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For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code:
int focal_length(int R, char Mirror)
{
int f=R/2;
if((R&1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code: def focal_length(R,Mirror):
f=R/2;
if(Mirror == ')'):
f=-f
if R%2==1:
f=f-1
return int(f)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, N is an even positive number. You have to make N/2 pairs among them such that each element is in exactly one pair. Score of a pair is the absolute difference between there values. Find the maximum sum of score of N/2 pairs.First Line of input contains N.
Second line of input contains N space seperated integers, denoting Arr.
Constraints:
1 <= N <= 100000
1 <= Arr[i] <= 1000000000
N will be evenPrint a single integer which is the maximum sum of score of N/2 pairs.Sample Input 1
2
1 2
Sample Output 1
1
Sample Input 2
4
1 4 2 4
Sample Output 2
5
Explanation: (1, 4) (2, 4) are the optimal pairs, I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
public static void main(String args[]) throws IOException {
BufferedReader inputMachine = new BufferedReader(new InputStreamReader(System.in));
int length = Integer.parseInt(inputMachine.readLine());
String[] arrText = inputMachine.readLine().trim().split(" ");
int[] nums = new int[length];
for (int i = 0; i < nums.length; i++) {
nums[i] = Integer.parseInt(arrText[i]);
}
Arrays.sort(nums);
int i = 0;
int j = length - 1;
long sum = 0;
while (i < j) {
sum += nums[j] - nums[i];
i++;
j--;
}
System.out.println(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, N is an even positive number. You have to make N/2 pairs among them such that each element is in exactly one pair. Score of a pair is the absolute difference between there values. Find the maximum sum of score of N/2 pairs.First Line of input contains N.
Second line of input contains N space seperated integers, denoting Arr.
Constraints:
1 <= N <= 100000
1 <= Arr[i] <= 1000000000
N will be evenPrint a single integer which is the maximum sum of score of N/2 pairs.Sample Input 1
2
1 2
Sample Output 1
1
Sample Input 2
4
1 4 2 4
Sample Output 2
5
Explanation: (1, 4) (2, 4) are the optimal pairs, I have written this Solution Code: n = int(input())
arr = list(map(int,input().split()))
arr.sort()
s = 0
for i in range(n//2):
s += arr[n-i-1]-arr[i]
print(s), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, N is an even positive number. You have to make N/2 pairs among them such that each element is in exactly one pair. Score of a pair is the absolute difference between there values. Find the maximum sum of score of N/2 pairs.First Line of input contains N.
Second line of input contains N space seperated integers, denoting Arr.
Constraints:
1 <= N <= 100000
1 <= Arr[i] <= 1000000000
N will be evenPrint a single integer which is the maximum sum of score of N/2 pairs.Sample Input 1
2
1 2
Sample Output 1
1
Sample Input 2
4
1 4 2 4
Sample Output 2
5
Explanation: (1, 4) (2, 4) are the optimal pairs, 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 a[n];
for(int i=0;i<n;++i){
cin>>a[i];
}
sort(a,a+n);
int ans=0;
for(int i=0;i<n/2;++i){
ans+=abs(a[i]-a[n-i-1]);
}
cout<<ans;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: 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: Joey and Chandler are super bored. So, Chandler makes up a game they can play. The game is called Chandy Game. Initially, Chandler has A candies and Joey has B candies.
In the first move Joey has to give Chandler 1 candy.
In the second move Chandler has to give Joey 2 candies.
In the third move Joey has to give Chandler 3 candies.
In the fourth move Chandler has to give Joey 4 candies.
In the fifth move Joey has to give Chandler 5 candy.
... and so on.
The game continues till one of the player can not make a move. The player who cannot make a move loses. Help them find who wins the game.Input contains two integers A and B.
Constraints:
0 <= A, B <= 10<sup>15</sup>Print "Chandler" (without quotes) if Chandler wins the game and "Joey" (without quotes) if Joey wins the game.Sample Input
2 1
Sample Output
Chandler
Explanation:
In first move Joey gives Chandler 1 candy so, Chandler has 3 candies and Joey has 0.
In second move Chandler gives Joey 2 candies so, Chandler has 1 candy and Joey has 2.
In third move Joey has to give Chandler 3 candies but he has only 2 candies so he loses., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main
{
public static void main(String args[])throws Exception
{
BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
StringBuilder sb=new StringBuilder();
String s[]=bu.readLine().split(" ");
long a=Long.parseLong(s[0]),b=Long.parseLong(s[1]);
if(a>=b) sb.append("Chandler");
else sb.append("Joey");
System.out.print(sb);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joey and Chandler are super bored. So, Chandler makes up a game they can play. The game is called Chandy Game. Initially, Chandler has A candies and Joey has B candies.
In the first move Joey has to give Chandler 1 candy.
In the second move Chandler has to give Joey 2 candies.
In the third move Joey has to give Chandler 3 candies.
In the fourth move Chandler has to give Joey 4 candies.
In the fifth move Joey has to give Chandler 5 candy.
... and so on.
The game continues till one of the player can not make a move. The player who cannot make a move loses. Help them find who wins the game.Input contains two integers A and B.
Constraints:
0 <= A, B <= 10<sup>15</sup>Print "Chandler" (without quotes) if Chandler wins the game and "Joey" (without quotes) if Joey wins the game.Sample Input
2 1
Sample Output
Chandler
Explanation:
In first move Joey gives Chandler 1 candy so, Chandler has 3 candies and Joey has 0.
In second move Chandler gives Joey 2 candies so, Chandler has 1 candy and Joey has 2.
In third move Joey has to give Chandler 3 candies but he has only 2 candies so he loses., I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
template<class C> void mini(C&a4, C b4){a4=min(a4,b4);}
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define mod 1000000007ll
#define pii pair<int,int>
/////////////
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int a,b;
cin>>a>>b;
if(a>=b)
cout<<"Chandler";
else
cout<<"Joey";
#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: Joey and Chandler are super bored. So, Chandler makes up a game they can play. The game is called Chandy Game. Initially, Chandler has A candies and Joey has B candies.
In the first move Joey has to give Chandler 1 candy.
In the second move Chandler has to give Joey 2 candies.
In the third move Joey has to give Chandler 3 candies.
In the fourth move Chandler has to give Joey 4 candies.
In the fifth move Joey has to give Chandler 5 candy.
... and so on.
The game continues till one of the player can not make a move. The player who cannot make a move loses. Help them find who wins the game.Input contains two integers A and B.
Constraints:
0 <= A, B <= 10<sup>15</sup>Print "Chandler" (without quotes) if Chandler wins the game and "Joey" (without quotes) if Joey wins the game.Sample Input
2 1
Sample Output
Chandler
Explanation:
In first move Joey gives Chandler 1 candy so, Chandler has 3 candies and Joey has 0.
In second move Chandler gives Joey 2 candies so, Chandler has 1 candy and Joey has 2.
In third move Joey has to give Chandler 3 candies but he has only 2 candies so he loses., I have written this Solution Code: ch,jo=map(int,input().split())
if ch>=jo:
print('Chandler')
else:
print('Joey'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: John has N candies. He wants to crush all of them. He feels that it would be boring to crush the candies randomly, so he found a method to crush them. He divides these candies into a minimum number of groups such that no group contains more than 3 candies. He crushes one candy from each group. If there are G groups in a single step, then the cost incurred in crushing a single candy for that step is G dollars. After candy from each group is crushed, he takes all the remaining candies and repeats the process until he has no candies left. He hasn't started crushing yet, but he wants to know how much the total cost would be incurred. Can you help him?
You have to answer Q-independent queries.The first line of input contains a single integer, Q denoting the number of queries.
Next, Q lines contain a single integer N denoting the number of candies John has.
<b>Constraints</b>
1 <= Q <= 5 * 10^4
1 <= N <= 10^9Print Q lines containing total cost incurred for each query.Sample Input 1:
1
4
Sample Output 1:
6
<b>Explanation:</b>
Query 1: First step John divides the candies into two groups of 3 and 1 candy respectively. Crushing one-one candy from both groups would cost him 2x2 = 4 dollars. He is now left with 2 candies. He divides it into one group. He crushes one candy for 1 dollar. Now, he is left with 1 candy. He crushes the last candy for 1 dollar. So, the total cost incurred is 4+1+1 = 6 dollars., I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int cost(int n){
if(n == 0) return 0;
int g = (n-1)/3 + 1;
return g*g + cost(n-g);
}
signed main() {
IOS;
clock_t start = clock();
int q; cin >> q;
while(q--){
int n;
cin >> n;
cout << cost(n) << endl;
}
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: John has N candies. He wants to crush all of them. He feels that it would be boring to crush the candies randomly, so he found a method to crush them. He divides these candies into a minimum number of groups such that no group contains more than 3 candies. He crushes one candy from each group. If there are G groups in a single step, then the cost incurred in crushing a single candy for that step is G dollars. After candy from each group is crushed, he takes all the remaining candies and repeats the process until he has no candies left. He hasn't started crushing yet, but he wants to know how much the total cost would be incurred. Can you help him?
You have to answer Q-independent queries.The first line of input contains a single integer, Q denoting the number of queries.
Next, Q lines contain a single integer N denoting the number of candies John has.
<b>Constraints</b>
1 <= Q <= 5 * 10^4
1 <= N <= 10^9Print Q lines containing total cost incurred for each query.Sample Input 1:
1
4
Sample Output 1:
6
<b>Explanation:</b>
Query 1: First step John divides the candies into two groups of 3 and 1 candy respectively. Crushing one-one candy from both groups would cost him 2x2 = 4 dollars. He is now left with 2 candies. He divides it into one group. He crushes one candy for 1 dollar. Now, he is left with 1 candy. He crushes the last candy for 1 dollar. So, the total cost incurred is 4+1+1 = 6 dollars., 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 q = Long.parseLong(br.readLine());
while(q-->0)
{
long N = Long.parseLong(br.readLine());
System.out.println(candyCrush(N,0,0));
}
}
static long candyCrush(long N, long cost,long group)
{
if(N==0)
{
return cost;
}
if(N%3==0)
{
group = N/3;
cost = cost + (group*group);
return candyCrush(N-group,cost,0);
}
else
{
group = (N/3)+1;
cost = cost + (group*group);
return candyCrush(N-group,cost,0);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a NxN matrix. You need to find the <a href = "https://en.wikipedia.org/wiki/Transpose">transpose</a> of the matrix.
The matrix is of form:
a b c ...
d e f ...
g h i ...
...........
There are N elements in each row.The first line of the input contains an integer N denoting the size of the square matrix.
The next N lines contain N single-spaced integers.
