The first line of the input contains two integers N and X.
The second line of the input contains N integers denoting Arr.
Constraints
1 <= N <= 100000
1 <= Arr[i] < 1000000007Print a single integer denoting the initial position of the chamber.Sample Input
5 480
1 4 4 3 1
Sample Output
10
Explanation: Initial position = 10
After first repositioning position = (10*1)%1000000007 = 10
After second repositioning position = (10*4)%1000000007 = 40
After third repositioning position = (40*4)%1000000007 = 160
After fourth repositioning position = (160*3)%1000000007 = 480
After fifth repositioning position = (480*1)%1000000007 = 480, 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>
/////////////
int power_mod(int a,int b,int mod){
int ans = 1;
while(b){
if(b&1)
ans = (ans*a)%mod;
b = b/2;
a = (a*a)%mod;
}
return ans;
}
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int x;
cin>>x;
int mo=1000000007;
int mu=1;
for(int i=0;i<n;++i){
int d;
cin>>d;
mu=(mu*d)%mo;
}
cout<<(x*power_mod(mu,mo-2,mo))%mo;
#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: Harry is trying to find the chamber of secrets but the chamber of secrets is well protected by dark spells. To find the chamber Harry needs to find the initial position of the chamber. Harry knows the final position of the chamber is X. He also knows that in total, the chamber has been repositioned N times. During the i<sup>th</sup> repositioning, the position of the chamber is changed to (previous position * Arr[i]) % 1000000007.
Given X and Arr, help Harry find the initial position of the chamber.
Note:- The initial position of the chamber is less than 1000000007.The first line of the input contains two integers N and X.
The second line of the input contains N integers denoting Arr.
Constraints
1 <= N <= 100000
1 <= Arr[i] < 1000000007Print a single integer denoting the initial position of the chamber.Sample Input
5 480
1 4 4 3 1
Sample Output
10
Explanation: Initial position = 10
After first repositioning position = (10*1)%1000000007 = 10
After second repositioning position = (10*4)%1000000007 = 40
After third repositioning position = (40*4)%1000000007 = 160
After fourth repositioning position = (160*3)%1000000007 = 480
After fifth repositioning position = (480*1)%1000000007 = 480, I have written this Solution Code: import java.io.*;import java.util.*;import java.math.*;
public class Main
{
long mod=1000000007l;int max=Integer.MAX_VALUE,min=Integer.MIN_VALUE;
long maxl=Long.MAX_VALUE,minl=Long.MIN_VALUE;
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st;StringBuilder sb;
public void tq()throws Exception
{
st=new StringTokenizer(br.readLine());
int tq=1;
o:
while(tq-->0)
{
int n=i();
long k=l();
long ar[]=arl(n);
long v=1l;
for(long x:ar)v=(v*x)%mod;
v=(k*(mul(v,mod-2,mod)))%mod;
pl(v);
}
}
public static void main(String[] a)throws Exception{new Main().tq();}
int[] so(int ar[]){Integer r[]=new Integer[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];
Arrays.sort(r);for(int x=0;x<ar.length;x++)ar[x]=r[x];return ar;}
long[] so(long ar[]){Long r[]=new Long[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];
Arrays.sort(r);for(int x=0;x<ar.length;x++)ar[x]=r[x];return ar;}
char[] so(char ar[])
{Character r[]=new Character[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];
Arrays.sort(r);for(int x=0;x<ar.length;x++)ar[x]=r[x];return ar;}
void s(String s){sb.append(s);}void s(int s){sb.append(s);}void s(long s){sb.append(s);}
void s(char s){sb.append(s);}void s(double s){sb.append(s);}
void ss(){sb.append(' ');}void sl(String s){sb.append(s);sb.append("\n");}
void sl(int s){sb.append(s);sb.append("\n");}
void sl(long s){sb.append(s);sb.append("\n");}void sl(char s){sb.append(s);sb.append("\n");}
void sl(double s){sb.append(s);sb.append("\n");}void sl(){sb.append("\n");}
int min(int a,int b){return a<b?a:b;}
int min(int a,int b,int c){return a<b?a<c?a:c:b<c?b:c;}
int max(int a,int b){return a>b?a:b;}
int max(int a,int b,int c){return a>b?a>c?a:c:b>c?b:c;}
long min(long a,long b){return a<b?a:b;}
long min(long a,long b,long c){return a<b?a<c?a:c:b<c?b:c;}
long max(long a,long b){return a>b?a:b;}
long max(long a,long b,long c){return a>b?a>c?a:c:b>c?b:c;}
int abs(int a){return Math.abs(a);}
long abs(long a){return Math.abs(a);}
int sq(int a){return (int)Math.sqrt(a);}long sq(long a){return (long)Math.sqrt(a);}
long gcd(long a,long b){return b==0l?a:gcd(b,a%b);}
boolean pa(String s,int i,int j){while(i<j)if(s.charAt(i++)!=s.charAt(j--))return false;return true;}
boolean[] si(int n)
{boolean bo[]=new boolean[n+1];bo[0]=true;bo[1]=true;for(int x=4;x<=n;x+=2)bo[x]=true;
for(int x=3;x*x<=n;x+=2){if(!bo[x]){int vv=(x<<1);for(int y=x*x;y<=n;y+=vv)bo[y]=true;}}
return bo;}long mul(long a,long b,long m)
{long r=1l;a%=m;while(b>0){if((b&1)==1)r=(r*a)%m;b>>=1;a=(a*a)%m;}return r;}
int i()throws IOException{if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
return Integer.parseInt(st.nextToken());}
long l()throws IOException{if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
return Long.parseLong(st.nextToken());}String s()throws IOException
{if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());return st.nextToken();}
double d()throws IOException{if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
return Double.parseDouble(st.nextToken());}void p(Object p){System.out.print(p);}
void p(String p){System.out.print(p);}void p(int p){System.out.print(p);}
void p(double p){System.out.print(p);}void p(long p){System.out.print(p);}
void p(char p){System.out.print(p);}void p(boolean p){System.out.print(p);}
void pl(Object p){System.out.println(p);}void pl(String p){System.out.println(p);}
void pl(int p){System.out.println(p);}void pl(char p){System.out.println(p);}
void pl(double p){System.out.println(p);}void pl(long p){System.out.println(p);}
void pl(boolean p){System.out.println(p);}void pl(){System.out.println();}
void s(int a[]){for(int e:a){sb.append(e);sb.append(' ');}sb.append("\n");}
