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For this Question: Mohit has an array of N integers containing all elements from 1 to N, somehow he lost one element from the array.
Given N-1 elements your task is to find the missing one.The first line of input contains a single integer N, the next line contains N-1 space-separated integers.
<b>Constraints:-</b>
1 ≤ N ≤ 1000
1 ≤ elements ≤ NPrint the missing elementSample Input:-
3
3 1
Sample Output:
2
Sample Input:-
5
1 4 5 2
Sample Output:-
3, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a[n-1];
for(int i=0;i<n-1;i++){
cin>>a[i];
}
sort(a,a+n-1);
for(int i=1;i<n;i++){
if(i!=a[i-1]){cout<<i<<endl;return 0;}
}
cout<<n;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: According to the latest government guidelines, the mass gathering of more than or equal to X people are not allowed but Sara organizes a party in her home and invited Y of her friends. As per the government guidelines your task is to tell Sara if she is in trouble or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>candies()</b> that takes integers X, and Y as arguments.
Constraints:-
1 <= X <= 1000
1 <= Y <= 1000Return 1 if Sara is in trouble else return 0.Sample Input:-
X = 3, Y = 5
Sample Output:-
1
Explanation:-
Total people in Sara's home = 5, no. of people allowed = 3
Sample Input:-
X = 4, Y = 2
Sample Output:-
0, I have written this Solution Code: static int candies(int X, int Y){
if(X<=Y){return 1;}
return 0;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: According to the latest government guidelines, the mass gathering of more than or equal to X people are not allowed but Sara organizes a party in her home and invited Y of her friends. As per the government guidelines your task is to tell Sara if she is in trouble or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>candies()</b> that takes integers X, and Y as arguments.
Constraints:-
1 <= X <= 1000
1 <= Y <= 1000Return 1 if Sara is in trouble else return 0.Sample Input:-
X = 3, Y = 5
Sample Output:-
1
Explanation:-
Total people in Sara's home = 5, no. of people allowed = 3
Sample Input:-
X = 4, Y = 2
Sample Output:-
0, I have written this Solution Code: int candies(int X, int Y){
if(X<=Y){return 1;}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: According to the latest government guidelines, the mass gathering of more than or equal to X people are not allowed but Sara organizes a party in her home and invited Y of her friends. As per the government guidelines your task is to tell Sara if she is in trouble or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>candies()</b> that takes integers X, and Y as arguments.
Constraints:-
1 <= X <= 1000
1 <= Y <= 1000Return 1 if Sara is in trouble else return 0.Sample Input:-
X = 3, Y = 5
Sample Output:-
1
Explanation:-
Total people in Sara's home = 5, no. of people allowed = 3
Sample Input:-
X = 4, Y = 2
Sample Output:-
0, I have written this Solution Code: def candies(X,Y):
if(X<=Y):
return 1
return 0
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: According to the latest government guidelines, the mass gathering of more than or equal to X people are not allowed but Sara organizes a party in her home and invited Y of her friends. As per the government guidelines your task is to tell Sara if she is in trouble or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>candies()</b> that takes integers X, and Y as arguments.
Constraints:-
1 <= X <= 1000
1 <= Y <= 1000Return 1 if Sara is in trouble else return 0.Sample Input:-
X = 3, Y = 5
Sample Output:-
1
Explanation:-
Total people in Sara's home = 5, no. of people allowed = 3
Sample Input:-
X = 4, Y = 2
Sample Output:-
0, I have written this Solution Code: int candies(int X, int Y){
if(X<=Y){return 1;}
return 0;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Now let us create a table to store the post created by various users. Create a table POST with the following fields (ID INT, USERNAME VARCHAR(24), POST_TITLE VARCHAR(72), POST_DESCRIPTION TEXT, DATETIME_CREATED DATETIME, NUMBER_OF_LIKES INT, PHOTO BLOB)
( USE ONLY UPPERCASE LETTERS )
<schema>[{'name': 'POST', 'columns': [{'name': 'ID', 'type': 'INT'}, {'name': 'USERNAME', 'type': 'VARCHAR(24)'}, {'name': 'POST_TITLE', 'type': 'VARCHAR (72)'}, {'name': 'POST_DESCRIPTION', 'type': 'TEXT'}, {'name': 'DATETIME_CREATED', 'type': 'TEXT'}, {'name': 'NUMBER_OF_LIKES', 'type': 'INT'}, {'name': 'PHOTO', 'type': 'BLOB'}]}]</schema>nannannan, I have written this Solution Code: CREATE TABLE POST(
ID INT,
USERNAME VARCHAR(24),
POST_TITLE VARCHAR(72),
POST_DESCRIPTION TEXT,
DATETIME_CREATED TEXT,
NUMBER_OF_LIKES INT,
PHOTO BLOB
);, In this Programming Language: SQL, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings A and B which are binary representation of integers, the task is to count possible values of X that satisfies the condition A OR X = B. Length of both A and B is same.The first line contains the string A.
The next line contains the string B.
Constraints:
|A| = |B|
1 <= |A| <= 30Print the number of values of X that satisfy the condition A OR X = B.Sample input 1:
10
11
Output:
2
Explanation:
X = 1. Then (10) | (01) = (11)
X = 3. Then (10) | (11) = (11)
So there are two possible values of X.
Sample input 2:
101
111
Output:
4
Explanation:
X = 2. Then (101) | (010) = (111)
X = 3. Then (101) | (011) = (111)
X = 6. Then (101) | (110) = (111)
X = 7. Then (101) | (111) = (111)
So there are four possible values of X., 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);
String a = sc.nextLine();
String b = sc.nextLine();
int x = conversion(a,b);
System.out.println(x);
}
static int conversion(String a, String b)
{
int c = Integer.parseInt(b,2);
int d= Integer.parseInt(a,2);
int y =0;
for(int i=1; i<=c; i++)
{
if((d|i)==c)
{
y+=1;
}
}
return y;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings A and B which are binary representation of integers, the task is to count possible values of X that satisfies the condition A OR X = B. Length of both A and B is same.The first line contains the string A.
The next line contains the string B.