<b>Constraints</b>
1 <= N <= 100
1 <=Ai <= 100000Output the transpose of the matrix in similar format as that of the input.Sample Input
2
1 3
2 2
Sample Output
1 2
3 2
Sample Input:
1 2
3 4
Sample Output:
1 3
2 4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main
{
public static void main (String[] args) throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
String arr[][]=new String[n][n];
String transpose[][]=new String[n][n];
int row;
int cols;
for(row=0;row<n;row++)
{
String rowNum=br.readLine();
String rowVals[]=rowNum.split(" ");
for(cols=0; cols<n;cols++)
{
arr[row][cols]=rowVals[cols];
}
}
for(row=0;row<n;row++)
{
for(cols=0; cols<n;cols++)
{
transpose[row][cols]=arr[cols][row];
System.out.print(transpose[row][cols]+" ");
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a NxN matrix. You need to find the <a href = "https://en.wikipedia.org/wiki/Transpose">transpose</a> of the matrix.
The matrix is of form:
a b c ...
d e f ...
g h i ...
...........
There are N elements in each row.The first line of the input contains an integer N denoting the size of the square matrix.
The next N lines contain N single-spaced integers.
<b>Constraints</b>
1 <= N <= 100
1 <=Ai <= 100000Output the transpose of the matrix in similar format as that of the input.Sample Input
2
1 3
2 2
Sample Output
1 2
3 2
Sample Input:
1 2
3 4
Sample Output:
1 3
2 4, I have written this Solution Code: x=int(input())
l1=[]
for i in range(x):
a1=list(map(int,input().split()))
l1.append(a1)
l4=[]
for j in range(x):
l3=[]
for i in range(x):
l3.append(l1[i][j])
l4.append(l3)
for i in range(x):
for j in range(x):
print(l4[i][j], end=" ")
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a NxN matrix. You need to find the <a href = "https://en.wikipedia.org/wiki/Transpose">transpose</a> of the matrix.
The matrix is of form:
a b c ...
d e f ...
g h i ...
...........
There are N elements in each row.The first line of the input contains an integer N denoting the size of the square matrix.
The next N lines contain N single-spaced integers.
<b>Constraints</b>
1 <= N <= 100
1 <=Ai <= 100000Output the transpose of the matrix in similar format as that of the input.Sample Input
2
1 3
2 2
Sample Output
1 2
3 2
Sample Input:
1 2
3 4
Sample Output:
1 3
2 4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a[n][n];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cin>>a[j][i];
}
}
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cout<<a[i][j]<<" ";
}
cout<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The call stack of javascript based on the LIFO format i. e. last in first out.
Complete the functions <b>greetEnglish</b> and <b>greetFrench</b> so as the greetings are printed in the order : Hindi, English and French.Name of the person as stringGreetings are printed in the order : Hindi, English, French<b>Sample Input :</b>
John
<b>Sample Output :</b>
Namaste John!
Hello John!
Bonjour John!, I have written this Solution Code: function greetHindi(person) {
console.log(`Namaste ${person}!`)
}
function greetEnglish(person) {
greetHindi(person)
console.log(`Hello ${person}!`)
}
function greetFrench(person) {
greetEnglish(person)
console.log(`Bonjour ${person}!`)
}
function Greetings(person) {
greetFrench(person)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: import java.io.*;
import java.util.*;
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 sc= new FastReader();
String str= sc.nextLine();
String a="Apple";
if(a.equals(str)){
System.out.println("Gravity");
}
else{
System.out.println("Space");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
//Work
int main()
{
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen ("INPUT.txt" , "r" , stdin);
//freopen ("OUTPUT.txt" , "w" , stdout);
}
#endif
//-----------------------------------------------------------------------------------------------------------//
string S;
cin>>S;
if(S=="Apple")
{
cout<<"Gravity"<<endl;
}
else
{
cout<<"Space"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: n=input()
if n=='Apple':print('Gravity')
else:print('Space'), 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: 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: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: a="Hello World"
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
public static void main(String args[]){
System.out.println("Hello World");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: As always, Sid has come up with a new problem on Number Theory. Since it is too tricky to solve, can you help him. A bonus, you'll get a chapo for solving it right.
Given an integer n, let F(n) be the product of all positive integers i less or equal than n such that n and i are co-prime.
Now you're given an integer X, you need to find the sum of (F(i) modulo i) for all positive integers i less than or equal to X. If the answer is A, report the value of (A % 1000000007).The only line of input contains a single integer X.
Constraints
2 <= X <= 500000000Output a single integer, the required output for the problem.Sample Input
5
Sample Output
10
Sample Input
1000
Sample Output
128825, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define int long long
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define MOD 1000000007
#define INF 1000000000000000007LL
const int N = 100005;
#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
const int M = 1000000;
int Lower[M+1], Higher[M+1];
char Used[M+1];
bool prime[100001];
void sieve(){
memset(prime, true, sizeof(prime));
for(int i=2; i<=100000; i++){
if(!prime[i])
continue;
for(int j=i*i; j<=100000; j+=i)
prime[j]=false;
}
}
long long PrimePi(long long n)
{
long long v = sqrt(n+1e-9), p, temp, q, j, end, i, d, t;
for(int i=0;i<=M;i++)Used[i]=0;
Higher[1] = n - 1;
for(p = 2; p <= v; p++) {
Lower[p] = p - 1;
Higher[p] = (n/p) - 1;
}
for(p = 2; p <= v; p++) {
if(Lower[p] == Lower[p-1])
continue;
temp = Lower[p-1];
q = p * p;
Higher[1] -= Higher[p] - temp;
j = 1 + (p & 1);
end = (v <= n/q) ? v : n/q;
for(i = p + j; i <= 1 + end; i += j) {
if(Used[i])
continue;
d = i * p;
if(d <= v)
Higher[i] -= Higher[d] - temp;
else {
t = n/d;
Higher[i] -= Lower[t] - temp;
}
}
if(q <= v)
for(i = q; i <= end; i += p*j)
Used[i] = 1;
for(i = v; i >= q; i--) {
t = i/p;
Lower[i] -= Lower[t] - temp;
}
}
return Higher[1];
}
int psum(int n){
unordered_map<int, int> s;
vector<int> v;
int r = sqrt(n);
for(int i=1; i<=r; i++){
int x = n/i;
v.pb(x);
}
int y = n/r;
for(int i=y-1; i>0; i--){
v.pb(i);
}
for(int i: v){
s[i] = (i*(i+1))/2 - 1;
}
for(int p=2; p<=r; p++){
if(s[p] > s[p-1]){
int sp = s[p-1];
int p2 = p*p;
for(int i: v){
if(i<p2)
break;
s[i] -= (p * (s[i/p] - sp));
}
}
}
return s[n];
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int n; cin>>n;
int ans = psum(n)-2*PrimePi(n)+psum(n/2)*2-2*PrimePi(n/2)+n-1;
sieve();
for(int i=3; i<30000; i++){
if(!prime[i])
continue;
int x = i*i;
while(x <= n){
// trace(x);
ans += (x-2);
x *= i;
}
x = i*i*2;
while(x <= n){
// trace(x);
ans += (x-2);
x *= i;
}
ans %= MOD;
}
cout<<ans;
end_routine();
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 , your task is to print the lowercase English word corresponding to the number if it is <=5 else print "Greater than 5".
Numbers <=5 and their corresponding words :
1 = one
2 = two
3 = three
4 = four
5 = fiveThe input contains a single integer N.
Constraint:
1 <= n <= 100Print a string consisting of the lowercase English word corresponding to the number if it is <=5 else print the string "Greater than 5"Sample Input:
4
Sample Output
four
Sample Input:
6
Sample Output:
Greater than 5, I have written this Solution Code: N = int(input())
if N > 5:
print("Greater than 5")
elif(N == 1):
print("one")
elif(N == 2):
print("two")
elif(N == 3):
print("three")
elif(N == 4):
print("four")
elif(N == 5):
print("five"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n , your task is to print the lowercase English word corresponding to the number if it is <=5 else print "Greater than 5".
Numbers <=5 and their corresponding words :
1 = one
2 = two
3 = three
4 = four
5 = fiveThe input contains a single integer N.
Constraint:
1 <= n <= 100Print a string consisting of the lowercase English word corresponding to the number if it is <=5 else print the string "Greater than 5"Sample Input:
4
Sample Output
four
Sample Input:
6
Sample Output:
Greater than 5, I have written this Solution Code: import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int side = scanner.nextInt();
String area = conditional(side);
System.out.println(area);
}static String conditional(int n){
if(n==1){return "one";}
else if(n==2){return "two";}
else if(n==3){return "three";}
else if(n==4){return "four";}
else if(n==5){return "five";}
else{
return "Greater than 5";}
}}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code: static int focal_length(int R, char Mirror)
{
int f=R/2;
if((R%2==1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code:
int focal_length(int R, char Mirror)
{
int f=R/2;
if((R&1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code:
int focal_length(int R, char Mirror)
{
int f=R/2;
if((R&1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code: def focal_length(R,Mirror):
f=R/2;
if(Mirror == ')'):
f=-f
if R%2==1:
f=f-1
return int(f)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix <b>A</b> of dimensions <b>n x m</b>. The task is to perform <b>boundary traversal</b> on the matrix in clockwise manner.The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase has two lines of input. The first line contains dimensions of the matrix A, n and m. The second line contains n*m elements separated by spaces.
Constraints:
1 <= T <= 100
1 <= n, m <= 30
0 <= A[i][j] <= 100For each testcase, in a new line, print the boundary traversal of the matrix A.Input:
4
4 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
3 4
12 11 10 9 8 7 6 5 4 3 2 1
1 4
1 2 3 4
4 1
1 2 3 4
Output:
1 2 3 4 8 12 16 15 14 13 9 5
12 11 10 9 5 1 2 3 4 8
1 2 3 4
1 2 3 4
Explanation:
Testcase1: The matrix is:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
The boundary traversal is 1 2 3 4 8 12 16 15 14 13 9 5
Testcase 2: Boundary Traversal will be 12 11 10 9 5 1 2 3 4 8.
Testcase 3: Boundary Traversal will be 1 2 3 4.
Testcase 4: Boundary Traversal will be 1 2 3 4., I have written this Solution Code:
def boundaryTraversal(matrix, N, M): #N = 3, M = 4
start_row = 0
start_col = 0
count = 0
if N != 1 and M != 1:
total = 2 * (N - 1) + 2 * (M - 1)
else:
total = max(M, N)
#First row
for i in range(start_col, M, 1): #0,1, 2, 3
count += 1
print(matrix[start_row][i], end = " ")
if count == total:
return
start_row += 1
#Last column
for i in range(start_row, N, 1): # 1, 2
count += 1
print(matrix[i][M - 1], end = " ")
if count == total:
return
M -= 1
#Last Row
for i in range(M - 1, start_col - 1, -1): # 2, 1, 0
count += 1
print(matrix[N - 1][i], end = " ")
if count == total:
return
#First Column
N -= 1 # 2
for i in range(N - 1, start_row - 1, -1):
count += 1
print(matrix[i][start_col], end = " ")
start_col += 1
testcase = int(input().strip())
for test in range(testcase):
dim = input().strip().split(" ")
N = int(dim[0])
M = int(dim[1])
matrix = []
elements = input().strip().split(" ") #N * M
start = 0
for i in range(N):
matrix.append(elements[start : start + M]) # (0, M - 1) | (M, 2*m-1)
start = start + M
boundaryTraversal(matrix, N, M)
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix <b>A</b> of dimensions <b>n x m</b>. The task is to perform <b>boundary traversal</b> on the matrix in clockwise manner.The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase has two lines of input. The first line contains dimensions of the matrix A, n and m. The second line contains n*m elements separated by spaces.