void s(long a[]){for(long e:a){sb.append(e);sb.append(' ');}sb.append("\n");}
void s(int ar[][]){for(int a[]:ar){for(int e:a){sb.append(e);sb.append(' ');}sb.append("\n");}}
void s(char a[]){for(char e:a){sb.append(e);sb.append(' ');}sb.append("\n");}
void s(char ar[][]){for(char a[]:ar){for(char e:a){sb.append(e);sb.append(' ');}sb.append("\n");}}
int[] ari(int n)throws IOException
{int ar[]=new int[n];if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int x=0;x<n;x++)ar[x]=Integer.parseInt(st.nextToken());return ar;}
int[][] ari(int n,int m)throws IOException
{int ar[][]=new int[n][m];for(int x=0;x<n;x++){if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int y=0;y<m;y++)ar[x][y]=Integer.parseInt(st.nextToken());}return ar;}
long[] arl(int n)throws IOException
{long ar[]=new long[n];if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int x=0;x<n;x++) ar[x]=Long.parseLong(st.nextToken());return ar;}
long[][] arl(int n,int m)throws IOException
{long ar[][]=new long[n][m];for(int x=0;x<n;x++)
{if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int y=0;y<m;y++)ar[x][y]=Long.parseLong(st.nextToken());}return ar;}
String[] ars(int n)throws IOException
{String ar[]=new String[n];if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int x=0;x<n;x++) ar[x]=st.nextToken();return ar;}
double[] ard(int n)throws IOException
{double ar[]=new double[n];if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int x=0;x<n;x++)ar[x]=Double.parseDouble(st.nextToken());return ar;}
double[][] ard(int n,int m)throws IOException
{double ar[][]=new double[n][m];for(int x=0;x<n;x++)
{if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int y=0;y<m;y++)ar[x][y]=Double.parseDouble(st.nextToken());}return ar;}
char[] arc(int n)throws IOException{char ar[]=new char[n];if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int x=0;x<n;x++)ar[x]=st.nextToken().charAt(0);return ar;}
char[][] arc(int n,int m)throws IOException{char ar[][]=new char[n][m];
for(int x=0;x<n;x++){String s=br.readLine();for(int y=0;y<m;y++)ar[x][y]=s.charAt(y);}return ar;}
void p(int ar[]){StringBuilder sb=new StringBuilder(2*ar.length);
for(int a:ar){sb.append(a);sb.append(' ');}System.out.println(sb);}
void p(int ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);
for(int a[]:ar){for(int aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}
void p(long ar[]){StringBuilder sb=new StringBuilder(2*ar.length);
for(long a:ar){sb.append(a);sb.append(' ');}System.out.println(sb);}
void p(long ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);
for(long a[]:ar){for(long aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}
void p(String ar[]){int c=0;for(String s:ar)c+=s.length()+1;StringBuilder sb=new StringBuilder(c);
for(String a:ar){sb.append(a);sb.append(' ');}System.out.println(sb);}
void p(double ar[]){StringBuilder sb=new StringBuilder(2*ar.length);
for(double a:ar){sb.append(a);sb.append(' ');}System.out.println(sb);}
void p(double ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);
for(double a[]:ar){for(double aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}
void p(char ar[]){StringBuilder sb=new StringBuilder(2*ar.length);
for(char aa:ar){sb.append(aa);sb.append(' ');}System.out.println(sb);}
void p(char ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);
for(char a[]:ar){for(char aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: <i> Why Geometry?? ? </i>
You are given 4*N+1 distinct points on the cartesian plane. Out of these points, 4*N points represent the end points of N rectangles (<b>with axis parallel to co-ordinate axis</b>), while one point does not belong to any of the rectangles. Report the coordinates of the point that does not belong to any of the N rectangles.The first line of the input contains an integer N.
The next 4*N+1 lines contain two integers X and Y, the coordinates of the given points.
Constraints
1 <= N <= 100000
0 <= X, Y <= 1000000
The given points always represent N rectangles and an extra pointOutput space separated X and Y coordinates of the extra point.Samle Input
1
1 3
1 1
3 1
1 4
3 3
Sample Output
1 4
Explanation: (1, 1), (1, 3), (3, 1), and (3, 3) represent the end of a rectangle, while (1, 4) is the extra point., I have written this Solution Code: n=int(input())
x,y=0,0
for i in range(4*n+1):
a1,b1=map(int,input().split())
x=x^a1
y=y^b1
print(x,y), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: <i> Why Geometry?? ? </i>
You are given 4*N+1 distinct points on the cartesian plane. Out of these points, 4*N points represent the end points of N rectangles (<b>with axis parallel to co-ordinate axis</b>), while one point does not belong to any of the rectangles. Report the coordinates of the point that does not belong to any of the N rectangles.The first line of the input contains an integer N.
The next 4*N+1 lines contain two integers X and Y, the coordinates of the given points.
Constraints
1 <= N <= 100000
0 <= X, Y <= 1000000
The given points always represent N rectangles and an extra pointOutput space separated X and Y coordinates of the extra point.Samle Input
1
1 3
1 1
3 1
1 4
3 3
Sample Output
1 4
Explanation: (1, 1), (1, 3), (3, 1), and (3, 3) represent the end of a rectangle, while (1, 4) is the extra point., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
n = n*4+1;
int x = 0;
int y = 0;
For(i, 0, n){
int a, b; cin>>a>>b;
x^=a;
y^=b;
}
cout<<x<<" "<<y<<"\n";
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: <i> Why Geometry?? ? </i>
You are given 4*N+1 distinct points on the cartesian plane. Out of these points, 4*N points represent the end points of N rectangles (<b>with axis parallel to co-ordinate axis</b>), while one point does not belong to any of the rectangles. Report the coordinates of the point that does not belong to any of the N rectangles.The first line of the input contains an integer N.
The next 4*N+1 lines contain two integers X and Y, the coordinates of the given points.