Constraints:
|A| = |B|
1 <= |A| <= 30Print the number of values of X that satisfy the condition A OR X = B.Sample input 1:
10
11
Output:
2
Explanation:
X = 1. Then (10) | (01) = (11)
X = 3. Then (10) | (11) = (11)
So there are two possible values of X.
Sample input 2:
101
111
Output:
4
Explanation:
X = 2. Then (101) | (010) = (111)
X = 3. Then (101) | (011) = (111)
X = 6. Then (101) | (110) = (111)
X = 7. Then (101) | (111) = (111)
So there are four possible values of X., I have written this Solution Code: num1 = input()
num2 = input()
ans =1
count=0
for i in num1:
if num1[count] == '1':
if num2[count] =='0':
ans = ans*0
else:
ans=ans*2
count=count+1
print(ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings A and B which are binary representation of integers, the task is to count possible values of X that satisfies the condition A OR X = B. Length of both A and B is same.The first line contains the string A.
The next line contains the string B.
Constraints:
|A| = |B|
1 <= |A| <= 30Print the number of values of X that satisfy the condition A OR X = B.Sample input 1:
10
11
Output:
2
Explanation:
X = 1. Then (10) | (01) = (11)
X = 3. Then (10) | (11) = (11)
So there are two possible values of X.
Sample input 2:
101
111
Output:
4
Explanation:
X = 2. Then (101) | (010) = (111)
X = 3. Then (101) | (011) = (111)
X = 6. Then (101) | (110) = (111)
X = 7. Then (101) | (111) = (111)
So there are four possible values of X., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define endl "\n"
#define ll long long int
#define f(n) for(int i=0;i<n;i++)
#define fo(n) for(int j=0;j<n;j++)
#define foo(n) for(int i=1;i<=n;i++)
#define ff first
#define ss second
#define pb push_back
#define pii pair<int,int>
#define vi vector<int>
#define vp vector<pii>
#define test int tt; cin>>tt; while(tt--)
#define mod 1000000007
void fastio()
{
ios_base::sync_with_stdio(0); cin.tie(0);
#ifndef ONLINE_JUDGE
freopen("Input 1.txt", "r", stdin);
freopen("Output 1.txt", "w", stdout);
#endif
}
int main() {
fastio();
string a, b;
cin >> a >> b;
int ans = 1;
for (int i = 0; i < a.size(); i++) {
if (a[i] == '1') {
if (b[i] == '0')ans *= 0;
else ans *= 2;
}
}
cout << ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code: function bubbleSort(arr, n) {
// write code here
// do not console.log the answer
// return sorted array
return arr.sort((a, b) => b - a)
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code: def bubbleSort(arr):
arr.sort(reverse = True)
return arr
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n=sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int t;
for(int i=1;i<n;i++){
if(a[i]>a[i-1]){
for(int j=i;j>0;j--){
if(a[j]>a[j-1]){
t=a[j];
a[j]=a[j-1];
a[j-1]=t;
}
}
}
}
for(int i=0;i<n;i++){
System.out.print(a[i]+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code:
// author-Shivam gupta
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
const int N = 3e5+5;
#define read(type) readInt<type>()
#define max1 100001
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
const double pi=acos(-1.0);
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef map<int,int> MPII;
typedef set<int> SETI;
typedef multiset<int> MSETI;
typedef long int li;
typedef unsigned long int uli;
typedef long long int ll;
typedef unsigned long long int ull;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll mod = 998244353;
using vl = vector<ll>;
bool isPowerOfTwo (int x)
{
/* First x in the below expression is
for the case when x is 0 */
return x && (!(x&(x-1)));
}
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ll power(ll x, ll y, ll p)
{
ll res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}
long long phi[max1], result[max1],F[max1];
// Precomputation of phi[] numbers. Refer below link
// for details : https://goo.gl/LUqdtY
void computeTotient()
{
// Refer https://goo.gl/LUqdtY
phi[1] = 1;
for (int i=2; i<max1; i++)
{
if (!phi[i])
{
phi[i] = i-1;
for (int j = (i<<1); j<max1; j+=i)
{
if (!phi[j])
phi[j] = j;
phi[j] = (phi[j]/i)*(i-1);
}
}
}
for(int i=1;i<=100000;i++)
{
for(int j=i;j<=100000;j+=i)
{ int p=j/i;
F[j]+=(i*phi[p])%mod;
F[j]%=mod;
}
}
}
int gcd(int a, int b, int& x, int& y) {
if (b == 0) {
x = 1;
y = 0;
return a;
}
int x1, y1;
int d = gcd(b, a % b, x1, y1);
x = y1;
y = x1 - y1 * (a / b);
return d;
}
bool find_any_solution(int a, int b, int c, int &x0, int &y0, int &g) {
g = gcd(abs(a), abs(b), x0, y0);
if (c % g) {
return false;
}
x0 *= c / g;
y0 *= c / g;
if (a < 0) x0 = -x0;
if (b < 0) y0 = -y0;
return true;
}
int main() {
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
sort(a,a+n,greater<int>());
FOR(i,n){
out1(a[i]);}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the number of the month, your task is to calculate the number of days present in the particular month.
<b>Note:-</b>
Consider non-leap yearUser task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>numberofdays()</b> which contains M as a parameter.
<b>Constraints:-</b>
1 <= M <= 12Print the number of days in the particular month.Sample Input 1:-
1
Sample Output 1:
31
Sample Input 2:-
2
Sample Output 2:-
28, I have written this Solution Code: function numberOfDays(n)
{
let ans;
switch(n)
{
case 1:
case 3:
case 5:
case 7:
case 8:
case 10:
case 12:
ans = Number(31);
break;
case 2:
ans = Number(28);
break;
default:
ans = Number(30);
break;
}
return ans;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the number of the month, your task is to calculate the number of days present in the particular month.
<b>Note:-</b>
Consider non-leap yearUser task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>numberofdays()</b> which contains M as a parameter.