Constraints:
1 <= T <= 100
1 <= n, m <= 30
0 <= A[i][j] <= 100For each testcase, in a new line, print the boundary traversal of the matrix A.Input:
4
4 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
3 4
12 11 10 9 8 7 6 5 4 3 2 1
1 4
1 2 3 4
4 1
1 2 3 4
Output:
1 2 3 4 8 12 16 15 14 13 9 5
12 11 10 9 5 1 2 3 4 8
1 2 3 4
1 2 3 4
Explanation:
Testcase1: The matrix is:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
The boundary traversal is 1 2 3 4 8 12 16 15 14 13 9 5
Testcase 2: Boundary Traversal will be 12 11 10 9 5 1 2 3 4 8.
Testcase 3: Boundary Traversal will be 1 2 3 4.
Testcase 4: Boundary Traversal will be 1 2 3 4., I have written this Solution Code: import java.io.*;
import java.lang.*;
import java.util.*;
class Main
{
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-- > 0)
{
int n1 = sc.nextInt();
int m1 = sc.nextInt();
int arr1[][] = new int[n1][m1];
for(int i = 0; i < n1; i++)
{
for(int j = 0; j < m1; j++)
arr1[i][j] = sc.nextInt();
}
boundaryTraversal(n1, m1,arr1);
System.out.println();
}
}
static void boundaryTraversal( int n1, int m1, int arr1[][])
{
// base cases
if(n1 == 1)
{
int i = 0;
while(i < m1)
System.out.print(arr1[0][i++] + " ");
}
else if(m1 == 1)
{
int i = 0;
while(i < n1)
System.out.print(arr1[i++][0]+" ");
}
else
{
// traversing the first row
for(int j=0;j<m1;j++)
{
System.out.print(arr1[0][j]+" ");
}
// traversing the last column
for(int j=1;j<n1;j++)
{
System.out.print(arr1[j][m1-1]+ " ");
}
// traversing the last row
for(int j=m1-2;j>=0;j--)
{
System.out.print(arr1[n1-1][j]+" ");
}
// traversing the first column
for(int j=n1-2;j>=1;j--)
{
System.out.print(arr1[j][0]+" ");
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a 2D grid. You are given two integers X and Y.
Consider a set "S" of points (x, y) on the cartesian plane such that 1 <= x <= N and 1 <= y <=M where x, y are positive integers.
You need to find number of line segments whose end points lies in set S such that the coordinates of the mid point also lies in the set S.The first line contains one integers – T (number of test cases).
The next T lines contains two integers N, M.
<b> Constraints: </b>
1 ≤ T ≤ 1000
1 ≤ N, M ≤ 1000Output T lines each containg a single integer denoting the number of such line segments.INPUT
2
3 3
6 7
OUTPUT
8
204, I have written this Solution Code: //HEADER FILES AND NAMESPACES
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#pragma GCC target("popcnt")
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
using cd = complex<double>;
const double PI = acos(-1);
// DEFINE STATEMENTS
const long long infty = 1e18;
#define num1 1000000007
#define num2 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define REPd(i,a,n) for(ll i=a; i>=n; i--)
#define pb push_back
#define pob pop_back
#define fr first
#define sc second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define all(x) x.begin(), x.end()
#define M_PI 3.14159265358979323846
#define epsilon (double)(0.000000001)
#define popcount __builtin_popcountll
#define fileio(x) freopen("input.txt", "r", stdin); freopen(x, "w", stdout);
#define out(x) cout << ((x) ? "Yes\n" : "No\n")
#define sz(x) x.size()
typedef long long ll;
typedef long long unsigned int llu;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
// DEBUG FUNCTIONS
#ifndef ONLINE_JUDGE
template<typename T>
void __p(T a) {
cout<<a;
}
template<typename T, typename F>
void __p(pair<T, F> a) {
cout<<"{";
__p(a.first);
cout<<",";
__p(a.second);
cout<<"}";
}
template<typename T>
void __p(std::vector<T> a) {
cout<<"{";
for(auto it=a.begin(); it<a.end(); it++)
__p(*it),cout<<",}"[it+1==a.end()];
}
template<typename T>
void __p(std::set<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T>
void __p(std::multiset<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T, typename F>
void __p(std::map<T,F> a) {
cout<<"{\n";
for(auto it=a.begin(); it!=a.end();++it)
{
__p(it->first);
cout << ": ";
__p(it->second);
cout<<"\n";
}
cout << "}\n";
}
template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) {
__p(a1);
__p(a...);
}
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
cout<<name<<" : ";
__p(arg1);
cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
int bracket=0,i=0;
for(;; i++)
if(names[i]==','&&bracket==0)
break;
else if(names[i]=='(')
bracket++;
else if(names[i]==')')
bracket--;
const char *comma=names+i;
cout.write(names,comma-names)<<" : ";
__p(arg1);
cout<<" | ";
__f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<" ", __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...)
#define error(...)
#endif
ll nc2(ll n){
return (n*(n-1))/2;
}
void solve(){
ll x,y; cin >> x >> y;
ll a = x/2, b = (x+1)/2;
ll c = y/2, d = (y+1)/2;
// trace(a,b,c,d);
ll ans = 2* (nc2(a) + nc2(b)) * (nc2(c) + nc2(d));
// trace(ans);
ans += (x)*(nc2(c) + nc2(d));
// trace(ans);
ans += (nc2(a) + nc2(b))*y;
// trace(ans);
cout << ans << "\n";
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
// cout.tie(NULL);
#ifdef LOCALFLAG
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
ll t = 1;
cin >> t;
while(t--){
solve();
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given push(), pop() and empty() operations of the stack of size N, remove the middle element of stack.
Note:
1. Please do not use any other data structure.
2. Middle - ceil(stack_size/2)<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>deleteMid()</b> that takes the stack and the integer N as parameters.
Constraints:
1 <= N <= 100
1 <= elements of stack <= 1000
<b>For Custom Input:-</b>
The first line of input should contains the size of stack N, the next line should contains N space separated integers containing elements of the stack.Return the modified stack i.e. after removing middle element from original stack. The driver code will take care of printing the elements of modified stackSample Input:
5
1 2 3 4 5
Sample Output:-
1 2 4 5
Sample Input:-
7
1 2 3 4 5 6 7
Sample Output:-
1 2 3 5 6 7, I have written this Solution Code: public static Stack<Integer> deleteMid1(Stack<Integer> s, int N, int count)
{
// if current is equal to half of size of stack
if(count == N/2)
{
s.pop();
return s;
}
int x=s.peek();
s.pop();
count += 1;
// recursively call deleteMid
s = deleteMid1(s,N,count);
s.push(x);
return s;
}
public static Stack<Integer> deleteMid(Stack<Integer> s, int N)
{
return deleteMid1(s,N,0);
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S. Find a string R which is an anagram of S and the hamming distance between S and R is maximum.
An anagram of a string is another string that contains the same characters, only the order of characters can be different.
Hamming distance between two strings of equal length is the number of positions at which the corresponding character is different.The first and the only line of input contains a single string S.
Constraints:
1 <= |S| <= 100000
S contains only lowercase letters of the English alphabet.Print the maximum hamming distance between S and R.Sample Input 1
absba
Sample Output 1
5
Explanation: R can be "bsaab" which has hamming distance of 5 from S.
Sample Input 2
aaa
Sample Output 2
0
Explanation: R can be "aaa" which has hamming distance of 0 from S., 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));
String s = br.readLine();
int[] arr = new int[27];
int total_sum = s.length();
for(int i = 0;i<s.length();i++){
arr[(int)(s.charAt(i))-97]++;
}
int count = 0;
int diff = 0;
for(int i = 0;i<27;i++){
diff = total_sum - arr[i];
if(arr[i] != 0 && diff>= arr[i]){
count+= arr[i];
}
else if( arr[i] != 0 && diff < arr[i]){
count+=diff;
}
}
System.out.println(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S. Find a string R which is an anagram of S and the hamming distance between S and R is maximum.
An anagram of a string is another string that contains the same characters, only the order of characters can be different.
Hamming distance between two strings of equal length is the number of positions at which the corresponding character is different.The first and the only line of input contains a single string S.
Constraints:
1 <= |S| <= 100000
S contains only lowercase letters of the English alphabet.Print the maximum hamming distance between S and R.Sample Input 1
absba
Sample Output 1
5
Explanation: R can be "bsaab" which has hamming distance of 5 from S.
Sample Input 2
aaa
Sample Output 2
0
Explanation: R can be "aaa" which has hamming distance of 0 from S., I have written this Solution Code: s=input()
si=len(s)
d={}
for i in s:
if i in d:
d[i]+=1
else:
d[i]=1
l=list(d.values())
k = list(d.keys())
a=max(l)
if(si>=(2*a)):
print(si)
else:
ans=si-a
print(2*ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S. Find a string R which is an anagram of S and the hamming distance between S and R is maximum.
An anagram of a string is another string that contains the same characters, only the order of characters can be different.
Hamming distance between two strings of equal length is the number of positions at which the corresponding character is different.The first and the only line of input contains a single string S.
Constraints:
1 <= |S| <= 100000
S contains only lowercase letters of the English alphabet.Print the maximum hamming distance between S and R.Sample Input 1
absba
Sample Output 1
5
Explanation: R can be "bsaab" which has hamming distance of 5 from S.
Sample Input 2
aaa
Sample Output 2
0
Explanation: R can be "aaa" which has hamming distance of 0 from S., 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
string s;
cin>>s;
int n=s.length();
int f[26]={};
int ma=0;
for(auto r:s){
f[r-'a']++;
ma=max(ma,f[r-'a']);
}
sort(s.begin(),s.end());
string r=s;
for(int i=0;i<n;++i)
r[i]=s[(i+ma)%n];
int ans=0;
for(int i=0;i<n;++i)
if(s[i]!=r[i])
++ans;
cout<<ans;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: We have provided you with an <code> addAndSubtract</code> object that contains <code>num1</code> and <code>num2</code> properties and <code>add()</code> and <code>subtract()</code> methods which add and subtract <code>num1</code> and <code>num2</code> respectively.
Create a <strong>calculator</strong> object that contains two functions <strong>product()</strong> and <strong>divide()</strong>. It should contain functions named the <code>product</code> and <code>divide</code> which return the product and division results of num1 and num2 respectively that is <code>num1* num2</code> and <code>num1/num2 </code>. But num1 and num2 are not defined in the calculator object.
Inherit the <code>addAndSubtract</code> object in the <code>calculator</code> object so that it can use num1 and num2 defined in the <code>addAndSubtract</code> object.