Constraints
1 <= N <= 100000
0 <= X, Y <= 1000000
The given points always represent N rectangles and an extra pointOutput space separated X and Y coordinates of the extra point.Samle Input
1
1 3
1 1
3 1
1 4
3 3
Sample Output
1 4
Explanation: (1, 1), (1, 3), (3, 1), and (3, 3) represent the end of a rectangle, while (1, 4) is the extra point., I have written this Solution Code: import java.io.*;
import java.io.IOException;
import java.math.BigInteger;
import java.util.*;
public class Main {
public static long mod = 1000000007 ;
public static double epsilon=0.00000000008854;
public static InputReader sc = new InputReader(System.in);
public static PrintWriter pw = new PrintWriter(System.out);
public static long pow(long x,long y,long mod){
long ans=1;
x%=mod;
while(y>0){
if((y&1)==1){
ans=(ans*x)%mod;
}
y=y>>1;
x=(x*x)%mod;
}
return ans;
}
public static void main(String[] args) throws Exception {
int n=sc.nextInt();
n=4*n+1;
int x=0,y=0;
for(int i=0;i<n;i++){
int a=sc.nextInt();
x^=a;
int b=sc.nextInt();
y^=b;
}
pw.println(x+" "+y);
pw.flush();
pw.close();
}
public static Comparator<Integer[]> MOquery(int block){
return
new Comparator<Integer[]>(){
@Override
public int compare(Integer a[],Integer b[]){
int a1=a[0]/block;
int b1=b[0]/block;
if(a1==b1){
if((a1&1)==1)
return a[1].compareTo(b[1]);
else{
return b[1].compareTo(a[1]);
}
}
return a1-b1;
}
};
}
public static Comparator<Long[]> column(int i){
return
new Comparator<Long[]>() {
@Override
public int compare(Long[] o1, Long[] o2) {
return o1[i].compareTo(o2[i]);
}
};
}
public static Comparator<Integer[]> column(){
return
new Comparator<Integer[]>() {
@Override
public int compare(Integer[] o1, Integer[] o2) {
long a1=o1[0];
long a2=o2[1];
long b1=o2[0];
long b2=o1[1];
int ans=0;
if(a1*a2>b1*b2){
ans=1;
}
else ans=-1;
return ans;
}
};
}
public static Comparator<Integer[]> col(int i){
return
new Comparator<Integer[]>() {
@Override
public int compare(Integer[] o1, Integer[] o2) {
if(o1[i]-o2[i]!=0)
return o1[i].compareTo(o2[i]);
return o1[i+1].compareTo(o2[i+1]);
}
};
}
public static Comparator<Integer> des(){
return
new Comparator<Integer>() {
@Override
public int compare(Integer o1, Integer o2) {
return o2.compareTo(o1);
}
};
}
public static String reverseString(String s){
StringBuilder input1 = new StringBuilder();
input1.append(s);
input1 = input1.reverse();
return input1.toString();
}
public static int[] scanArray(int n){
int a[]=new int [n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
return a;
}
public static long[] scanLongArray(int n){
long a[]=new long [n];
for(int i=0;i<n;i++)
a[i]=sc.nextLong();
return a;
}
public static String [] scanStrings(int n){
String a[]=new String [n];
for(int i=0;i<n;i++)
a[i]=sc.nextLine();
return a;
}
public static class InputReader {
private final InputStream stream;
private final byte[] buf = new byte[8192];
private int curChar, snumChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int snext() {
if (snumChars == -1)
throw new InputMismatchException();
if (curChar >= snumChars) {
curChar = 0;
try {
snumChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (snumChars <= 0)
return -1;
}
return buf[curChar++];
}
public int nextInt() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = snext();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = snext();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = snext();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = snext();
} while (!isSpaceChar(c));
return res * sgn;
}
public int[] nextIntArray(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt();
}
return a;
}
public String readString() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = snext();
} while (!isSpaceChar(c));
return res.toString();
}
public String nextLine() {
int c = snext();
while (isSpaceChar(c))
c = snext();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = snext();
} while (!isEndOfLine(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, if the year is a multiple of 100 and not a multiple of 400, then it is not a leap year.<b>User Task:</b>
Complete the function <b>LeapYear()</b> that takes integer n as a parameter.
<b>Constraint:</b>
1 <= n <= 5000If it is a leap year then print <b>YES</b> and if it is not a leap year, then print <b>NO</b>Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code: n = int(input())
if (n%4==0 and n%100!=0 or n%400==0):
print("YES")
elif n==0:
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, if the year is a multiple of 100 and not a multiple of 400, then it is not a leap year.<b>User Task:</b>
Complete the function <b>LeapYear()</b> that takes integer n as a parameter.
<b>Constraint:</b>
1 <= n <= 5000If it is a leap year then print <b>YES</b> and if it is not a leap year, then print <b>NO</b>Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code:
import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int n = scanner.nextInt();
LeapYear(n);
}
static void LeapYear(int year){
if(year%400==0 || (year%100 != 0 && year%4==0)){System.out.println("YES");}
else {
System.out.println("NO");}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ved is a salesman. He needs to earn at least W rupees in N days for his livelihood. However, on a given day i(1 ≤ i ≤ N), he can only earn A<sub>i</sub> rupees by working on that day.
Ved, being a lazy salesman, wants to take as many holidays as possible. It is known that on a holiday, Ved does not work and thus does not earn anything. Help maximize the number of days on which Ved can take a holiday. It is guaranteed that Ved can always earn at least W rupees by working on all days.The first line contains a single integer T - the number of test cases. Then the test cases follow. The first line of each test case contains two integers N and W - the size of the array A and the money Ved has to earn in those N days. The second line of each test case contains N space- separated integers A<sub>1</sub>, A<sub>2</sub>,...., A<sub>N</sub> denoting the money which Ved can earn on each day.
<b>Constraints</b>
1 ≤ T ≤ 100
1 ≤ W ≤ 10000
1 ≤ N ≤ 100
1 ≤ A<sub>i</sub> ≤ 100
It is guaranteed that Ved can always earn at least W rupees by working on all days.For each test case, output the maximum number of holidays Ved can take.Sample Input :
3
5 10
3 2 6 1 3
4 15
3 2 4 6
6 8
9 12 6 5 2 2
Sample Output :
2
0
5
Explanation
Ved can work on 2-nd, 3-rd and 5-th day earning 2 + 6 + 3 = 11 rupees ≥W rupees. Thus he can take 2 holidays on the 1-st and the 4-th day., 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();
while (t>0){
int n = sc.nextInt();
int w = sc.nextInt();
int [] arr = new int [n];
for(int i=0; i<n; i++){
arr[i]= sc.nextInt();
}
Arrays.sort(arr);
int count =0;
int sum = 0;
for(int i=n-1; i>=0; i--){
sum = sum + arr[i];
count ++;
if(sum>=w){
break;
}
}
int ans;
ans = n - count;
System.out.println(ans);
t--;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ved is a salesman. He needs to earn at least W rupees in N days for his livelihood. However, on a given day i(1 ≤ i ≤ N), he can only earn A<sub>i</sub> rupees by working on that day.
Ved, being a lazy salesman, wants to take as many holidays as possible. It is known that on a holiday, Ved does not work and thus does not earn anything. Help maximize the number of days on which Ved can take a holiday. It is guaranteed that Ved can always earn at least W rupees by working on all days.The first line contains a single integer T - the number of test cases. Then the test cases follow. The first line of each test case contains two integers N and W - the size of the array A and the money Ved has to earn in those N days. The second line of each test case contains N space- separated integers A<sub>1</sub>, A<sub>2</sub>,...., A<sub>N</sub> denoting the money which Ved can earn on each day.
<b>Constraints</b>
1 ≤ T ≤ 100
1 ≤ W ≤ 10000
1 ≤ N ≤ 100
1 ≤ A<sub>i</sub> ≤ 100
It is guaranteed that Ved can always earn at least W rupees by working on all days.For each test case, output the maximum number of holidays Ved can take.Sample Input :
3
5 10
3 2 6 1 3
4 15
3 2 4 6
6 8
9 12 6 5 2 2
Sample Output :
2
0
5
Explanation
Ved can work on 2-nd, 3-rd and 5-th day earning 2 + 6 + 3 = 11 rupees ≥W rupees. Thus he can take 2 holidays on the 1-st and the 4-th day., I have written this Solution Code: tc = int(input())
for x in range(tc):
n, w = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
a.reverse()
sum = 0
i = 0
while sum < w:
sum = sum + a[i]
i = i+1
print(n-i), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ved is a salesman. He needs to earn at least W rupees in N days for his livelihood. However, on a given day i(1 ≤ i ≤ N), he can only earn A<sub>i</sub> rupees by working on that day.