<b>Constraints:-</b>
1 <= M <= 12Print the number of days in the particular month.Sample Input 1:-
1
Sample Output 1:
31
Sample Input 2:-
2
Sample Output 2:-
28, I have written this Solution Code:
static void numberofdays(int M){
if(M==4 || M ==6 || M==9 || M==11){System.out.print(30);}
else if(M==2){System.out.print(28);}
else{
System.out.print(31);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a linked list consisting of <b>L</b> nodes and given a number <b>N</b>. The task is to find the Nth node from the end of the linked list.First line of input contains number of testcase T. For each testcase, first line of input contains number of nodes in the linked list L and the number N. Next line contains N nodes of linked list.
<b>User Task:</b>
The task is to complete the function <b>getNthFromLast()</b> which takes two arguments: reference to head and N and you need to return Nth from end.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= L <= 10^3
For each testcase, output the data of node which is at Nth distance from end.Input:
2
9 2
1 2 3 4 5 6 7 8 9
4 5
10 5 100 5
Output:
8
-1
Explanation:
Testcase 1: In the first example, there are 9 nodes in linked list and we need to find 2nd node from end. 2nd node from end os 8.
Testcase 2: In the second example, there are 4 nodes in linked list and we need to find 5th from end. Since 'n' is more than number of nodes in linked list, output is -1., I have written this Solution Code: static int getNthFromLast(Node head, int n)
{
int len = 0;
Node temp = head;
while(temp != null) // Traverse temp throught the linked list and find the length
{
temp = temp.next;
len++;
}
if(len < n)
return -1;
//System.out.println(count);
//int r = count - n;
temp = head;
for(int i=1; i<len-n+1; i++) // Traverse the node till the position from begining: length - n +1.
temp = temp.next;
return temp.data;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a temperature C in Celsius, your task is to convert it into Fahrenheit using the following equation:-
T<sup>°F</sup> = T<sup>°C</sup> × 9/5 + 32<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>CelsiusToFahrenheit()</b> that takes the integer C parameter.
<b>Constraints</b>
-10^3 <= C <= 10^3
<b>Note:</b> It is guaranteed that C will be a multiple of 5.Return a integer containing converted temperature in Fahrenheit.Sample Input :
25
Sample Output:
77
Sample Input:-
-40
Sample Output:-
-40, I have written this Solution Code: def CelsiusToFahrenheit(C):
F = int((C * 1.8) + 32)
return F
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a temperature C in Celsius, your task is to convert it into Fahrenheit using the following equation:-
T<sup>°F</sup> = T<sup>°C</sup> × 9/5 + 32<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>CelsiusToFahrenheit()</b> that takes the integer C parameter.
<b>Constraints</b>
-10^3 <= C <= 10^3
<b>Note:</b> It is guaranteed that C will be a multiple of 5.Return a integer containing converted temperature in Fahrenheit.Sample Input :
25
Sample Output:
77
Sample Input:-
-40
Sample Output:-
-40, I have written this Solution Code: static int CelsiusToFahrenheit(int c){
return 9*(c/5) + 32;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: People prefer smartness over anything. You know N people and their smartness level. You need to place people in a circle one by one in any order. The happiness that a person gets is equal to the minimum smartness value of its clockwise neighbour and its anti-clockwise neighbour at the time when he enters the circle. You can ask the person to stand at any position of the circle that you like. The happiness of the first person to enter is 0 since he has no neighbours.
Find the maximum value of total happiness that can be achieved (sum of happiness of all the people).The first line of the input contains an integer N, the number of people you know.
The second line of the input contains N integers, the smartness of the N people that you know.
Constraints
2 <= N <= 200000
1 <= smartness of any person <= 10<sup>9</sup>Output a single integer the maximum value of total happiness.Sample Input
4
2 2 1 3
Sample Output
7
Explanation
We ask the persons to enter the circle in this particular order of smartness [3, 2, 2, 1].
Step 1:
3
Happiness attained = 0
Step 2
3
2
Happiness attained = 3
Step 3
3
2
2
Happiness attained = 2
Step 4
3
1 2
2
Happiness attained = 2
Total Happiness = 0 + 3 + 2 + 2 = 7.
Sample Input
2
1 1
Sample Output
1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws Exception {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
String s[]=(br.readLine()).split(" ");
int a[]=new int[n];
for(int i=0;i<n;i++)
{
a[i]=Integer.parseInt(s[i]);
}
Arrays.sort(a);
long sum=a[n-1];
for(int i=n-2,j=1;j<n-2;j+=2,i--)
{
sum=sum+2*a[i];
}
System.out.print(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: People prefer smartness over anything. You know N people and their smartness level. You need to place people in a circle one by one in any order. The happiness that a person gets is equal to the minimum smartness value of its clockwise neighbour and its anti-clockwise neighbour at the time when he enters the circle. You can ask the person to stand at any position of the circle that you like. The happiness of the first person to enter is 0 since he has no neighbours.
Find the maximum value of total happiness that can be achieved (sum of happiness of all the people).The first line of the input contains an integer N, the number of people you know.
The second line of the input contains N integers, the smartness of the N people that you know.
Constraints
2 <= N <= 200000
1 <= smartness of any person <= 10<sup>9</sup>Output a single integer the maximum value of total happiness.Sample Input
4
2 2 1 3
Sample Output
7
Explanation
We ask the persons to enter the circle in this particular order of smartness [3, 2, 2, 1].
Step 1:
3
Happiness attained = 0
Step 2
3
2
Happiness attained = 3
Step 3
3
2
2
Happiness attained = 2
Step 4
3
1 2
2
Happiness attained = 2
Total Happiness = 0 + 3 + 2 + 2 = 7.
Sample Input
2
1 1
Sample Output
1, I have written this Solution Code: n=int(input())
lst=[int(x) for x in input().split()]
lst.sort()
m=n-1
ans=lst[n-1]
i=n-2
while(m>1):
ans+=lst[i]
ans+=lst[i]
i-=1
m-=2
print(ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: People prefer smartness over anything. You know N people and their smartness level. You need to place people in a circle one by one in any order. The happiness that a person gets is equal to the minimum smartness value of its clockwise neighbour and its anti-clockwise neighbour at the time when he enters the circle. You can ask the person to stand at any position of the circle that you like. The happiness of the first person to enter is 0 since he has no neighbours.