Note:- You won't be able to test it by generating outputTakes no argumentReturn a numberconst addAndSubtract = {
num1: 6,
num2: 3,
add() {
return . . .;
},
subtract() {
return . . .;
},
}
const calculator = {
//... your code
}
console.log(calculator.num1); // 6
console.log(calculator.subtract()); // 3
console.log(calculator.product()); // 18
<strong>Note: Give 6 and 3 as num1 and num2 respectively in the input section separated by a comma and then run the code to check the correct output</strong>, I have written this Solution Code: const addAndSubtract = {
num1: 6,
num2: 3,
add() {
return (this.num1 + this.num2);
},
subtract() {
return (this.num1 - this.num2);
},
}
const calculator = {
__proto__: addAndSubtract,
product() {
return (this.num1 * this.num2);
},
divide() {
return (this.num1 / this.num2);
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a large integer N. Find the sum of its digits. Eg:- if the integer is 1234, the answer is 1+2+3+4=10.The first and only line of input contains the integer N.
Constraints
The number of digits in N won't exceed 100000.Output a single integer, the sum of digits in N.Sample Input
1234
Sample Output
10
Sample Input
11111111111111111111
Sample Output
20, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String str=br.readLine();
int sum=0;
for(int i=0;i<str.length();i++){
char c=str.charAt(i);
int k=c-'0';
sum+=k;}
System.out.println(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a large integer N. Find the sum of its digits. Eg:- if the integer is 1234, the answer is 1+2+3+4=10.The first and only line of input contains the integer N.
Constraints
The number of digits in N won't exceed 100000.Output a single integer, the sum of digits in N.Sample Input
1234
Sample Output
10
Sample Input
11111111111111111111
Sample Output
20, 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(){
string s; cin>>s;
int ans = 0;
For(i, 0, sz(s)){
ans += (s[i]-'0');
}
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 are given a large integer N. Find the sum of its digits. Eg:- if the integer is 1234, the answer is 1+2+3+4=10.The first and only line of input contains the integer N.
Constraints
The number of digits in N won't exceed 100000.Output a single integer, the sum of digits in N.Sample Input
1234
Sample Output
10
Sample Input
11111111111111111111
Sample Output
20, I have written this Solution Code: s = input()
count = 0
for x in s:count+=int(x)
print(count) ;, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string(1-indexed). Print all the characters of the string at odd positions.The first line of the input contains a string S. String contains only lowercase english letters.
Constraints:-
1 <= |S| <= 100
The output should contain the character's at odd positions seperated by space.Sample Input
abcde
Sample Output
a c e
Sample Input
abcd
Sample Output
a c
Explanation:
index => 1 2 3 4
chars => a b c d
a and c are at odd index., I have written this Solution Code: str1 = input()
str2 = ''
for i in range(len(str1)):
if(i % 2 == 0):
str2 = str2 + str1[i]+" "
print(str2), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string(1-indexed). Print all the characters of the string at odd positions.The first line of the input contains a string S. String contains only lowercase english letters.
Constraints:-
1 <= |S| <= 100
The output should contain the character's at odd positions seperated by space.Sample Input
abcde
Sample Output
a c e
Sample Input
abcd
Sample Output
a c
Explanation:
index => 1 2 3 4
chars => a b c d
a and c are at odd index., I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
String s = sc.next();
for(int i = 0;i<s.length();i++){
if(i%2==0){
System.out.print(s.charAt(i)+" ");
}
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string(1-indexed). Print all the characters of the string at odd positions.The first line of the input contains a string S. String contains only lowercase english letters.
Constraints:-
1 <= |S| <= 100
The output should contain the character's at odd positions seperated by space.Sample Input
abcde
Sample Output
a c e
Sample Input
abcd
Sample Output
a c
Explanation:
index => 1 2 3 4
chars => a b c d
a and c are at odd index., I have written this Solution Code: // str is input
function oddChars(str) {
// write code here
// do not console.log
// return the output as a string
return str.split('').filter((v,idx)=> idx % 2 === 0).join(' ')
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string(1-indexed). Print all the characters of the string at odd positions.The first line of the input contains a string S. String contains only lowercase english letters.
Constraints:-
1 <= |S| <= 100
The output should contain the character's at odd positions seperated by space.Sample Input
abcde
Sample Output
a c e
Sample Input
abcd
Sample Output
a c
Explanation:
index => 1 2 3 4
chars => a b c d
a and c are at odd index., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
string s;
cin>>s;
for(int i=0;i<s.length();i++){
if(!(i&1)){cout<<s[i]<<" ";}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a positive integer N. Print the square of N.The input consists of a single integer N.
<b> Constraints: </b>
1 ≤ N ≤ 50Print a single integer – the value of N<sup>2</sup>.Sample Input 1:
5
Sample Output 1:
25, I have written this Solution Code:
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
int n = in.nextInt();
out.println(n*n);
out.flush();
}
static FastScanner in = new FastScanner();
static PrintWriter out = new PrintWriter(System.out);
static int oo = Integer.MAX_VALUE;
static long ooo = Long.MAX_VALUE;
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
void ini() throws FileNotFoundException {
br = new BufferedReader(new FileReader("input.txt"));
}
String next() {
while(!st.hasMoreTokens())
try { st = new StringTokenizer(br.readLine()); }
catch(IOException e) {}
return st.nextToken();
}
String nextLine(){
try{ return br.readLine(); }
catch(IOException e) { } return "";
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int a[] = new int[n];
for(int i=0;i<n;i++) a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
long[] readArrayL(int n) {
long a[] = new long[n];
for(int i=0;i<n;i++) a[i] = nextLong();
return a;
}
double nextDouble() {
return Double.parseDouble(next());
}
double[] readArrayD(int n) {
double a[] = new double[n];
for(int i=0;i<n;i++) a[i] = nextDouble();
return a;
}
}
static final Random random = new Random();
static void ruffleSort(int[] a){
int n = a.length;
for(int i=0;i<n;i++){
int j = random.nextInt(n), temp = a[j];
a[j] = a[i]; a[i] = temp;
}
Arrays.sort(a);
}
static void ruffleSortL(long[] a){
int n = a.length;
for(int i=0;i<n;i++){
int j = random.nextInt(n);
long temp = a[j];
a[j] = a[i]; a[i] = temp;
}
Arrays.sort(a);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a positive integer N. Print the square of N.The input consists of a single integer N.
<b> Constraints: </b>
1 ≤ N ≤ 50Print a single integer – the value of N<sup>2</sup>.Sample Input 1:
5
Sample Output 1:
25, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
cout<<(n*n)<<'\n';
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a positive integer N. Print the square of N.The input consists of a single integer N.
<b> Constraints: </b>
1 ≤ N ≤ 50Print a single integer – the value of N<sup>2</sup>.Sample Input 1:
5
Sample Output 1:
25, I have written this Solution Code: n=int(input())
print(n*n), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.Math.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
String[] st = bf.readLine().split(" ");
if(Integer.parseInt(st[1])==0)
System.out.print(-1);
else {
int f = (Integer.parseInt(st[0])/Integer.parseInt(st[1]));
System.out.print(f);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: D,Q = input().split()
D = int(D)
Q = int(Q)
if(0<=D and Q<=100 and Q >0):
print(int(D/Q))
else:
print('-1'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: #include <iostream>
using namespace std;
int main(){
int n,m;
cin>>n>>m;
if(m==0){cout<<-1;return 0;}
cout<<n/m;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: a="Hello World"
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
public static void main(String args[]){
System.out.println("Hello World");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Being told that an unsorted array contains (n - 1) of n consecutive numbers (where the bounds are defined), find the missing number in O(n) time.
The bounds tell that all the numbers between the lower bound and the upper bound are present in the array except one number which is missing. You have to find that missing number.The function takes three arguments , the first argument is the array of integers, the second argument is the upper bound and the third argument is the lower bound.
All the numbers between the lower and the upper bounds are present in the array (inclusive of both upper and lower bound) except one number which is missing.
Input is provided in the form of an array which would have 3 elements. The first element is the array of integers, the second element is the upper bound and the third element is the lower bound. All three elements are used internally to call the function.
Example: [[1, 4, 3] ,4, 1]
Here, [1,4,3] is the array of integers
4 is the upper bound
1 is the lower boundThe function should print the missing number in the console.const input = [[1, 4, 3] ,4, 1];
const arr = input[0];
const upper_bound = input[1];
const lower_bound = input[2];
findMissingNumber(arr, upper_bound, lower_bound); //prints 2
// Explanation: From numbers 1 to 4, only 2 is missing from the array, I have written this Solution Code:
function findMissingNumber(arrayOfIntegers, upperBound, lowerBound) {
// Iterate through array to find the sum of the numbers
let sumOfIntegers = 0;
for (let i = 0; i < arrayOfIntegers.length; i++) {
sumOfIntegers += arrayOfIntegers[i];
}
// Find theoretical sum of the consecutive numbers using a variation of Gauss Sum.
// Formula: [(N * (N + 1)) / 2] - [(M * (M - 1)) / 2];
// N is the upper bound and M is the lower bound
upperLimitSum = (upperBound * (upperBound + 1)) / 2;
lowerLimitSum = (lowerBound * (lowerBound - 1)) / 2;
theoreticalSum = upperLimitSum - lowerLimitSum;
console.log(theoreticalSum - sumOfIntegers);
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: It's Solo's 1st birthday and everyone is gifting her chocolates. There are N guests invited, and the i<sup>th</sup> guest gives Solo C<sub>i</sub> chocolates.
Find the total number of chocolates that Solo receives.The first line of the input contains an integer N, the number of guests.
The second line of the input contains N integers C<sub>1</sub>, C<sub>2</sub>, ....C<sub>N</sub>
Constraints
1 <= N <= 100
1 <= C<sub>i</sub> <= 100Output a single integer, the total number of chocolates that Solo receives.Sample Input
5
1 2 4 3 2
Sample Output
12
Explanation: Solo receives a total of 1+2+4+3+2 = 12 chocolates.
Sample Input
1
2
Sample Output
2, 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;
int ans = 0;
For(i, 0, n){
int a; cin>>a;
ans += a;
}
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: It's Solo's 1st birthday and everyone is gifting her chocolates. There are N guests invited, and the i<sup>th</sup> guest gives Solo C<sub>i</sub> chocolates.
Find the total number of chocolates that Solo receives.The first line of the input contains an integer N, the number of guests.
The second line of the input contains N integers C<sub>1</sub>, C<sub>2</sub>, ....C<sub>N</sub>
Constraints
1 <= N <= 100
1 <= C<sub>i</sub> <= 100Output a single integer, the total number of chocolates that Solo receives.Sample Input
5
1 2 4 3 2
Sample Output
12
Explanation: Solo receives a total of 1+2+4+3+2 = 12 chocolates.
Sample Input
1
2
Sample Output
2, I have written this Solution Code: n = int(input())
chocolates = list(map(int, input().strip().split(" ")))
count = 0
for val in chocolates:
count += val
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: It's Solo's 1st birthday and everyone is gifting her chocolates. There are N guests invited, and the i<sup>th</sup> guest gives Solo C<sub>i</sub> chocolates.
Find the total number of chocolates that Solo receives.The first line of the input contains an integer N, the number of guests.