Ved, being a lazy salesman, wants to take as many holidays as possible. It is known that on a holiday, Ved does not work and thus does not earn anything. Help maximize the number of days on which Ved can take a holiday. It is guaranteed that Ved can always earn at least W rupees by working on all days.The first line contains a single integer T - the number of test cases. Then the test cases follow. The first line of each test case contains two integers N and W - the size of the array A and the money Ved has to earn in those N days. The second line of each test case contains N space- separated integers A<sub>1</sub>, A<sub>2</sub>,...., A<sub>N</sub> denoting the money which Ved can earn on each day.
<b>Constraints</b>
1 ≤ T ≤ 100
1 ≤ W ≤ 10000
1 ≤ N ≤ 100
1 ≤ A<sub>i</sub> ≤ 100
It is guaranteed that Ved can always earn at least W rupees by working on all days.For each test case, output the maximum number of holidays Ved can take.Sample Input :
3
5 10
3 2 6 1 3
4 15
3 2 4 6
6 8
9 12 6 5 2 2
Sample Output :
2
0
5
Explanation
Ved can work on 2-nd, 3-rd and 5-th day earning 2 + 6 + 3 = 11 rupees ≥W rupees. Thus he can take 2 holidays on the 1-st and the 4-th day., I have written this Solution Code: #include <sys/time.h>
#include <algorithm>
#include <cassert>
#include <climits>
#include <cmath>
#include <cstdio>
#include <functional>
#include <iostream>
#include <list>
#include <map>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <vector>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
#define FOR(i, s, e, t) for ((i) = (s); (i) < (e); (i) += (t))
#define REP(i, e) for (int i = 0; i < (e); ++i)
#define REP1(i, s, e) for (int i = (s); i < (e); ++i)
#define RREP(i, e) for (int i = (e); i >= 0; --i)
#define RREP1(i, e, s) for (int i = (e); i >= (s); --i)
ll inf = 1e18;
int __FAST_IO__ = []() {
std::ios::sync_with_stdio(0);
std::cin.tie(0);
std::cout.tie(0);
return 0;
}();
int main() {
int t;
cin >> t;
while (t--) {
int N, W;
cin >> N >> W;
vector<int> v(N);
REP(i, N)
cin >> v[i];
sort(v.begin(), v.end(), greater<int>());
int sum = 0, id = 0;
while (sum < W) {
sum += v[id];
id++;
}
printf("%d\n", N - id);
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N your task is to print its first two digits in reverse order. For eg:- If the given number is 123 then the output will be 21.<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>firstTwo()</b> that takes integer N argument.
Constraints:-
111 <= N <= 9999
Note:- It is guaranteed that the given number will not contain any 0.Return the first two digits of the given number in reverse order.Sample Input:-
3423
Sample Output:-
43
Sample Input:-
1234
Sample Output:-
21, I have written this Solution Code: def firstTwo(N):
while(N>99):
N=N//10
return (N%10)*10 + N//10
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N your task is to print its first two digits in reverse order. For eg:- If the given number is 123 then the output will be 21.<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>firstTwo()</b> that takes integer N argument.
Constraints:-
111 <= N <= 9999
Note:- It is guaranteed that the given number will not contain any 0.Return the first two digits of the given number in reverse order.Sample Input:-
3423
Sample Output:-
43
Sample Input:-
1234
Sample Output:-
21, I have written this Solution Code:
int firstTwo(int N){
while(N>99){
N/=10;
}
int ans = (N%10)*10 + N/10;
return ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N your task is to print its first two digits in reverse order. For eg:- If the given number is 123 then the output will be 21.<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>firstTwo()</b> that takes integer N argument.
Constraints:-
111 <= N <= 9999
Note:- It is guaranteed that the given number will not contain any 0.Return the first two digits of the given number in reverse order.Sample Input:-
3423
Sample Output:-
43
Sample Input:-
1234
Sample Output:-
21, I have written this Solution Code:
int firstTwo(int N){
while(N>99){
N/=10;
}
int ans = (N%10)*10 + N/10;
return ans;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N your task is to print its first two digits in reverse order. For eg:- If the given number is 123 then the output will be 21.<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>firstTwo()</b> that takes integer N argument.
Constraints:-
111 <= N <= 9999
Note:- It is guaranteed that the given number will not contain any 0.Return the first two digits of the given number in reverse order.Sample Input:-
3423
Sample Output:-
43
Sample Input:-
1234
Sample Output:-
21, I have written this Solution Code: static int firstTwo(int N){
while(N>99){
N/=10;
}
int ans = (N%10)*10 + N/10;
return ans;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings s and t of lengths m and n respectively, Print the length of the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, Print 0.The first line of the input contains the string s.
The next line of the input contains the string t.
Constraints
1 <= m, n <= 10<sup>5</sup>
s and t consist of uppercase and lowercase English letters.Print the length of minimum window substring.Sample Input
ADOBECODEBANC
ABC
Sample Output
BANC
Explanation
The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t., I have written this Solution Code: /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Main
{
public static String minWindow(String s, String t) {
if (s.length() == 0 || t.length() == 0) {
return "";
}
// Dictionary which keeps a count of all the unique characters in t.
Map<Character, Integer> dictT = new HashMap<Character, Integer>();
for (int i = 0; i < t.length(); i++) {
int count = dictT.getOrDefault(t.charAt(i), 0);
dictT.put(t.charAt(i), count + 1);
}
// Number of unique characters in t, which need to be present in the desired window.
int required = dictT.size();
// Left and Right pointer
int l = 0, r = 0;
// formed is used to keep track of how many unique characters in t
// are present in the current window in its desired frequency.
// e.g. if t is "AABC" then the window must have two A's, one B and one C.
// Thus formed would be = 3 when all these conditions are met.
int formed = 0;
// Dictionary which keeps a count of all the unique characters in the current window.
Map<Character, Integer> windowCounts = new HashMap<Character, Integer>();
// ans list of the form (window length, left, right)
int[] ans = {-1, 0, 0};
while (r < s.length()) {
// Add one character from the right to the window
char c = s.charAt(r);
int count = windowCounts.getOrDefault(c, 0);
windowCounts.put(c, count + 1);
// If the frequency of the current character added equals to the
// desired count in t then increment the formed count by 1.
if (dictT.containsKey(c) && windowCounts.get(c).intValue() == dictT.get(c).intValue()) {
formed++;
}
// Try and contract the window till the point where it ceases to be 'desirable'.
while (l <= r && formed == required) {
c = s.charAt(l);
// Save the smallest window until now.
if (ans[0] == -1 || r - l + 1 < ans[0]) {
ans[0] = r - l + 1;
ans[1] = l;
ans[2] = r;
}
// The character at the position pointed by the
// `Left` pointer is no longer a part of the window.
windowCounts.put(c, windowCounts.get(c) - 1);
if (dictT.containsKey(c) && windowCounts.get(c).intValue() < dictT.get(c).intValue()) {
formed--;
}
// Move the left pointer ahead, this would help to look for a new window.
l++;
}
// Keep expanding the window once we are done contracting.
r++;
}
return ans[0] == -1 ? "" : s.substring(ans[1], ans[2] + 1);
}
public static void main (String[] args) throws java.lang.Exception
{
Scanner inp = new Scanner(System.in);
String s = inp.nextLine();
String t = inp.nextLine();
String res = minWindow(s,t);
System.out.println(res.length());
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Take an integer as input and print it.The first line contains integer as input.