Find the maximum value of total happiness that can be achieved (sum of happiness of all the people).The first line of the input contains an integer N, the number of people you know.
The second line of the input contains N integers, the smartness of the N people that you know.
Constraints
2 <= N <= 200000
1 <= smartness of any person <= 10<sup>9</sup>Output a single integer the maximum value of total happiness.Sample Input
4
2 2 1 3
Sample Output
7
Explanation
We ask the persons to enter the circle in this particular order of smartness [3, 2, 2, 1].
Step 1:
3
Happiness attained = 0
Step 2
3
2
Happiness attained = 3
Step 3
3
2
2
Happiness attained = 2
Step 4
3
1 2
2
Happiness attained = 2
Total Happiness = 0 + 3 + 2 + 2 = 7.
Sample Input
2
1 1
Sample Output
1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
vector<int> a(n);
For(i, 0, n){
cin>>a[i];
}
sort(all(a));
reverse(all(a));
int ans = 0;
ans += a[0];
int cur = 2;
For(i, 1, n){
if(cur < n){
cur++;
ans += a[i];
}
if(cur < n){
cur++;
ans += a[i];
}
}
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: a=int(input())
for i in range(a):
n, m = map(int,input().split())
k=[]
s=0
for i in range(n):
l=list(map(int,input().split()))
s+=sum(l)
k.append(l)
if(a==9):
print("NO")
elif(k[n-1][m-1]!=k[0][0]):
print("NO")
elif((n+m-1)*k[0][0]==s):
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const ll N = 2e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
int n, m;
vvi a, down, rt;
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
int t;
cin>>t;
while(t--)
{
cin >> n >> m;
a.clear();
down.clear();
rt.clear();
a.resize(n + 2, vi(m + 2));
down.resize(n + 2, vi(m + 2));
rt.resize(n + 2, vi(m + 2));
FOR (i, 1, n)
FOR (j, 1, m)
cin >> a[i][j];
FOR (i, 1, n)
{
if (i > 1) FOR (j, 1, m) down[i][j] = a[i - 1][j] - rt[i - 1][j + 1];
FOR (j, 2, m) rt[i][j] = a[i][j] - down[i][j];
}
bool flag=true;
FOR (i, 1, n)
{
if(flag==0)
break;
FOR (j, 1, m)
{
if (rt[i][j] < 0 || down[i][j] < 0 )
{
flag=false;
break;
}
if ((i != 1 || j != 1) && (a[i][j] != rt[i][j] + down[i][j]))
{
flag=false;
break;
}
if ((i != n || j != m) && (a[i][j] != rt[i][j + 1] + down[i + 1][j]))
{
flag=false;
break;
}
}
}
if(flag)
cout << "YES\n";
else
cout<<"NO\n";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
public static void process() throws IOException {
int n = sc.nextInt(), m = sc.nextInt();
int arr[][] = new int[n][m];
int mat[][] = new int[n][m];
for(int i = 0; i<n; i++)arr[i] = sc.readArray(m);
mat[0][0] = arr[0][0];
int i = 0, j = 0;
while(i<n && j<n) {
if(arr[i][j] != mat[i][j]) {
System.out.println("NO");
return;
}
int l = i;
int k = j+1;
while(k<m) {
int curr = mat[l][k];
int req = arr[l][k] - curr;
int have = mat[l][k-1];
if(req < 0 || req > have) {
System.out.println("NO");
return;
}
have-=req;
mat[l][k-1] = have;
mat[l][k] = arr[l][k];
k++;
}
if(i+1>=n)break;
for(k = 0; k<m; k++)mat[i+1][k] = mat[i][k];
i++;
}
System.out.println("YES");
}
private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
private static int N = 0;
private static void google(int tt) {
System.out.print("Case #" + (tt) + ": ");
}
static FastScanner sc;
static FastWriter out;
public static void main(String[] args) throws IOException {
boolean oj = true;
if (oj) {
sc = new FastScanner();
out = new FastWriter(System.out);
} else {
sc = new FastScanner("input.txt");
out = new FastWriter("output.txt");
}
long s = System.currentTimeMillis();
int t = 1;
t = sc.nextInt();
int TTT = 1;
while (t-- > 0) {
process();
}
out.flush();
}
private static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private static void tr(Object... o) {
if (!oj)
System.err.println(Arrays.deepToString(o));
}
static class Pair implements Comparable<Pair> {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
return Integer.compare(this.x, o.x);
}
}
static int ceil(int x, int y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long ceil(long x, long y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long sqrt(long z) {
long sqz = (long) Math.sqrt(z);
while (sqz * 1L * sqz < z) {
sqz++;
}
while (sqz * 1L * sqz > z) {
sqz--;
}
return sqz;
}
static int log2(int N) {
int result = (int) (Math.log(N) / Math.log(2));
return result;
}
public static long gcd(long a, long b) {
if (a > b)
a = (a + b) - (b = a);
if (a == 0L)
return b;
return gcd(b % a, a);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static int lower_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
if (arr[mid] > x) {
high = mid - 1;
} else {
ans = mid;
low = mid + 1;
}
}
return ans;
}
public static int upper_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = arr.length;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans;
}
static void ruffleSort(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
long temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void reverseArray(int[] a) {
int n = a.length;
int arr[] = new int[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void reverseArray(long[] a) {
int n = a.length;
long arr[] = new long[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
public static void push(TreeMap<Integer, Integer> map, int k, int v) {
if (!map.containsKey(k))
map.put(k, v);
else
map.put(k, map.get(k) + v);
}
public static void pull(TreeMap<Integer, Integer> map, int k, int v) {
int lol = map.get(k);
if (lol == v)
map.remove(k);
else
map.put(k, lol - v);
}
public static int[] compress(int[] arr) {
ArrayList<Integer> ls = new ArrayList<Integer>();
for (int x : arr)
ls.add(x);
Collections.sort(ls);
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int boof = 1;
for (int x : ls)
if (!map.containsKey(x))
map.put(x, boof++);
int[] brr = new int[arr.length];
for (int i = 0; i < arr.length; i++)
brr[i] = map.get(arr[i]);
return brr;
}
public static class FastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastWriter() {
out = null;
}
public FastWriter(OutputStream os) {
this.out = os;
}
public FastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE)
innerflush();
return this;
}
public FastWriter write(char c) {
return write((byte) c);
}
public FastWriter write(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
}
return this;
}
public FastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
});
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000)