The second line of the input contains N integers C<sub>1</sub>, C<sub>2</sub>, ....C<sub>N</sub>
Constraints
1 <= N <= 100
1 <= C<sub>i</sub> <= 100Output a single integer, the total number of chocolates that Solo receives.Sample Input
5
1 2 4 3 2
Sample Output
12
Explanation: Solo receives a total of 1+2+4+3+2 = 12 chocolates.
Sample Input
1
2
Sample Output
2, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int A[] = new int[n];
for(int i=0;i<n;i++){
A[i]=sc.nextInt();
}
int Total=0;
for(int i=0;i<n;i++){
Total+=A[i];
}
System.out.print(Total);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<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>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<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>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: def DivisorProblem(N):
ans=0
while N>1:
cnt=2
while N%cnt!=0:
cnt=cnt+1
N = N//cnt
ans=ans+1
return ans
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<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>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: static int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<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>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a side of a square, your task is to calculate its area.<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>Area()</b> that takes the side of the square as a parameter.Return the area of the squareSample Input:-
3
Sample Output:-
9
Sample Input:-
6
Sample Output:-
36
, I have written this Solution Code: static int Area(int side){
return side*side;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number (greater than 2), print two prime numbers whose sum will be equal to given number, else print -1 if no such number exists.
NOTE: A solution will always exist if the number is even. Read Goldbach’s conjecture.
If [a, b] is one solution with a <= b, and [c, d] is another solution with c <= d, and a < c then print [a, b] only
and not all possible solutions.The first line contains an integer T, depicting total number of test cases. Then following T lines contains an integer N.
Constraints:
1 ≤ T ≤ 100
2 ≤ N ≤ 1000000Print the two prime numbers in a single line with space in between if exist else print -1.Sample Input:
2
8
3
Sample Output:
3 5
-1, I have written this Solution Code:
// author-Shivam gupta
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1073741824
#define read(type) readInt<type>()
#define max1 1000001
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
const double pi=acos(-1.0);
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef map<int,int> MPII;
typedef set<int> SETI;
typedef multiset<int> MSETI;
typedef long int li;
typedef unsigned long int uli;
typedef long long int ll;
typedef unsigned long long int ull;
bool isPowerOfTwo (int x)
{
/* First x in the below expression is
for the case when x is 0 */
return x && (!(x&(x-1)));
}
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ll power(ll x, ll y, ll p)
{
ll res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}
// Returns n^(-1) mod p
ll modInverse(ll n, ll p)
{
return power(n, p-2, p);
}
// Returns nCr % p using Fermat's little
// theorem.
ll ncr(ll n, ll r,ll p)
{
// Base case
if (r==0)
return 1;
// Fill factorial array so that we
// can find all factorial of r, n
// and n-r
ll fac[n+1];
fac[0] = 1;
for (ll i=1 ; i<=n; i++)
fac[i] = fac[i-1]*i%p;
return (fac[n]* modInverse(fac[r], p) % p *
modInverse(fac[n-r], p) % p) % p;
}
ll fastexp (ll a, ll b, ll n) {
ll res = 1;
while (b) {
if (b & 1) res = res*a%n;
a = a*a%n;
b >>= 1;
}
return res;
}
bool a[max1];
int main() {
FOR(i,max1){
a[i]=false;}
for(int i=2;i<max1;i++){
if(a[i]==false){
for(int j=i+i;j<max1;j+=i){
a[j]=true;
}
}}
vector<int> v;
map<int,int> m;
for(int i=2;i<max1;i++){
if(a[i]==false){
v.EB(i);
m[i]++;
}
}
int t;
cin>>t;
while(t--){
int n;
cin>>n;
bool win=false;
for(int i=0;i<v.size();i++){
if(m.find(n-v[i])!=m.end()){
if(v[i]>n){break;}
cout<<v[i]<<" "<<n-v[i]<<endl;win=true;break;
}
}
if(win==false){out(-1);}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number (greater than 2), print two prime numbers whose sum will be equal to given number, else print -1 if no such number exists.
NOTE: A solution will always exist if the number is even. Read Goldbach’s conjecture.
If [a, b] is one solution with a <= b, and [c, d] is another solution with c <= d, and a < c then print [a, b] only
and not all possible solutions.The first line contains an integer T, depicting total number of test cases. Then following T lines contains an integer N.
Constraints:
1 ≤ T ≤ 100
2 ≤ N ≤ 1000000Print the two prime numbers in a single line with space in between if exist else print -1.Sample Input:
2
8
3
Sample Output:
3 5
-1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static boolean isPrime(int n)
{
boolean ans=true;
if(n<=1)
{
ans=false;
}
else if(n==2 || n==3)
{
ans=true;
}
else if(n%2==0 || n%3==0)
{
ans=false;
}
else
{
for(int i=5;i*i<=n;i+=6)
{
if(n%i==0 || n%(i+2)==0)
{
ans=false;
break;
}
}
}
return ans;
}
public static void main (String[] args) throws IOException {
BufferedReader scan=new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(scan.readLine());
while(t-->0)
{
int n=Integer.parseInt(scan.readLine());
int a=0,b=0;
if(n%2==1)
{
if(isPrime(n-2))
{
a=2;b=n-2;
}
}
else
{
for(int i=2;i<=n/2;i++)
{
if(isPrime(i))
{
if(isPrime(n-i))
{
a=i;b=n-i;
break;
}
}
}
}
if(a!=0 && b!=0)
{
System.out.println(a+" "+b);
}
else
{
System.out.println(-1);
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number (greater than 2), print two prime numbers whose sum will be equal to given number, else print -1 if no such number exists.
NOTE: A solution will always exist if the number is even. Read Goldbach’s conjecture.
If [a, b] is one solution with a <= b, and [c, d] is another solution with c <= d, and a < c then print [a, b] only
and not all possible solutions.The first line contains an integer T, depicting total number of test cases. Then following T lines contains an integer N.
Constraints:
1 ≤ T ≤ 100
2 ≤ N ≤ 1000000Print the two prime numbers in a single line with space in between if exist else print -1.Sample Input:
2
8
3
Sample Output:
3 5
-1, I have written this Solution Code: def isprime(num):
if(num==1 or num==0):
return False
for i in range(2,int(num**0.5)+1):
if(num%i==0):
return False
return True
T = int(input())
for test in range(T):
N = int(input())
value = -1
if(N%2==0):
for i in range(2,int(N/2)+1):
if(isprime(i)):
if(isprime(N-i)):
value = i
break
else:
if(isprime(N-2)):
value = 2
if(value==-1):
print(value)
else:
print(str(value)+" "+str(N-value)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer n. The task is to find the sum of product of all possible pairs of (x, y) where x and y will be represented as such y = floor(n/x), x varies from 1 to n.The first line of input contains an integer T. Each test case contains an integer N.
Constraints:
1 <= T <= 100
1 <= N <= 10^6For each test case in a new line print the required output.Input:
2
5
10
Output:
21
87
Explanation:
Testcase 1:
Following are the possible pairs of (x, y): (1, 5), (2, 2), (3, 1), (4, 1), (5, 1).
So, 1*5 + 2*2 + 3*1 + 4*1 + 5*1
= 5 + 4 + 3 + 4 + 5
= 21., I have written this Solution Code: import math
def sumOfRange(a, b):
i = (a * (a + 1)) >> 1;
j = (b * (b + 1)) >> 1;
return (i - j);
def sumofproduct(n):
sum = 0;
root = int(math.sqrt(n));
for i in range(1, root + 1):
up = int(n / i);
low = max(int(n / (i + 1)), root);
sum += (i * sumOfRange(up, low));
sum += (i * int(n / i));
return sum;
T = int(input())
for i in range(T):
n = int(input())
print(sumofproduct(n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer n. The task is to find the sum of product of all possible pairs of (x, y) where x and y will be represented as such y = floor(n/x), x varies from 1 to n.The first line of input contains an integer T. Each test case contains an integer N.
Constraints:
1 <= T <= 100
1 <= N <= 10^6For each test case in a new line print the required output.Input:
2
5
10
Output:
21
87
Explanation:
Testcase 1:
Following are the possible pairs of (x, y): (1, 5), (2, 2), (3, 1), (4, 1), (5, 1).
So, 1*5 + 2*2 + 3*1 + 4*1 + 5*1
= 5 + 4 + 3 + 4 + 5
= 21., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int p[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int ans = 0;
int n; cin >> n;
for(int i = 1; i <= n; i++){
ans += i*(n/i);
}
cout << ans << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer n. The task is to find the sum of product of all possible pairs of (x, y) where x and y will be represented as such y = floor(n/x), x varies from 1 to n.The first line of input contains an integer T. Each test case contains an integer N.
Constraints:
1 <= T <= 100
1 <= N <= 10^6For each test case in a new line print the required output.Input:
2
5
10
Output:
21
87
Explanation:
Testcase 1:
Following are the possible pairs of (x, y): (1, 5), (2, 2), (3, 1), (4, 1), (5, 1).
So, 1*5 + 2*2 + 3*1 + 4*1 + 5*1
= 5 + 4 + 3 + 4 + 5
= 21., I have written this Solution Code: import java.io.*;
import java.util.*;
import java.math.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
while(t!=0){
int n=Integer.parseInt(br.readLine());
long res=0;
for(int i=1;i<=n;i++){
res=res+(i*(int)Math.floor(n/i));
}
System.out.println(res);
t--;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: import java.io.*;
import java.util.*;
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 sc= new FastReader();
String str= sc.nextLine();
String a="Apple";
if(a.equals(str)){
System.out.println("Gravity");
}
else{
System.out.println("Space");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
//Work
int main()
{
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen ("INPUT.txt" , "r" , stdin);
//freopen ("OUTPUT.txt" , "w" , stdout);
}
#endif
//-----------------------------------------------------------------------------------------------------------//
string S;
cin>>S;
if(S=="Apple")
{
cout<<"Gravity"<<endl;
}
else
{
cout<<"Space"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: n=input()
if n=='Apple':print('Gravity')
else:print('Space'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers, find two numbers such that they add up to a specific target number k.
Output the indices of the elements that add up to the sum (array is 1 indexed).
If multiple solutions exist, output the one where index2 is minimum. If there are multiple solutions with the minimum index2, choose the one with minimum index1 out of them.
For example:
array A[] = 1 1 3 2 2
k = 3
Output: 1 4
Explanation: pair of indices which gives target k are: (1, 4), (1, 5), (2, 4), (2, 5). Among all such pairs (1, 4) satisfies above condition.The first line contain integers N and k, the number of elements in the array and the target sum.
The second line of the input contains N singly spaces integers.
Constraints:-
2 <= N <= 100000
1 <= A[i] <= 1000000000
1 <= k <= 2000000000Output two integers the indices of the two elements. It is guaranteed that the ans will always existSample Input
5 3
1 1 3 2 2
Sample Output
1 4
Explanation: The satisfying pairs are (1, 4), (2, 4), (1, 5), (2, 5).