<b>Constraints</b>
1 <= N <= 10Print the input integer in a single lineSample Input:-
2
Sample Output:-
2
Sample Input:-
4
Sample Output:-
4, I have written this Solution Code: /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Main
{
public static void printVariable(int variable){
System.out.println(variable);
}
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int num = sc.nextInt();
printVariable(num);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of size N containing 0s and 1s only. The task is to count the subarrays having an equal number of 0s and 1s.The first line of the input contains an integer N denoting the size of the array and the second line contains N space-separated 0s and 1s.
Constraints:-
1 <= N <= 10^6
0 <= A[i] <= 1For each test case, print the count of required sub-arrays in new line.Sample Input
7
1 0 0 1 0 1 1
Sample Output
8
The index range for the 8 sub-arrays are:
(0, 1), (2, 3), (0, 3), (3, 4), (4, 5)
(2, 5), (0, 5), (1, 6), I have written this Solution Code: size = int(input())
givenList = list(map(int,input().split()))
hs = {}
sm = 0
ct = 0
for i in givenList:
if i == 0:
i = -1
sm = sm + i
if sm == 0:
ct += 1
if sm not in hs.keys():
hs[sm] = 1
else:
freq = hs[sm]
ct = ct +freq
hs[sm] = freq + 1
print(ct), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of size N containing 0s and 1s only. The task is to count the subarrays having an equal number of 0s and 1s.The first line of the input contains an integer N denoting the size of the array and the second line contains N space-separated 0s and 1s.
Constraints:-
1 <= N <= 10^6
0 <= A[i] <= 1For each test case, print the count of required sub-arrays in new line.Sample Input
7
1 0 0 1 0 1 1
Sample Output
8
The index range for the 8 sub-arrays are:
(0, 1), (2, 3), (0, 3), (3, 4), (4, 5)
(2, 5), (0, 5), (1, 6), I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max1 1000001
int a[max1];
int main(){
int n;
cin>>n;
for(int i=0;i<n;i++){
cin>>a[i];
if(a[i]==0){a[i]=-1;}
}
long sum=0;
unordered_map<long,int> m;
long cnt=0;
for(int i=0;i<n;i++){
sum+=a[i];
if(sum==0){cnt++;}
cnt+=m[sum];
m[sum]++;
}
cout<<cnt;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of size N containing 0s and 1s only. The task is to count the subarrays having an equal number of 0s and 1s.The first line of the input contains an integer N denoting the size of the array and the second line contains N space-separated 0s and 1s.
Constraints:-
1 <= N <= 10^6
0 <= A[i] <= 1For each test case, print the count of required sub-arrays in new line.Sample Input
7
1 0 0 1 0 1 1
Sample Output
8
The index range for the 8 sub-arrays are:
(0, 1), (2, 3), (0, 3), (3, 4), (4, 5)
(2, 5), (0, 5), (1, 6), I have written this Solution Code: import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
// don't change the name of this class
// you can add inner classes if needed
class Main {
public static void main (String[] args) {
// Your code here
Scanner sc = new Scanner(System.in);
int arrSize = sc.nextInt();
long arr[] = new long[arrSize];
for(int i = 0; i < arrSize; i++)
arr[i] = sc.nextInt();
System.out.println(countSubarrays(arr, arrSize));
}
static long countSubarrays(long arr[], int arrSize)
{
for(int i = 0; i < arrSize; i++)
{
if(arr[i] == 0)
arr[i] = -1;
}
long ans = 0;
long sum = 0;
HashMap<Long, Integer> hash = new HashMap<>();
for(int i = 0; i < arrSize; i++)
{
sum += arr[i];
if(sum == 0)
ans++;
if(hash.containsKey(sum) == true)
{
ans += hash.get(sum);
int freq = hash.get(sum);
hash.put(sum, freq+1);
}
else hash.put(sum, 1);
}
return ans;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given four positive integers A, B, C, and D, can a rectangle have these integers as its sides?The input consists of a single line containing four space-separated integers A, B, C and D.
<b> Constraints: </b>
1 ≤ A, B, C, D ≤ 1000Output "Yes" if a rectangle is possible; otherwise, "No" (without quotes).Sample Input 1:
3 4 4 3
Sample Output 1:
Yes
Sample Explanation 1:
A rectangle is possible with 3, 4, 3, and 4 as sides in clockwise order.
Sample Input 2:
3 4 3 5
Sample Output 2:
No
Sample Explanation 2:
No rectangle can be made with these side lengths., I have written this Solution Code: #include<iostream>
using namespace std;
int main(){
int a,b,c,d;
cin >> a >> b >> c >> d;
if((a==c) &&(b==d)){
cout << "Yes\n";
}else if((a==b) && (c==d)){
cout << "Yes\n";
}else if((a==d) && (b==c)){
cout << "Yes\n";
}else{
cout << "No\n";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Newton wants to take revenge from two apples fallen on his head. So, he applies force F<sub>1</sub> on first apple (mass M<sub>1</sub>) resulting in acceleration of A<sub>1</sub> and F<sub>2</sub> on second apple (mass M<sub>2</sub>) resulting in acceleration of A<sub>2</sub>. Given M<sub>1</sub>, A<sub>1</sub>, M<sub>2</sub>, A<sub>2</sub>. Calculate total force applied by him on two apples.
<b>Note:</b> F = M*A is the equation of relation between force, mass and acceleration.First line contains four integers M<sub>1</sub>, A<sub>1</sub>, M<sub>2</sub>, A<sub>2</sub>.
1 <= M<sub>1</sub>, A<sub>1</sub>, M<sub>2</sub>, A<sub>2</sub> <= 100Output total force applied by Newton.INPUT:
1 2 3 4
OUTPUT:
14
Explanation:
Total force is equal to 1*2 + 3*4 = 14., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main(){
int m1,a1,m2,a2;
cin >> m1 >> a1 >> m2 >> a2;
cout << (m1*a1)+(m2*a2) << endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have infinite supply of coins, of given denominations. Can you figure out the minimum number of coins required so that coins sum- up to a certain required value. We'll use dynamic programmic to solve this question.
You are given an amount denoted by value. You are also given an array of coins. The array contains the denominations of the given coins. You need to find the minimum number of coins to make the change for value using the coins of the given denominations. Also, keep in mind that you have infinite supply of the coins.