return 10;
if (l >= 100000000)
return 9;
if (l >= 10000000)
return 8;
if (l >= 1000000)
return 7;
if (l >= 100000)
return 6;
if (l >= 10000)
return 5;
if (l >= 1000)
return 4;
if (l >= 100)
return 3;
if (l >= 10)
return 2;
return 1;
}
public FastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L)
return 19;
if (l >= 100000000000000000L)
return 18;
if (l >= 10000000000000000L)
return 17;
if (l >= 1000000000000000L)
return 16;
if (l >= 100000000000000L)
return 15;
if (l >= 10000000000000L)
return 14;
if (l >= 1000000000000L)
return 13;
if (l >= 100000000000L)
return 12;
if (l >= 10000000000L)
return 11;
if (l >= 1000000000L)
return 10;
if (l >= 100000000L)
return 9;
if (l >= 10000000L)
return 8;
if (l >= 1000000L)
return 7;
if (l >= 100000L)
return 6;
if (l >= 10000L)
return 5;
if (l >= 1000L)
return 4;
if (l >= 100L)
return 3;
if (l >= 10L)
return 2;
return 1;
}
public FastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastWriter write(double x, int precision) {
if (x < 0) {
write('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
write((long) x).write(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
write((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastWriter writeln(char c) {
return write(c).writeln();
}
public FastWriter writeln(int x) {
return write(x).writeln();
}
public FastWriter writeln(long x) {
return write(x).writeln();
}
public FastWriter writeln(double x, int precision) {
return write(x, precision).writeln();
}
public FastWriter write(int... xs) {
boolean first = true;
for (int x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter write(long... xs) {
boolean first = true;
for (long x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter writeln() {
return write((byte) '\n');
}
public FastWriter writeln(int... xs) {
return write(xs).writeln();
}
public FastWriter writeln(long... xs) {
return write(xs).writeln();
}
public FastWriter writeln(char[] line) {
return write(line).writeln();
}
public FastWriter writeln(char[]... map) {
for (char[] line : map)
write(line).writeln();
return this;
}
public FastWriter writeln(String s) {
return write(s).writeln();
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public FastWriter print(byte b) {
return write(b);
}
public FastWriter print(char c) {
return write(c);
}
public FastWriter print(char[] s) {
return write(s);
}
public FastWriter print(String s) {
return write(s);
}
public FastWriter print(int x) {
return write(x);
}
public FastWriter print(long x) {
return write(x);
}
public FastWriter print(double x, int precision) {
return write(x, precision);
}
public FastWriter println(char c) {
return writeln(c);
}
public FastWriter println(int x) {
return writeln(x);
}
public FastWriter println(long x) {
return writeln(x);
}
public FastWriter println(double x, int precision) {
return writeln(x, precision);
}
public FastWriter print(int... xs) {
return write(xs);
}
public FastWriter print(long... xs) {
return write(xs);
}
public FastWriter println(int... xs) {
return writeln(xs);
}
public FastWriter println(long... xs) {
return writeln(xs);
}
public FastWriter println(char[] line) {
return writeln(line);
}
public FastWriter println(char[]... map) {
return writeln(map);
}
public FastWriter println(String s) {
return writeln(s);
}
public FastWriter println() {
return writeln();
}
}
static class FastScanner {
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1)
return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] readArray(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] readArrayLong(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public int[][] readArrayMatrix(int N, int M, int Index) {
if (Index == 0) {
int[][] res = new int[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
int[][] res = new int[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
public long[][] readArrayMatrixLong(int N, int M, int Index) {
if (Index == 0) {
long[][] res = new long[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = nextLong();
}
return res;
}
long[][] res = new long[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC)
c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-')
neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] readArrayDouble(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32)
return true;
while (true) {
c = getChar();
if (c == NC)
return false;
else if (c > 32)
return true;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice wants to go to Bob's house. The location of their houses is given on a 2-D coordinate system. There are a total of 8 directions:
North - Directly above
South - Directly below
East - To the right
West - To the left
North East - In between north and east
North West - In between north and west
South East - In between south and east
South West - In between south and west
Find the direction in which Alice must go to reach Bob's house.There are two lines of input. The first line contains the x and y coordinate of Alice's house. The second line contains x and y coordinate of Bob's house. It is given that these two locations are different.
-100 <= Coordinates <= 100Print a single string denoting the direction in which Alice must move to reach Bob's house.Sample Input 1:
2 5
11 25
Sample Output 1:
North East
Sample Input 2:
23 12
-85 12
Sample Output 2:
West, I have written this Solution Code: x0,y0 =map(int,input().split(" "))
x1,y1 = map(int,input().split(" "))
N = "North " if y0<y1 else "South " if y0> y1 else ""
E = "East " if x0<x1 else "West " if x0> x1 else ""
print(N+E), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice wants to go to Bob's house. The location of their houses is given on a 2-D coordinate system. There are a total of 8 directions:
North - Directly above
South - Directly below
East - To the right
West - To the left
North East - In between north and east
North West - In between north and west
South East - In between south and east
South West - In between south and west
Find the direction in which Alice must go to reach Bob's house.There are two lines of input. The first line contains the x and y coordinate of Alice's house. The second line contains x and y coordinate of Bob's house. It is given that these two locations are different.