Sample Input:
2 2
1 1
Sample Output:
1 2, I have written this Solution Code: import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.HashMap;
import java.util.InputMismatchException;
import static java.lang.Math.pow;
class InputReader {
private boolean finished = false;
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public static InputReader getInputReader(boolean readFromTextFile) throws FileNotFoundException {
return ((readFromTextFile) ? new InputReader(new FileInputStream("src/input.txt"))
: new InputReader(System.in));
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int peek() {
if (numChars == -1) {
return -1;
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
return -1;
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String nextString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private String readLine0() {
StringBuilder buf = new StringBuilder();
int c = read();
while (c != '\n' && c != -1) {
if (c != '\r') {
buf.appendCodePoint(c);
}
c = read();
}
return buf.toString();
}
public String readLine() {
String s = readLine0();
while (s.trim().length() == 0) {
s = readLine0();
}
return s;
}
public String readLine(boolean ignoreEmptyLines) {
if (ignoreEmptyLines) {
return readLine();
} else {
return readLine0();
}
}
public BigInteger readBigInteger() {
try {
return new BigInteger(nextString());
} catch (NumberFormatException e) {
throw new InputMismatchException();
}
}
public char nextCharacter() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
return (char) c;
}
public double nextDouble() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E') {
return res * pow(10, nextInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c)) {
if (c == 'e' || c == 'E') {
return res * pow(10, nextInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public boolean isExhausted() {
int value;
while (isSpaceChar(value = peek()) && value != -1) {
read();
}
return value == -1;
}
public String next() {
return nextString();
}
public SpaceCharFilter getFilter() {
return filter;
}
public void setFilter(SpaceCharFilter filter) {
this.filter = filter;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
public int[] nextIntArray(int n) {
int[] array = new int[n];
for (int i = 0; i < n; ++i) array[i] = nextInt();
return array;
}
public int[] nextSortedIntArray(int n) {
int array[] = nextIntArray(n);
Arrays.sort(array);
return array;
}
public int[] nextSumIntArray(int n) {
int[] array = new int[n];
array[0] = nextInt();
for (int i = 1; i < n; ++i) array[i] = array[i - 1] + nextInt();
return array;
}
public long[] nextLongArray(int n) {
long[] array = new long[n];
for (int i = 0; i < n; ++i) array[i] = nextLong();
return array;
}
public long[] nextSumLongArray(int n) {
long[] array = new long[n];
array[0] = nextInt();
for (int i = 1; i < n; ++i) array[i] = array[i - 1] + nextInt();
return array;
}
public long[] nextSortedLongArray(int n) {
long array[] = nextLongArray(n);
Arrays.sort(array);
return array;
}
}
class Main {
public static void main (String[] args) throws FileNotFoundException {
InputReader sc = InputReader.getInputReader(false);
int n = sc.nextInt();
long target = sc.nextLong();
long[] arr = new long[n+1];
HashMap<Long ,Integer> mp = new HashMap<>();
for(int i=1;i<=n;i++){
arr[i] = sc.nextLong();
}
for(int i=1;i<=n;i++){
if (mp.containsKey(target-arr[i])){
System.out.println(mp.get(target-arr[i])+ " "+ i);
break;
}
if(!mp.containsKey(arr[i])){
mp.put(arr[i] , i);
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers, find two numbers such that they add up to a specific target number k.
Output the indices of the elements that add up to the sum (array is 1 indexed).
If multiple solutions exist, output the one where index2 is minimum. If there are multiple solutions with the minimum index2, choose the one with minimum index1 out of them.
For example:
array A[] = 1 1 3 2 2
k = 3
Output: 1 4
Explanation: pair of indices which gives target k are: (1, 4), (1, 5), (2, 4), (2, 5). Among all such pairs (1, 4) satisfies above condition.The first line contain integers N and k, the number of elements in the array and the target sum.
The second line of the input contains N singly spaces integers.
Constraints:-
2 <= N <= 100000
1 <= A[i] <= 1000000000
1 <= k <= 2000000000Output two integers the indices of the two elements. It is guaranteed that the ans will always existSample Input
5 3
1 1 3 2 2
Sample Output
1 4
Explanation: The satisfying pairs are (1, 4), (2, 4), (1, 5), (2, 5).
Sample Input:
2 2
1 1
Sample Output:
1 2, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
long long n,k;
cin>>n>>k;
long long a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
unordered_map<long long ,int > m;
long long x;
for(int i=0;i<n;i++){
x=k-a[i];
if(m.find(x)!=m.end()){
cout<<m[x]+1<<" "<<i+1;
return 0;
}
if(m.find(a[i])==m.end()){m[a[i]]=i;}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N and F(N) find value of F(1), if, F(i)=(F(i-1) + F(i-1))%1000000007 and 0 <= F(1) < 1000000007.First and the only line of input contains two integers N and F(N).
Constraints:
1 <= N <= 1000000000
0 <= F(N) < 1000000007Print a single integer, F(1).Sample Input 1
2 6
Sample Output 1
3
Exlpanation: F(1) = 3, F(2)=(3+3)%1000000007 = 6.
Sample Input 2
3 6
Sample Input 2
500000005
Explanation:
F(1) = 500000005
F(2) = (500000005+500000005)%1000000007 = 3
F(3)= (3+3)%1000000007 = 6, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws NumberFormatException, IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer(br.readLine());
int n=Integer.parseInt(st.nextToken());
int fn=Integer.parseInt(st.nextToken());
long ans=fn;
int P=1000000007;
long inv2n=(long)pow(pow(2,n-1,P)%P,P-2,P);
ans=((ans%P)*(inv2n%P))%P;
System.out.println((ans)%P);
}
static long pow(long x, long y,long P)
{
long res = 1l;
while (y > 0)
{
if ((y & 1) == 1)
res = (res * x)%P;
y = y >> 1;
x = (x * x)%P;
}
return res;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N and F(N) find value of F(1), if, F(i)=(F(i-1) + F(i-1))%1000000007 and 0 <= F(1) < 1000000007.First and the only line of input contains two integers N and F(N).
Constraints:
1 <= N <= 1000000000
0 <= F(N) < 1000000007Print a single integer, F(1).Sample Input 1
2 6
Sample Output 1
3
Exlpanation: F(1) = 3, F(2)=(3+3)%1000000007 = 6.
Sample Input 2
3 6
Sample Input 2
500000005
Explanation:
F(1) = 500000005
F(2) = (500000005+500000005)%1000000007 = 3
F(3)= (3+3)%1000000007 = 6, I have written this Solution Code: n,f=map(int,input().strip().split())
mod=10**9+7
m1=pow(2,n-1,mod)
m=pow(m1,mod-2,mod)
print(f*m%mod), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N and F(N) find value of F(1), if, F(i)=(F(i-1) + F(i-1))%1000000007 and 0 <= F(1) < 1000000007.First and the only line of input contains two integers N and F(N).
Constraints:
1 <= N <= 1000000000
0 <= F(N) < 1000000007Print a single integer, F(1).Sample Input 1
2 6
Sample Output 1
3
Exlpanation: F(1) = 3, F(2)=(3+3)%1000000007 = 6.
Sample Input 2
3 6
Sample Input 2
500000005
Explanation:
F(1) = 500000005
F(2) = (500000005+500000005)%1000000007 = 3
F(3)= (3+3)%1000000007 = 6, 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>
/////////////
long long powerm(long long x, unsigned long long y, long long p)
{
long long res = 1;
x = x % p;
while (y > 0)
{
if (y & 1)
res = (res*x) % p;
y = y>>1;
x = (x*x) % p;
}
return res;
}
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n,fn;
cin>>n>>fn;
int mo=1000000007;
cout<<(fn*powerm(powerm(2,n-1,mo),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: Ram is given an array <b>A</b> of length N, Ram can right-rotate the array any number of times (possibly zero times also). His task is to find out the maximum value of A<sub>1</sub> + A<sub>N</sub>.
<b>Note</b>: Right rotation of array [A<sub>1</sub>, A<sub>2</sub>,. , A<sub>N</sub>] is [A<sub>N</sub>, A<sub>1</sub>,. , A<sub>N-1</sub>].The first line of the input contains a single integer T, denoting the number of test cases.
The first line of each test case contains a single integer N, denoting the size of an array A.
The second line of each test case contains N space- separated integers denoting array A.
<b>Constraints</b>
1 ≤ T ≤ 1000
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ A<sub>i</sub> ≤ 10<sup>9</sup>For each test case output on a new line denoting the maximum value of A<sub>1</sub> + A<sub>N</sub>.Sample Input
3
2
5 8
3
5 10 15
4
4 4 4 4
Sample Output
13
25
8, 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 t = sc.nextInt();
for(int i=0;i<t;i++){
int n = sc.nextInt();
int a[] = new int[n];
int max=0;
for(int j=0;j<n;j++){
a[j] = sc.nextInt();
}
for(int j=0;j<n-1;j++){
if(max<a[j]+a[j+1]){
max=a[j]+a[j+1];
}
}
if(max<a[0]+a[n-1]){
max=a[0]+a[n-1];
}
System.out.println(max);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ram is given an array <b>A</b> of length N, Ram can right-rotate the array any number of times (possibly zero times also). His task is to find out the maximum value of A<sub>1</sub> + A<sub>N</sub>.
<b>Note</b>: Right rotation of array [A<sub>1</sub>, A<sub>2</sub>,. , A<sub>N</sub>] is [A<sub>N</sub>, A<sub>1</sub>,. , A<sub>N-1</sub>].The first line of the input contains a single integer T, denoting the number of test cases.
The first line of each test case contains a single integer N, denoting the size of an array A.
The second line of each test case contains N space- separated integers denoting array A.
<b>Constraints</b>
1 ≤ T ≤ 1000
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ A<sub>i</sub> ≤ 10<sup>9</sup>For each test case output on a new line denoting the maximum value of A<sub>1</sub> + A<sub>N</sub>.Sample Input
3
2
5 8
3
5 10 15
4
4 4 4 4
Sample Output
13
25
8, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main() {
auto start = std::chrono::high_resolution_clock::now();
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
int tt;
cin >> tt;
int total = 0;
while (tt--) {
int n;
cin >> n;
total += n;
vector<int> a(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
int maxi = a[0] + a[n - 1];
for (int i = 0; i < n - 1; i++) {
maxi = max(a[i] + a[i + 1], maxi);
}
cout << maxi << "\n";
}
assert(total <= 1e5);
return 0;
}, 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: It is a moment of triumph in Stepstones, as they have captured all the ships from Westeros. To celebrate this occasion, they have prepared a problem for you to solve:
You are given an integer R, and an array A consisting of positive integers. You can do this operation at most once (possibly zero times):
• Choose any element of the array. Replace it with any integer X, such that 1 ≤ X ≤ R.
Find the maximum possible GCD of the entire array.The first line consists of two space-separated integers N and R.
The second line consists of N space-separated integers – A<sub>1</sub>, A<sub>2</sub> ... A<sub>N</sub>.