If making the change is not possible then print "Not Possible" without quotes.The first line of input contains one integer T - the number of testcases. Then T testcases follow.
Each testcase contains three lines of input. The first line contains value for which you need to make the change. The second line contains the Size of the array. The third line contains the denominations of the coins.
Constraints:
1 <= T <= 100
1 <= value <= 1000
1 <= numberOfCoins <= 1000
1 <= denominationOfCoins <= 1000For each testcase, in a new line, print the minimum number of coins required.
If making the change is not possible then print "Not Possible" without quotes.Input:
2
10
4
2 5 3 6
5
3
3 6 3
Output:
2
Not Possible
Explanation:
Testcase1:
There are five ways to make 10 using [2, 5, 3, 6].
{2, 2, 2, 2, 2}
{2, 2, 3, 3}
{2, 2, 6}
{2, 3, 5}
{5, 5}
Minimum number of coins is required when we use two 5 coins
Thus answer is 2
Testcase2:
There are no ways to make 5 using [3, 6, 3]
Thus answer is Not Possible., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int testCases = Integer.parseInt(br.readLine());
while(testCases-- > 0){
int value = Integer.parseInt(br.readLine());
int numOfCoins = Integer.parseInt(br.readLine());
int[] coinsArr = new int[numOfCoins];
String str = br.readLine();
String[] s1 = str.split(" ");
for(int i = 0 ; i < numOfCoins; i++){
coinsArr[i] = Integer.parseInt(s1[i]);
}
if(minCoin(coinsArr, coinsArr.length,value) == -1){
System.out.println("Not Possible");
}else{
System.out.println(minCoin(coinsArr,coinsArr.length, value));
}
}
}
public static int minCoin(int[] coins, int m, int v){
int dp[] = new int[v + 1];
dp[0] = 0;
for(int i = 1; i <= v; i++){
dp[i] = Integer.MAX_VALUE;
}
for(int i = 0; i <= v; i++){
for(int j = 0; j < m; j++){
if(coins[j] <= i){
int subres = dp[i - coins[j]];
if(subres != Integer.MAX_VALUE && subres + 1 < dp[i])
dp[i] = subres + 1;
}
}
}
if(dp[v] == Integer.MAX_VALUE){
return -1;
}
return dp[v];
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have infinite supply of coins, of given denominations. Can you figure out the minimum number of coins required so that coins sum- up to a certain required value. We'll use dynamic programmic to solve this question.
You are given an amount denoted by value. You are also given an array of coins. The array contains the denominations of the given coins. You need to find the minimum number of coins to make the change for value using the coins of the given denominations. Also, keep in mind that you have infinite supply of the coins.
If making the change is not possible then print "Not Possible" without quotes.The first line of input contains one integer T - the number of testcases. Then T testcases follow.
Each testcase contains three lines of input. The first line contains value for which you need to make the change. The second line contains the Size of the array. The third line contains the denominations of the coins.
Constraints:
1 <= T <= 100
1 <= value <= 1000
1 <= numberOfCoins <= 1000
1 <= denominationOfCoins <= 1000For each testcase, in a new line, print the minimum number of coins required.
If making the change is not possible then print "Not Possible" without quotes.Input:
2
10
4
2 5 3 6
5
3
3 6 3
Output:
2
Not Possible
Explanation:
Testcase1:
There are five ways to make 10 using [2, 5, 3, 6].
{2, 2, 2, 2, 2}
{2, 2, 3, 3}
{2, 2, 6}
{2, 3, 5}
{5, 5}
Minimum number of coins is required when we use two 5 coins
Thus answer is 2
Testcase2:
There are no ways to make 5 using [3, 6, 3]
Thus answer is Not Possible., I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 10001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int coinChange(vector<int> coins, int amount) {
int n = coins.size();
int dp[n+1][amount+2];
for(int i=0;i<n+1;i++){
for(int j=0;j<amount+2;j++){
dp[i][j] = INT_MAX;
}
}
for(int i=0;i<n+1;i++){
dp[i][0]=0;
}
for(int i=1;i<n+1;i++){
for(int j = 1;j<=amount;j++){
if(j<coins[i-1]){
dp[i][j] = dp[i-1][j];
}else{
if(dp[i][j-coins[i-1]]!=INT_MAX)
dp[i][j] = min(dp[i-1][j], 1 + dp[i][j-coins[i-1]]);
else dp[i][j] = dp[i-1][j];
}
}
}
if(dp[n][amount]==INT_MAX) return -1;
return dp[n][amount];
}
signed main()
{
int t;
cin>>t;
while(t--){
int a,n;
cin>>a>>n;
vector<int> v(n);
FOR(i,n){
cin>>v[i];}
int x = coinChange(v,a);
if(x==-1){out("Not Possible");}
else{
out(x);
}
}
}
, In this Programming Language: C++, 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: While doing homework, Nobita stuck on a problem and asks for your help.
Problem statement:-
Given three integers X, Y, and N. Your task is to check if it is possible to form N by using only combinations of X and Y.
i.e check if there exist any P and Q such that P*X + Q*Y = N
Note:- P and Q can be negativeThe input contains only three integers X, Y, and N.
Constraints:-
1 <= X, Y, N <= 100000Print "YES" if it is possible to form N else print "NO".Sample Input:-
3 5 18
Sample Output:-
YES
Explanation:-
1 * 3 + 3 * 5 = 18 (P = 1, Q = 3)
Sample Input:-
4 8 15
Sample Output:-
NO, 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 input[] = br.readLine().trim().split(" ");
int x = Integer.parseInt(input[0]),
y = Integer.parseInt(input[1]),
n = Integer.parseInt(input[2]);
boolean flag = false;
for(int q = 1; q <= 100000; q++) {
if((n - (q * y)) % x == 0) {
System.out.println("YES");
flag = true;
break;
}
}
if(!flag)
System.out.println("NO");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While doing homework, Nobita stuck on a problem and asks for your help.
Problem statement:-
Given three integers X, Y, and N. Your task is to check if it is possible to form N by using only combinations of X and Y.
i.e check if there exist any P and Q such that P*X + Q*Y = N
Note:- P and Q can be negativeThe input contains only three integers X, Y, and N.
Constraints:-
1 <= X, Y, N <= 100000Print "YES" if it is possible to form N else print "NO".Sample Input:-
3 5 18
Sample Output:-
YES
Explanation:-
1 * 3 + 3 * 5 = 18 (P = 1, Q = 3)
Sample Input:-
4 8 15
Sample Output:-
NO, I have written this Solution Code: l,m,z=input().split()
x=int(l)
y=int(m)
n=int(z)
flag=0
for i in range(1,n):
sum1=n-i*y
sum2=n+i*y
if(sum1%x==0 or sum2%y==0):
flag=1
break
if flag==1:
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While doing homework, Nobita stuck on a problem and asks for your help.