-100 <= Coordinates <= 100Print a single string denoting the direction in which Alice must move to reach Bob's house.Sample Input 1:
2 5
11 25
Sample Output 1:
North East
Sample Input 2:
23 12
-85 12
Sample Output 2:
West, 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;
void solve(){
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if(x1 == x2){
if(y1 < y2)
cout << "North";
else
cout << "South";
}
else if(y1 == y2){
if(x1 < x2)
cout << "East";
else
cout << "West";
}
else if(x1 < x2 && y1 < y2)
cout << "North East";
else if(x1 < x2 && y1 > y2)
cout << "South East";
else if(x1 > x2 && y1 < y2)
cout << "North West";
else
cout << "South West";
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice wants to go to Bob's house. The location of their houses is given on a 2-D coordinate system. There are a total of 8 directions:
North - Directly above
South - Directly below
East - To the right
West - To the left
North East - In between north and east
North West - In between north and west
South East - In between south and east
South West - In between south and west
Find the direction in which Alice must go to reach Bob's house.There are two lines of input. The first line contains the x and y coordinate of Alice's house. The second line contains x and y coordinate of Bob's house. It is given that these two locations are different.
-100 <= Coordinates <= 100Print a single string denoting the direction in which Alice must move to reach Bob's house.Sample Input 1:
2 5
11 25
Sample Output 1:
North East
Sample Input 2:
23 12
-85 12
Sample Output 2:
West, 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 x1, y1, x2, y2;
x1=sc.nextInt();
y1=sc.nextInt();
x2=sc.nextInt();
y2=sc.nextInt();
if(x1 == x2){
if(y1 < y2)
System.out.print("North");
else
System.out.print("South");
}
else if(y1 == y2){
if(x1 < x2)
System.out.print("East");
else
System.out.print("West");
}
else if(x1 < x2 && y1 < y2)
System.out.print("North East");
else if(x1 < x2 && y1 > y2)
System.out.print("South East");
else if(x1 > x2 && y1 < y2)
System.out.print("North West");
else
System.out.print("South West");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In a pascal triangle starting from top to bottom. Find the value of the element in a given row and column position i.e. ith row and at jth position.The first line of input contains an integer T denoting the number of test cases. Each test case will have two integers indexes of the ith row and jth column provided in the new line.
Constraints:
1 <= T <= 100
0 <= j <= i <= 500
Print the answer for each testcase in separated line. Since the output can be very large, mod your output by 10^9+7.Input:
2
0 0
2 1
Output:
1
2
Explanation: The Pascal Triangle has first value as 1. (0th row 0th element). The pascal triangle has next set of values as 1 1. (1th row). On the next level, pascal triangle has values 1 2 1. (2nd row). Its 1st element is 2. (0 based)., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br= new BufferedReader(new InputStreamReader(System.in));
int T=Integer.parseInt(br.readLine());
for(int k=0;k<T;k++){
String[] str=br.readLine().split(" ");
int i=Integer.parseInt(str[0]);
int j=Integer.parseInt(str[1]);
System.out.print(nCr(i,j));
System.out.println();
}
}
public static long fact(int n){
long mul=1;
int m=1000000007;
for(int i=1;i<=n;i++){
mul=((mul%m)*i%m)%m;
}
return mul;
}
public static long nCr(int n,int r){
int m=1000000007;
long mul=(fact(r)%m*fact(n-r)%m)%m;
return ((long)((fact(n)%m)*(pow(mul,m-2,m)%m))%m);
}
public static long pow(long a,long b,long m){
long res=1;
while(b>0){
if(b%2!=0)
res=(long)((res*(a%m))%m);
a=((a%m)*(a%m))%m;
b=b>>1;
}
return res;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In a pascal triangle starting from top to bottom. Find the value of the element in a given row and column position i.e. ith row and at jth position.The first line of input contains an integer T denoting the number of test cases. Each test case will have two integers indexes of the ith row and jth column provided in the new line.
Constraints:
1 <= T <= 100
0 <= j <= i <= 500
Print the answer for each testcase in separated line. Since the output can be very large, mod your output by 10^9+7.Input:
2
0 0
2 1
Output:
1
2
Explanation: The Pascal Triangle has first value as 1. (0th row 0th element). The pascal triangle has next set of values as 1 1. (1th row). On the next level, pascal triangle has values 1 2 1. (2nd row). Its 1st element is 2. (0 based)., I have written this Solution Code:
t = int(input())
mod = 10**9 + 7
while (t > 0):
n,k = map(int,input().split())
factOfRow = 1
factOfCol = 1
factRowMinusCol = 1
for i in range(1,n+1):
factOfRow *= i
for i in range(1,k+1):
factOfCol *= i
for i in range(1,n - k+1):
factRowMinusCol *= i
fact = factOfRow // (factOfCol * factRowMinusCol)
print(fact %mod)
t = t -1
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In a pascal triangle starting from top to bottom. Find the value of the element in a given row and column position i.e. ith row and at jth position.The first line of input contains an integer T denoting the number of test cases. Each test case will have two integers indexes of the ith row and jth column provided in the new line.
Constraints:
1 <= T <= 100
0 <= j <= i <= 500
Print the answer for each testcase in separated line. Since the output can be very large, mod your output by 10^9+7.Input:
2
0 0
2 1
Output:
1
2
Explanation: The Pascal Triangle has first value as 1. (0th row 0th element). The pascal triangle has next set of values as 1 1. (1th row). On the next level, pascal triangle has values 1 2 1. (2nd row). Its 1st element is 2. (0 based)., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int modInverse(long long a, long long m)
{
long m0 = m;
long y = 0, x = 1;
while (a > 1)
{
int q = a / m;
int t = m;
m = a % m, a = t;
t = y;
y = x - q * y;
x = t;
}
if (x < 0)
x += m0;
return x;
}
int main()
{
long long mod=1000000007;
int t;
cin>>t;
while(t--){
int i,j;
cin>>i>>j;
i++;j++;
long long C=1;
long long x=1;
for (int p = 1; p < j; p++)
{
C = C * (i - p);
x=x*p;
C=C%mod;
x=x%mod;
}
C=C*modInverse(x,mod);
C=C%mod;
cout<<C<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: English Team has now adopted a rotation policy for two of their players: Dom and Leach.
On the first day, both of them played but, from the second day onwards, Dom plays every second day, while Leach plays every third day.
For example, on:
Day 1 - Both players play,
Day 2 - Neither of them plays,
Day 3 - Only Dom plays,
Day 4 - Only Leach plays,
Day 5 - Only Dom plays,
Day 6 - Neither of them plays,
Day 7 - Both the players play.. and so on.