<b>Constraints:</b>
2 ≤ N ≤ 10<sup>5</sup>
1 ≤ R ≤ 10<sup>5</sup>
1 ≤ A<sub>i</sub> ≤ 10<sup>5</sup>Print a single integer – the maximum possible GCD of the array.Sample Input 1:
3 10
2 3 4
Sample Output 1:
2
Sample Explanation 1:
We can change the second element from 3 to 2. Then the GCD of the array becomes gcd(2, 2, 4) = 2, which is the maximum possible.
Sample Input 2:
2 3
10 15
Sample Output 2:
5
Sample Explanation 2:
We do need to perform any operation. The GCD of the array is gcd(10, 15) = 5, which is the maximum possible.
Sample Input 3:
3 5
11 6 12
Sample Output 3:
3
Sample Explanation 3:
We can change the first element from 11 to 3. Then the GCD of the array becomes gcd(3, 6, 12) = 3, which is the maximum possible., I have written this Solution Code: def find_gcd(x, y):
while(y):
x, y = y, x % y
return x
nr = list(map(int,input().split()))
n = nr[0]
r = nr[1]
_ = list(map(int,input().split()))
num1 = _[0]
num2 = _[1]
gcd = find_gcd(num1, num2)
for i in range(2,n):
gcd = find_gcd(gcd, _[i])
if gcd>=r:
print(gcd)
else:
while r:
num = 0
idx = None
for i in range(n):
if _[i]%r != 0:
num += 1
idx = i
if num > 1:
break
if idx!= None:
temp = _[idx]
if num == 1:
_[idx] = r
num1 = _[0]
num2 = _[1]
tgcd = find_gcd(num1, num2)
for i in range(2,n):
tgcd = find_gcd(tgcd, _[i])
gcd = max(tgcd,gcd)
_[idx] = temp
break
r -= 1
print(gcd), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: It is a moment of triumph in Stepstones, as they have captured all the ships from Westeros. To celebrate this occasion, they have prepared a problem for you to solve:
You are given an integer R, and an array A consisting of positive integers. You can do this operation at most once (possibly zero times):
• Choose any element of the array. Replace it with any integer X, such that 1 ≤ X ≤ R.
Find the maximum possible GCD of the entire array.The first line consists of two space-separated integers N and R.
The second line consists of N space-separated integers – A<sub>1</sub>, A<sub>2</sub> ... A<sub>N</sub>.
<b>Constraints:</b>
2 ≤ N ≤ 10<sup>5</sup>
1 ≤ R ≤ 10<sup>5</sup>
1 ≤ A<sub>i</sub> ≤ 10<sup>5</sup>Print a single integer – the maximum possible GCD of the array.Sample Input 1:
3 10
2 3 4
Sample Output 1:
2
Sample Explanation 1:
We can change the second element from 3 to 2. Then the GCD of the array becomes gcd(2, 2, 4) = 2, which is the maximum possible.
Sample Input 2:
2 3
10 15
Sample Output 2:
5
Sample Explanation 2:
We do need to perform any operation. The GCD of the array is gcd(10, 15) = 5, which is the maximum possible.
Sample Input 3:
3 5
11 6 12
Sample Output 3:
3
Sample Explanation 3:
We can change the first element from 11 to 3. Then the GCD of the array becomes gcd(3, 6, 12) = 3, which is the maximum possible., I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main {
public static long mod = 1000000007;
public static long mod2 = 998244353;
public static void main(String[] args) throws java.lang.Exception {
Reader sc = new Reader();
FastNum in = new FastNum(System.in);
Writer out = new Writer(System.out);
int n = in.nextInt();
int r = in.nextInt();
int[] a = in.nextIntArray(n);
int ans = a[0];
for (int i = 1; i < n; i++) {
ans = gcd(ans, a[i]);
}
int max = MaxGCD(a, n);
for (int i = 1; i <= r; i++) {
ans = Math.max(ans, gcd(max, i));
}
out.println(ans);
out.flush();
}
static int MaxGCD(int a[], int n) {
int[] Prefix = new int[n + 2];
int[] Suffix = new int[n + 2];
Prefix[1] = a[0];
for (int i = 2; i <= n; i += 1) {
Prefix[i] = gcd(Prefix[i - 1],
a[i - 1]);
}
Suffix[n] = a[n - 1];
for (int i = n - 1; i >= 1; i -= 1) {
Suffix[i] = gcd(Suffix[i + 1],
a[i - 1]);
}
int ans = Math.max(Suffix[2], Prefix[n - 1]);
for (int i = 2; i < n; i += 1) {
ans = Math.max(ans, gcd(Prefix[i - 1],
Suffix[i + 1]));
}
return ans;
}
static long modSum(long p, long q) {
if (q > 0) {
return (p + q) % mod;
} else {
return (p + q + mod) % mod;
}
}
static long invMod(long p, long q, long m) {
long expo = m - 2;
while (expo != 0) {
if ((expo & 1) == 1) {
p = (p * q) % m;
}
q = (q * q) % m;
expo >>= 1;
}
return p;
}
public static int gcd(int a, int b) {
if (a == 0)
return b;
return gcd(b % a, a);
}
public static long power(long a, long b) {
if (b == 0)
return 1;
long answer = power(a, b / 2) % mod;
answer = (answer * answer) % mod;
if (b % 2 != 0)
answer = (answer * a) % mod;
return answer;
}
public static void swap(int x, int y) {
int t = x;
x = y;
y = t;
}
public static long min(long a, long b) {
if (a < b) return a;
return b;
}
public static long divide(long a, long b) {
return (a % mod * (power(b, mod - 2) % mod)) % mod;
}
public static long nCr(long n, long r) {
long answer = 1;
long k = min(r, n - r);
for (long i = 0; i < k; i++) {
answer = (answer % mod * (n - i) % mod) % mod;
answer = divide(answer, i + 1);
}
return answer % mod;
}
public static boolean plaindrome(String str) {
StringBuilder sb = new StringBuilder();
sb.append(str);
return (str.equals((sb.reverse()).toString()));
}
public static class Pair {
int a;
int b;
Pair(int s, int e) {
a = s;
b = e;
}
@Override
public boolean equals(Object o) {
if (this == o)
return true;
if (o == null || getClass() != o.getClass())
return false;
Pair pair = (Pair) o;
return a == pair.a && b == pair.b;
}
@Override
public int hashCode() {
return Objects.hash(a, b);
}
}
static class Assert {
static void check(boolean e) {
if (!e) {
throw new AssertionError();
}
}
}
static class FastNum implements AutoCloseable {
InputStream is;
byte buffer[] = new byte[1 << 16];
int size = 0;
int pos = 0;
FastNum(InputStream is) {
this.is = is;
}
int nextChar() {
if (pos >= size) {
try {
size = is.read(buffer);
} catch (IOException e) {
throw new IOError(e);
}
pos = 0;
if (size == -1) {
return -1;
}
}
Assert.check(pos < size);
int c = buffer[pos] & 0xFF;
pos++;
return c;
}
int nextInt() {
int c = nextChar();
while (c == ' ' || c == '\r' || c == '\n' || c == '\t') {
c = nextChar();
}
if (c == '-') {
c = nextChar();
Assert.check('0' <= c && c <= '9');
int n = -(c - '0');
c = nextChar();
while ('0' <= c && c <= '9') {
int d = c - '0';
c = nextChar();
Assert.check(n > Integer.MIN_VALUE / 10
|| n == Integer.MIN_VALUE / 10 && d <= -(Integer.MIN_VALUE % 10));
n = n * 10 - d;
}
return n;
} else {
Assert.check('0' <= c && c <= '9');
int n = c - '0';
c = nextChar();
while ('0' <= c && c <= '9') {
int d = c - '0';
c = nextChar();
Assert.check(
n < Integer.MAX_VALUE / 10 || n == Integer.MAX_VALUE / 10 && d <= Integer.MAX_VALUE % 10);
n = n * 10 + d;
}
return n;
}
}
char nextCh() {
int c = nextChar();
while (c == ' ' || c == '\r' || c == '\n' || c == '\t') {
c = nextChar();
}
return (char) c;
}
long nextLong() {
int c = nextChar();
while (c == ' ' || c == '\r' || c == '\n' || c == '\t') {
c = nextChar();
}
if (c == '-') {
c = nextChar();
Assert.check('0' <= c && c <= '9');
long n = -(c - '0');
c = nextChar();
while ('0' <= c && c <= '9') {
int d = c - '0';
c = nextChar();
Assert.check(n > Long.MIN_VALUE / 10
|| n == Long.MIN_VALUE / 10 && d <= -(Long.MIN_VALUE % 10));
n = n * 10 - d;
}
return n;
} else {
Assert.check('0' <= c && c <= '9');
long n = c - '0';
c = nextChar();
while ('0' <= c && c <= '9') {
int d = c - '0';
c = nextChar();
Assert.check(
n < Long.MAX_VALUE / 10 || n == Long.MAX_VALUE / 10 && d <= Long.MAX_VALUE % 10);
n = n * 10 + d;
}
return n;
}
}
int[] nextIntArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
}
return arr;
}
long[] nextLongArray(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
}
return arr;
}
char[] nextCharArray(int n) {
char[] arr = new char[n];
for (int i = 0; i < n; i++) {
arr[i] = nextCh();
}
return arr;
}
@Override
public void close() {
}
}
static class Writer extends PrintWriter {
public Writer(java.io.Writer out) {
super(out);
}
public Writer(java.io.Writer out, boolean autoFlush) {
super(out, autoFlush);
}
public Writer(OutputStream out) {
super(out);
}
public Writer(OutputStream out, boolean autoFlush) {
super(out, autoFlush);
}
public void printArray(int[] arr) {
for (int j : arr) {
print(j);
print(' ');
}
println();
}
public void printArray(long[] arr) {
for (long j : arr) {
print(j);
print(' ');
}
println();
}
}
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private final DataInputStream din;
private final byte[] buffer;
private int bufferPointer, bytesRead;
public Reader() {
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException {
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException {
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException {
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException {
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException {
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException {
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException {
if (din == null)
return;
din.close();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: It is a moment of triumph in Stepstones, as they have captured all the ships from Westeros. To celebrate this occasion, they have prepared a problem for you to solve:
You are given an integer R, and an array A consisting of positive integers. You can do this operation at most once (possibly zero times):
• Choose any element of the array. Replace it with any integer X, such that 1 ≤ X ≤ R.
Find the maximum possible GCD of the entire array.The first line consists of two space-separated integers N and R.
The second line consists of N space-separated integers – A<sub>1</sub>, A<sub>2</sub> ... A<sub>N</sub>.
<b>Constraints:</b>
2 ≤ N ≤ 10<sup>5</sup>
1 ≤ R ≤ 10<sup>5</sup>
1 ≤ A<sub>i</sub> ≤ 10<sup>5</sup>Print a single integer – the maximum possible GCD of the array.Sample Input 1:
3 10
2 3 4
Sample Output 1:
2
Sample Explanation 1:
We can change the second element from 3 to 2. Then the GCD of the array becomes gcd(2, 2, 4) = 2, which is the maximum possible.
Sample Input 2:
2 3
10 15
Sample Output 2:
5
Sample Explanation 2:
We do need to perform any operation. The GCD of the array is gcd(10, 15) = 5, which is the maximum possible.