Problem statement:-
Given three integers X, Y, and N. Your task is to check if it is possible to form N by using only combinations of X and Y.
i.e check if there exist any P and Q such that P*X + Q*Y = N
Note:- P and Q can be negativeThe input contains only three integers X, Y, and N.
Constraints:-
1 <= X, Y, N <= 100000Print "YES" if it is possible to form N else print "NO".Sample Input:-
3 5 18
Sample Output:-
YES
Explanation:-
1 * 3 + 3 * 5 = 18 (P = 1, Q = 3)
Sample Input:-
4 8 15
Sample Output:-
NO, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1000001
#define MOD 1000000000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int cnt[max1];
signed main(){
int t;t=1;
while(t--){
int x,y,n;
cin>>x>>y>>n;
int p=__gcd(x,y);
if(n%p==0){out("YES");}
else{
out("NO");
}
}
}
, 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 "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: n=int(input())
for i in range(1,n+1):
if i%3==0 and i%5==0:
print("NewtonSchool",end=" ")
elif i%3==0:
print("Newton",end=" ")
elif i%5==0:
print("School",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 "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
static void NewtonSchool(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("NewtonSchool ");}
else if(i%5==0){System.out.print("School ");}
else if(i%3==0){System.out.print("Newton ");}
else{System.out.print(i+" ");}
}
}
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int x= sc.nextInt();
NewtonSchool(x);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix <b>A</b> of dimensions <b>n x m</b>. The task is to perform <b>boundary traversal</b> on the matrix in clockwise manner.The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase has two lines of input. The first line contains dimensions of the matrix A, n and m. The second line contains n*m elements separated by spaces.
Constraints:
1 <= T <= 100
1 <= n, m <= 30
0 <= A[i][j] <= 100For each testcase, in a new line, print the boundary traversal of the matrix A.Input:
4
4 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
3 4
12 11 10 9 8 7 6 5 4 3 2 1
1 4
1 2 3 4
4 1
1 2 3 4
Output:
1 2 3 4 8 12 16 15 14 13 9 5
12 11 10 9 5 1 2 3 4 8
1 2 3 4
1 2 3 4
Explanation:
Testcase1: The matrix is:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
The boundary traversal is 1 2 3 4 8 12 16 15 14 13 9 5
Testcase 2: Boundary Traversal will be 12 11 10 9 5 1 2 3 4 8.
Testcase 3: Boundary Traversal will be 1 2 3 4.
Testcase 4: Boundary Traversal will be 1 2 3 4., I have written this Solution Code:
def boundaryTraversal(matrix, N, M): #N = 3, M = 4
start_row = 0
start_col = 0
count = 0
if N != 1 and M != 1:
total = 2 * (N - 1) + 2 * (M - 1)
else:
total = max(M, N)
#First row
for i in range(start_col, M, 1): #0,1, 2, 3
count += 1
print(matrix[start_row][i], end = " ")
if count == total:
return
start_row += 1
#Last column
for i in range(start_row, N, 1): # 1, 2
count += 1
print(matrix[i][M - 1], end = " ")
if count == total:
return
M -= 1
#Last Row
for i in range(M - 1, start_col - 1, -1): # 2, 1, 0
count += 1
print(matrix[N - 1][i], end = " ")
if count == total:
return
#First Column
N -= 1 # 2
for i in range(N - 1, start_row - 1, -1):
count += 1
print(matrix[i][start_col], end = " ")
start_col += 1
testcase = int(input().strip())
for test in range(testcase):
dim = input().strip().split(" ")
N = int(dim[0])
M = int(dim[1])
matrix = []
elements = input().strip().split(" ") #N * M
start = 0
for i in range(N):
matrix.append(elements[start : start + M]) # (0, M - 1) | (M, 2*m-1)
start = start + M
boundaryTraversal(matrix, N, M)
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix <b>A</b> of dimensions <b>n x m</b>. The task is to perform <b>boundary traversal</b> on the matrix in clockwise manner.The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase has two lines of input. The first line contains dimensions of the matrix A, n and m. The second line contains n*m elements separated by spaces.
Constraints:
1 <= T <= 100
1 <= n, m <= 30
0 <= A[i][j] <= 100For each testcase, in a new line, print the boundary traversal of the matrix A.Input:
4
4 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
3 4
12 11 10 9 8 7 6 5 4 3 2 1
1 4
1 2 3 4
4 1
1 2 3 4
Output:
1 2 3 4 8 12 16 15 14 13 9 5
12 11 10 9 5 1 2 3 4 8
1 2 3 4
1 2 3 4
Explanation:
Testcase1: The matrix is:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
The boundary traversal is 1 2 3 4 8 12 16 15 14 13 9 5
Testcase 2: Boundary Traversal will be 12 11 10 9 5 1 2 3 4 8.
Testcase 3: Boundary Traversal will be 1 2 3 4.
Testcase 4: Boundary Traversal will be 1 2 3 4., I have written this Solution Code: import java.io.*;
import java.lang.*;
import java.util.*;
class Main
{
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-- > 0)
{
int n1 = sc.nextInt();
int m1 = sc.nextInt();
int arr1[][] = new int[n1][m1];
for(int i = 0; i < n1; i++)
{
for(int j = 0; j < m1; j++)
arr1[i][j] = sc.nextInt();
}
boundaryTraversal(n1, m1,arr1);
System.out.println();
}
}
static void boundaryTraversal( int n1, int m1, int arr1[][])
{
// base cases
if(n1 == 1)
{
int i = 0;
while(i < m1)
System.out.print(arr1[0][i++] + " ");
}
else if(m1 == 1)
{
int i = 0;
while(i < n1)
System.out.print(arr1[i++][0]+" ");
}
else
{
// traversing the first row
for(int j=0;j<m1;j++)
{
System.out.print(arr1[0][j]+" ");
}
// traversing the last column
for(int j=1;j<n1;j++)
{
System.out.print(arr1[j][m1-1]+ " ");
}
// traversing the last row
for(int j=m1-2;j>=0;j--)
{
System.out.print(arr1[n1-1][j]+" ");
}
// traversing the first column
for(int j=n1-2;j>=1;j--)
{
System.out.print(arr1[j][0]+" ");
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: 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: 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: 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: Given a number N, you need to check whether the given number is Armstrong number or not. Now what is Armstrong number let us see below:
<b>A number is said to be Armstrong if it is equal to sum of cube of its digits. </b>User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isArmstrong()</b> which contains N as a parameter.
Constraints:
1 <= N <= 10^4You need to return "true" if the given number is an Armstrong number otherwise "false"Sample Input 1:
1
Sample Output 1:
true
Sample Input 2:
147
Sample Output 2:
false
Sample Input 3:
371
Sample Output 3:
true
, I have written this Solution Code: static boolean isArmstrong(int N)
{
int num = N;
int sum = 0;
while(N > 0)
{
int digit = N%10;
sum += digit*digit*digit;
N = N/10;
}
if(num == sum)
return true;
else return false;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you need to check whether the given number is Armstrong number or not. Now what is Armstrong number let us see below:
<b>A number is said to be Armstrong if it is equal to sum of cube of its digits. </b>User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isArmstrong()</b> which contains N as a parameter.