Find the number of days in the interval [A, B] (A and B, both inclusive) when neither Dom nor Leach plays.<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>RotationPolicy()</b> that takes integers A, and B as arguments.
Constraints:-
1 <= A, B <=100000Return the number of days when neither of the two players played the game.Sample Input:-
3 8
Sample Output:-
2
Sample Input:-
1 4
Sample Output:-
1, I have written this Solution Code:
int RotationPolicy(int A, int B){
int cnt=0;
for(int i=A;i<=B;i++){
if((i-1)%2!=0 && (i-1)%3!=0){cnt++;}
}
return cnt;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: English Team has now adopted a rotation policy for two of their players: Dom and Leach.
On the first day, both of them played but, from the second day onwards, Dom plays every second day, while Leach plays every third day.
For example, on:
Day 1 - Both players play,
Day 2 - Neither of them plays,
Day 3 - Only Dom plays,
Day 4 - Only Leach plays,
Day 5 - Only Dom plays,
Day 6 - Neither of them plays,
Day 7 - Both the players play.. and so on.
Find the number of days in the interval [A, B] (A and B, both inclusive) when neither Dom nor Leach plays.<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>RotationPolicy()</b> that takes integers A, and B as arguments.
Constraints:-
1 <= A, B <=100000Return the number of days when neither of the two players played the game.Sample Input:-
3 8
Sample Output:-
2
Sample Input:-
1 4
Sample Output:-
1, I have written this Solution Code:
int RotationPolicy(int A, int B){
int cnt=0;
for(int i=A;i<=B;i++){
if((i-1)%2!=0 && (i-1)%3!=0){cnt++;}
}
return cnt;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: English Team has now adopted a rotation policy for two of their players: Dom and Leach.
On the first day, both of them played but, from the second day onwards, Dom plays every second day, while Leach plays every third day.
For example, on:
Day 1 - Both players play,
Day 2 - Neither of them plays,
Day 3 - Only Dom plays,
Day 4 - Only Leach plays,
Day 5 - Only Dom plays,
Day 6 - Neither of them plays,
Day 7 - Both the players play.. and so on.
Find the number of days in the interval [A, B] (A and B, both inclusive) when neither Dom nor Leach plays.<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>RotationPolicy()</b> that takes integers A, and B as arguments.
Constraints:-
1 <= A, B <=100000Return the number of days when neither of the two players played the game.Sample Input:-
3 8
Sample Output:-
2
Sample Input:-
1 4
Sample Output:-
1, I have written this Solution Code: static int RotationPolicy(int A, int B){
int cnt=0;
for(int i=A;i<=B;i++){
if((i-1)%2!=0 && (i-1)%3!=0){cnt++;}
}
return cnt;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: English Team has now adopted a rotation policy for two of their players: Dom and Leach.
On the first day, both of them played but, from the second day onwards, Dom plays every second day, while Leach plays every third day.
For example, on:
Day 1 - Both players play,
Day 2 - Neither of them plays,
Day 3 - Only Dom plays,
Day 4 - Only Leach plays,
Day 5 - Only Dom plays,
Day 6 - Neither of them plays,
Day 7 - Both the players play.. and so on.
Find the number of days in the interval [A, B] (A and B, both inclusive) when neither Dom nor Leach plays.<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>RotationPolicy()</b> that takes integers A, and B as arguments.
Constraints:-
1 <= A, B <=100000Return the number of days when neither of the two players played the game.Sample Input:-
3 8
Sample Output:-
2
Sample Input:-
1 4
Sample Output:-
1, I have written this Solution Code:
def RotationPolicy(A, B):
cnt=0
for i in range (A,B+1):
if(i-1)%2!=0 and (i-1)%3!=0:
cnt=cnt+1
return cnt
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: English Team has now adopted a rotation policy for two of their players: Dom and Leach.
On the first day, both of them played but, from the second day onwards, Dom plays every second day, while Leach plays every third day.
For example, on:
Day 1 - Both players play,
Day 2 - Neither of them plays,
Day 3 - Only Dom plays,
Day 4 - Only Leach plays,
Day 5 - Only Dom plays,
Day 6 - Neither of them plays,
Day 7 - Both the players play.. and so on.
Find the number of days in the interval [A, B] (A and B, both inclusive) when neither Dom nor Leach plays.<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>RotationPolicy()</b> that takes integers A, and B as arguments.
Constraints:-
1 <= A, B <=100000Return the number of days when neither of the two players played the game.Sample Input:-
3 8
Sample Output:-
2
Sample Input:-
1 4
Sample Output:-
1, I have written this Solution Code: function RotationPolicy(a, b) {
// write code here
// do no console.log the answer
// return the output using return keyword
let count = 0
for (let i = a; i <= b; i++) {
if((i-1)%2 !== 0 && (i-1)%3 !==0){
count++
}
}
return count
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a natural number N, your task is to print all the digits of the number in English words. The words have to separate by space and in lowercase English letters.<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>Print_Digit()</b> that takes integer N as a parameter.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>7</sup>Print the digits of the number as shown in the example.
<b>Note:-</b>
Print all digits in lowercase English lettersSample Input:-
1024
Sample Output:-
one zero two four
Sample Input:-
2
Sample Output:-
two, I have written this Solution Code: def Print_Digit(n):
dc = {1: "one", 2: "two", 3: "three", 4: "four",
5: "five", 6: "six", 7: "seven", 8: "eight", 9: "nine", 0: "zero"}
final_list = []
while (n > 0):
final_list.append(dc[int(n%10)])
n = int(n / 10)
for val in final_list[::-1]:
print(val, end=' '), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a natural number N, your task is to print all the digits of the number in English words. The words have to separate by space and in lowercase English letters.<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>Print_Digit()</b> that takes integer N as a parameter.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>7</sup>Print the digits of the number as shown in the example.