Sample Input 3:
3 5
11 6 12
Sample Output 3:
3
Sample Explanation 3:
We can change the first element from 11 to 3. Then the GCD of the array becomes gcd(3, 6, 12) = 3, which is the maximum possible., I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const ll N = 1e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
vi divisor(N + 1);
readb(n, r);
FORD (i, r, 1)
{
for (int j = i; j <= N; j += i)
if (!divisor[j]) divisor[j] = i;
}
readarr(a, n);
int pre[n + 1] = {}, suf[n + 2] = {};
FOR (i, 1, n) pre[i] = __gcd(pre[i - 1], a[i]);
FORD (i, n, 1) suf[i] = __gcd(suf[i + 1], a[i]);
int ans = pre[n];
FOR (i, 1, n)
{
int x = __gcd(pre[i - 1], suf[i + 1]);
ans = max(ans, divisor[x]);
}
print(ans);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A pair is called lucky if its sum is even and positive. Given three numbers find if there exists a lucky pair or not.The only line contains three integers a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>
<b>Constraints:</b>
-10<sup>9</sup> <= a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub> <= 10<sup>9</sup>Print "YES" without quotes if there exists a lucky pair otherwise print "NO" without quotes.Sample Input 1:
23 32 12
Sample Output 1:
YES
Sample Input 2:
1 -1 2
Sample Output 2:
NO, I have written this Solution Code: //HEADER FILES AND NAMESPACES
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#pragma GCC target("popcnt")
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
// DEFINE STATEMENTS
const long long infty = 1e18;
#define num1 1000000007
#define num2 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define REPd(i,a,n) for(ll i=a; i>=n; i--)
#define pb push_back
#define pob pop_back
#define fr first
#define sc second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define all(x) x.begin(), x.end()
#define M_PI 3.14159265358979323846
#define epsilon (double)(0.000000001)
#define popcount __builtin_popcountll
#define fileio(x) freopen("input.txt", "r", stdin); freopen(x, "w", stdout);
#define out(x) cout << ((x) ? "Yes\n" : "No\n")
#define sz(x) x.size()
typedef long long ll;
typedef long long unsigned int llu;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
void solve(){
vector<ll>a(3);
for(ll i = 0;i<3;i++)cin >> a[i];
for(ll i = 0;i<3;i++){
for(ll j = i+1;j<3;j++){
if((a[i]+a[j])>0 && (a[i]+a[j])%2 == 0){
cout << "YES\n";
return;
}
}
}
cout << "NO\n";
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
// cout.tie(NULL);
#ifdef LOCALFLAG
freopen("Input.txt", "r", stdin);
freopen("Output2.txt", "w", stdout);
#endif
ll t = 1;
//cin >> t;
while(t--){
solve();
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of objects representing a group of students, each with a name and an array of test scores. Your task is to use <code>map</code>, <code>filter</code>, and <code>reduce</code> to calculate the average test score for each student, and then return an array of objects containing only the students who have an average score above <b>90</b>.
The array of objects that you will return should have the keys <code>"name"</code> and <code>"average"</code> which should contain the name and the average marks of the student if his average marks is greater than <b>90</b>.The <code>highPerformers</code> function takes an array of objects as argumentReturn an array of objects containing only the students who have an average score above 90. The objects in the array should have the keys "name" and "average" which should contain the name and the average marks of the student if his average score is greater than 90.const students = [
{ name: "Ram", scores: [80, 95, 60] },
{ name: "Mohan", scores: [85, 70, 90] },
{ name: "Sai", scores: [60, 70, 80] },
{ name: "Hemang", scores: [95, 90, 94] },
];
const res = highPerformers(students);
console.log(res);
// [ { name: 'Hemang', average: 93 } ], I have written this Solution Code: function highPerformers(students) {
const studentAverages = students.map((student) => {
const sum = student.scores.reduce((acc, score) => acc + score);
return { name: student.name, average: sum / student.scores.length };
});
const res = studentAverages.filter((student) => student.average > 90);
return res;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc =new Scanner(System.in);
int T= sc.nextInt();
for(int i=0;i<T;i++){
int arrsize=sc.nextInt();
int max=0,secmax=0,thirdmax=0,j;
for(int k=0;k<arrsize;k++){
j=sc.nextInt();
if(j>max){
thirdmax=secmax;
secmax=max;
max=j;
}
else if(j>secmax){
thirdmax=secmax;
secmax=j;
}
else if(j>thirdmax){
thirdmax=j;
}
if(k%10000==0){
System.gc();
}
}
System.out.println(max+" "+secmax+" "+thirdmax+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: t=int(input())
while t>0:
t-=1
n=int(input())
l=list(map(int,input().strip().split()))
li=[0,0,0]
for i in l:
x=i
for j in range(0,3):
y=min(x,li[j])
li[j]=max(x,li[j])
x=y
print(li[0],end=" ")
print(li[1],end=" ")
print(li[2]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin>>t;
while(t--){
long long n;
cin>>n;
vector<long> a(n);
long ans[3]={0};
long x,y;
for(int i=0;i<n;i++){
cin>>a[i];
x=a[i];
for(int j=0;j<3;j++){
y=min(x,ans[j]);
ans[j]=max(x,ans[j]);
// cout<<ans[j]<<" ";
x=y;
}
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
if(ans[2]<ans[1]){
swap(ans[1],ans[2]);
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
cout<<ans[2]<<" "<<ans[1]<<" "<<ans[0]<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: function maxNumbers(arr,n) {
// write code here
// do not console.log the answer
// return the answer as an array of 3 numbers
return arr.sort((a,b)=>b-a).slice(0,3)
};
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: John has N candies. He wants to crush all of them. He feels that it would be boring to crush the candies randomly, so he found a method to crush them. He divides these candies into a minimum number of groups such that no group contains more than 3 candies. He crushes one candy from each group. If there are G groups in a single step, then the cost incurred in crushing a single candy for that step is G dollars. After candy from each group is crushed, he takes all the remaining candies and repeats the process until he has no candies left. He hasn't started crushing yet, but he wants to know how much the total cost would be incurred. Can you help him?
You have to answer Q-independent queries.The first line of input contains a single integer, Q denoting the number of queries.
Next, Q lines contain a single integer N denoting the number of candies John has.
<b>Constraints</b>
1 <= Q <= 5 * 10^4
1 <= N <= 10^9Print Q lines containing total cost incurred for each query.Sample Input 1:
1
4
Sample Output 1:
6
<b>Explanation:</b>
Query 1: First step John divides the candies into two groups of 3 and 1 candy respectively. Crushing one-one candy from both groups would cost him 2x2 = 4 dollars. He is now left with 2 candies. He divides it into one group. He crushes one candy for 1 dollar. Now, he is left with 1 candy. He crushes the last candy for 1 dollar. So, the total cost incurred is 4+1+1 = 6 dollars., I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int cost(int n){
if(n == 0) return 0;
int g = (n-1)/3 + 1;
return g*g + cost(n-g);
}
signed main() {
IOS;
clock_t start = clock();
int q; cin >> q;
while(q--){
int n;
cin >> n;
cout << cost(n) << endl;
}
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: John has N candies. He wants to crush all of them. He feels that it would be boring to crush the candies randomly, so he found a method to crush them. He divides these candies into a minimum number of groups such that no group contains more than 3 candies. He crushes one candy from each group. If there are G groups in a single step, then the cost incurred in crushing a single candy for that step is G dollars. After candy from each group is crushed, he takes all the remaining candies and repeats the process until he has no candies left. He hasn't started crushing yet, but he wants to know how much the total cost would be incurred. Can you help him?
You have to answer Q-independent queries.The first line of input contains a single integer, Q denoting the number of queries.
Next, Q lines contain a single integer N denoting the number of candies John has.
<b>Constraints</b>
1 <= Q <= 5 * 10^4
1 <= N <= 10^9Print Q lines containing total cost incurred for each query.Sample Input 1:
1
4
Sample Output 1:
6
<b>Explanation:</b>
Query 1: First step John divides the candies into two groups of 3 and 1 candy respectively. Crushing one-one candy from both groups would cost him 2x2 = 4 dollars. He is now left with 2 candies. He divides it into one group. He crushes one candy for 1 dollar. Now, he is left with 1 candy. He crushes the last candy for 1 dollar. So, the total cost incurred is 4+1+1 = 6 dollars., 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 q = Long.parseLong(br.readLine());
while(q-->0)
{
long N = Long.parseLong(br.readLine());
System.out.println(candyCrush(N,0,0));
}
}
static long candyCrush(long N, long cost,long group)
{
if(N==0)
{
return cost;
}
if(N%3==0)
{
group = N/3;
cost = cost + (group*group);
return candyCrush(N-group,cost,0);
}
else
{
group = (N/3)+1;
cost = cost + (group*group);
return candyCrush(N-group,cost,0);
}
}
}, 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: Given an array <b>A</b> of N integers, For each i (1 ≤ i ≤ N) your task is to find the value of x+y,
where x is the largest number less than i such that A[x]>A[i] and (A[x] is the element present at position x.)
y is the smallest number greater than i such that A[y]>A[i]
If there is no x < i such that A[x] > A[i], then take x=−1. Similarly, if there is no y>i such that A[y]>A[i], then take y=−1.First line consists of a single integer denoting N.
Second line consists of N space separated integers denoting the array A.
Constraints:
1 ≤ N ≤ 1000000
1 ≤ A[i] ≤ 10000000000Print N space separated integers, denoting x+y for each i(1 ≤ i ≤ N)Sample Input
5
5 4 1 3 2
Sample Output
-2 0 6 1 3
Explanation:-
For element 5:- x=-1(No element exist greater than 5 in left), y=-1 (No element exist greater than 5 in right)
For element 4:- x=1, y=-1
For element 1:- x=2, y=4
For element 3:- x=2, y=-1
For element 2:- x=4, y=-1
Sample Input
5
6 4 6 8 2
Sample Output
3 4 3 -2 3, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static int maxleft(int a[],int pos)
{
if(pos==1)
{
return -1;
}
else
{int max=a[pos];int s=-1;
for(int i=pos-1;i>=0;i--)
{
if(max<a[i])
{
s=i;
break;
}
}
if(s==pos)
{
return -1;
}
else
{
return s;
}
}
}
static int maxright(int a[],int pos,int n)
{
int max=a[pos];
int s=-1;
if(pos==n)
{
return -1;
}
else
{ for(int i=pos+1;i<=n;i++)
{
if(max<a[i])
{
s=i;
break;
}
}
if(s==pos)
{
return -1;
}
else
{
return s;
}
}
}
public static void main (String[] args) throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
int a[]=new int[n+1];
StringTokenizer st=new StringTokenizer(br.readLine()," ");
for(int i=1;i<n+1;i++)
{
a[i]=Integer.parseInt(st.nextToken());
}
for(int i=1;i<n+1;i++)
{
int maxl=maxleft(a,i);
int maxr=maxright(a,i,n);
System.out.print((maxl+maxr)+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
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