Constraints:
1 <= N <= 10^4You need to return "true" if the given number is an Armstrong number otherwise "false"Sample Input 1:
1
Sample Output 1:
true
Sample Input 2:
147
Sample Output 2:
false
Sample Input 3:
371
Sample Output 3:
true
, I have written this Solution Code: function isArmstrong(n) {
// write code here
// do no console.log the answer
// return the output using return keyword
let sum = 0
let k = n;
while(k !== 0){
sum += Math.pow( k%10,3)
k = Math.floor(k/10)
}
return sum === n
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Al is given task to build a skyscraper of N floors. He can build 2**i (2 to the power i, where i can be any non-negative integer) floors in one day. Report the minimum number of days required to build the skyscraper.First and only line of input contains a single integer N.
Constraints :
1 <= N <= 1000000000000Print a single integer, the minimum number of days required to build the skyscraper.Sample Input:
5
Sample Output:
2
Sample Input:
1
Sample Output:
1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader buf =new BufferedReader(new InputStreamReader(System.in));
long t=Long.parseLong(buf.readLine().trim());
int ans=0;
for(int i=45;i>=0;i--)
{
if(((t>>i)&1)==1)
ans++;
}
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Al is given task to build a skyscraper of N floors. He can build 2**i (2 to the power i, where i can be any non-negative integer) floors in one day. Report the minimum number of days required to build the skyscraper.First and only line of input contains a single integer N.
Constraints :
1 <= N <= 1000000000000Print a single integer, the minimum number of days required to build the skyscraper.Sample Input:
5
Sample Output:
2
Sample Input:
1
Sample Output:
1, I have written this Solution Code: n = int(input())
print(bin(n).count('1')), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Al is given task to build a skyscraper of N floors. He can build 2**i (2 to the power i, where i can be any non-negative integer) floors in one day. Report the minimum number of days required to build the skyscraper.First and only line of input contains a single integer N.
Constraints :
1 <= N <= 1000000000000Print a single integer, the minimum number of days required to build the skyscraper.Sample Input:
5
Sample Output:
2
Sample Input:
1
Sample Output:
1, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
signed main()
{
int n;
cin>>n;
int cnt=0;
while(n>0)
{
int p=n%2LL;
cnt+=p;
n/=2LL;
}
cout<<cnt<<endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You went to shopping. You bought an item worth N rupees. What is the minimum change that you can get from the shopkeeper if you possess only 200 and 500 rupees notes.
Eg: If N = 678, the minimum change you can receive is 22, if you pay the shopkeeper a 500 and a 200 rupee note. You can show that no other combination can lead to a change lesser than this like (200, 200, 200, 200) or (500, 500).
Note: You have infinite number of 200 and 500 rupees notes. Enjoy, XD.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000000000Output a single integer, the minimum amount of change that you will receive.Sample Input
678
Sample Output
22
Sample Input
900
Sample Output
0, I have written this Solution Code: def sample(n):
if n<200:
print(200-n)
elif n<400:
print(400-n)
elif n<500:
print(500-n)
else:
div=n//100
if div*100==n:
print(0)
else:
print((div+1)*100-n)
n=int(input())
sample(n), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You went to shopping. You bought an item worth N rupees. What is the minimum change that you can get from the shopkeeper if you possess only 200 and 500 rupees notes.
Eg: If N = 678, the minimum change you can receive is 22, if you pay the shopkeeper a 500 and a 200 rupee note. You can show that no other combination can lead to a change lesser than this like (200, 200, 200, 200) or (500, 500).
Note: You have infinite number of 200 and 500 rupees notes. Enjoy, XD.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000000000Output a single integer, the minimum amount of change that you will receive.Sample Input
678
Sample Output
22
Sample Input
900
Sample Output
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
if(n <= 200){
cout<<200-n;
return;
}
if(n <= 400){
cout<<400-n;
return;
}
int ans = (100-n%100)%100;
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You went to shopping. You bought an item worth N rupees. What is the minimum change that you can get from the shopkeeper if you possess only 200 and 500 rupees notes.
Eg: If N = 678, the minimum change you can receive is 22, if you pay the shopkeeper a 500 and a 200 rupee note. You can show that no other combination can lead to a change lesser than this like (200, 200, 200, 200) or (500, 500).
Note: You have infinite number of 200 and 500 rupees notes. Enjoy, XD.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000000000Output a single integer, the minimum amount of change that you will receive.Sample Input
678
Sample Output
22
Sample Input
900
Sample Output
0, 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 ans=0;
if(n <= 200){
ans = 200-n;
}
else if(n <= 400){
ans=400-n;
}
else{
ans = (100-n%100)%100;
}
System.out.print(ans);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three positive integers A, B and C. If all the three integers are equal, print "Yes". Otherwise, print "No".
Note that the quotation marks are not part of the string, thus you should print your answer without the quotation marks.The input consists of three space separated integers A, B and C.
Constraints:
1 ≤ A, B, C ≤ 100Print a single word "Yes" if all the integers are equal, otherwise print "No" (without the quotes). Note that the output is case- sensitive.Sample Input 1:
1 3 1
Sample Output 1:
No
Sample Input 2:
5 5 5
Sample Output 2:
Yes, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
String[] input = in.readLine().split(" ");
int a = Integer.parseInt(input[0]);
int b = Integer.parseInt(input[1]);
int c = Integer.parseInt(input[2]);
if(a==b && b==c && c == a) {
System.out.println("Yes");
} else {
System.out.println("No");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three positive integers A, B and C. If all the three integers are equal, print "Yes". Otherwise, print "No".
Note that the quotation marks are not part of the string, thus you should print your answer without the quotation marks.The input consists of three space separated integers A, B and C.
Constraints:
1 ≤ A, B, C ≤ 100Print a single word "Yes" if all the integers are equal, otherwise print "No" (without the quotes). Note that the output is case- sensitive.Sample Input 1:
1 3 1
Sample Output 1:
No
Sample Input 2:
5 5 5
Sample Output 2:
Yes, I have written this Solution Code: a,b,c=map(int,input().split())
if a==b and b==c:
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three positive integers A, B and C. If all the three integers are equal, print "Yes". Otherwise, print "No".
Note that the quotation marks are not part of the string, thus you should print your answer without the quotation marks.The input consists of three space separated integers A, B and C.
Constraints:
1 ≤ A, B, C ≤ 100Print a single word "Yes" if all the integers are equal, otherwise print "No" (without the quotes). Note that the output is case- sensitive.Sample Input 1:
1 3 1
Sample Output 1:
No
Sample Input 2:
5 5 5
Sample Output 2:
Yes, I have written this Solution Code: #include <iostream>
using namespace std;
int main() {
int a, b, c;
cin >> a >> b >> c;
if (a == b && b == c) cout << "Yes";
else cout << "No";
}
, 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: 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 |
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