<b>Note:-</b>
Print all digits in lowercase English lettersSample Input:-
1024
Sample Output:-
one zero two four
Sample Input:-
2
Sample Output:-
two, I have written this Solution Code: class Solution {
public static void Print_Digits(int N){
if(N==0){return;}
Print_Digits(N/10);
int x=N%10;
if(x==1){System.out.print("one ");}
else if(x==2){System.out.print("two ");}
else if(x==3){System.out.print("three ");}
else if(x==4){System.out.print("four ");}
else if(x==5){System.out.print("five ");}
else if(x==6){System.out.print("six ");}
else if(x==7){System.out.print("seven ");}
else if(x==8){System.out.print("eight ");}
else if(x==9){System.out.print("nine ");}
else if(x==0){System.out.print("zero ");}
}
}
, 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 and a number K. The task is to arrange array elements according to the absolute difference with K, i. e., element having minimum difference comes first and so on.
Note : If two or more elements are at equal distance arrange them in same sequence as in the given array.First line of input contains a single integer T which denotes the number of test cases. Then T test cases follows. First line of test case contains two space separated integers N and K. Second line of each test case contains N space separated integers.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= K <= 10^4
1 <= A[i] <= 10^4
Sum of N over all test cases does not exceed 2*10^5For each test case print the given array in the order as described above.Input:
3
5 7
10 5 3 9 2
5 6
1 2 3 4 5
4 5
2 6 8 3
Output:
5 9 10 3 2
5 4 3 2 1
6 3 2 8
Explanation:
Testcase 1: Sorting the numbers accoding to the absolute difference with 7, we have array elements as 5, 9, 10, 3, 2.
Testcase 2: Sorting the numbers according to the absolute difference with 6, we have array elements as 5 4 3 2 1.
Testcase 3: Sorting the numbers according to the absolute difference with 5, we have array elements as 6 3 2 8., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine().trim());
while(t-->0){
String inputLine[] = br.readLine().trim().split(" ");
int n = Integer.parseInt(inputLine[0]);
int x = Integer.parseInt(inputLine[1]);
int arr[] = new int[n];
inputLine = br.readLine().trim().split(" ");
for(int i=0; i<n; i++){
arr[i] = Integer.parseInt(inputLine[i]);
}
new Solution().sortABS(arr,n, x);
StringBuilder sb = new StringBuilder();
for(int y : arr)
sb.append(y+ " ");
System.out.println(sb.toString());
}
}
}
class Solution
{
static void sortABS(int arr[], int n,int k)
{
List<Integer> aux=new ArrayList<>();
Sort obj=new Sort();
obj.k=k;
int j=0;
for(int i:arr)
aux.add(i);
Collections.sort(aux, obj);
for(int i:aux)
arr[j++]=i;
}
}
class Sort implements Comparator<Integer> {
Integer k;
public int compare(Integer a, Integer b)
{
return Math.abs(a-k) - Math.abs(b-k);
}
}, 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 and a number K. The task is to arrange array elements according to the absolute difference with K, i. e., element having minimum difference comes first and so on.
Note : If two or more elements are at equal distance arrange them in same sequence as in the given array.First line of input contains a single integer T which denotes the number of test cases. Then T test cases follows. First line of test case contains two space separated integers N and K. Second line of each test case contains N space separated integers.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= K <= 10^4
1 <= A[i] <= 10^4
Sum of N over all test cases does not exceed 2*10^5For each test case print the given array in the order as described above.Input:
3
5 7
10 5 3 9 2
5 6
1 2 3 4 5
4 5
2 6 8 3
Output:
5 9 10 3 2
5 4 3 2 1
6 3 2 8
Explanation:
Testcase 1: Sorting the numbers accoding to the absolute difference with 7, we have array elements as 5, 9, 10, 3, 2.
Testcase 2: Sorting the numbers according to the absolute difference with 6, we have array elements as 5 4 3 2 1.
Testcase 3: Sorting the numbers according to the absolute difference with 5, we have array elements as 6 3 2 8., I have written this Solution Code: def merge(arr,l,m,h,x):
n1=m-l+1
n2=h-m
L=arr[l:m+1]
R=arr[m+1:h+1]
i=j=0
k=l
while(i<n1 and j<n2):
if abs(x-L[i])<=abs(x-R[j]):
arr[k]=L[i]
i+=1
else:
arr[k]=R[j]
j+=1
k+=1
while(i<n1):
arr[k]=L[i]
i+=1
k+=1
while(j<n2):
arr[k]=R[j]
j+=1
k+=1
def mergeSort(arr,l,h,k):
if l<h:
m=l+(h-l)//2
mergeSort(arr,l,m,k)
mergeSort(arr,m+1,h,k)
merge(arr,l,m,h,k)
if __name__=="__main__":
t=int(input())
while t>0:
n,k=map(int,input().split())
arr=list(map(int,input().split()))
mergeSort(arr,0,len(arr)-1,k)
for i in arr:
print(i,end=" ")
print()
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements and a number K. The task is to arrange array elements according to the absolute difference with K, i. e., element having minimum difference comes first and so on.
Note : If two or more elements are at equal distance arrange them in same sequence as in the given array.First line of input contains a single integer T which denotes the number of test cases. Then T test cases follows. First line of test case contains two space separated integers N and K. Second line of each test case contains N space separated integers.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= K <= 10^4
1 <= A[i] <= 10^4
Sum of N over all test cases does not exceed 2*10^5For each test case print the given array in the order as described above.Input:
3
5 7
10 5 3 9 2
5 6
1 2 3 4 5
4 5
2 6 8 3
Output:
5 9 10 3 2
5 4 3 2 1
6 3 2 8
Explanation:
Testcase 1: Sorting the numbers accoding to the absolute difference with 7, we have array elements as 5, 9, 10, 3, 2.
Testcase 2: Sorting the numbers according to the absolute difference with 6, we have array elements as 5 4 3 2 1.
Testcase 3: Sorting the numbers according to the absolute difference with 5, we have array elements as 6 3 2 8., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int k;
bool cmp(pi a, pi b){
int x = abs(a.fi-k), y = abs(b.fi-k);
if(x == y)
return a.se < b.se;
return x < y;
}
pi a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n >> k;
for(int i = 1; i <= n; i++){
cin >> a[i].fi;
a[i].se = i;
}
sort(a + 1, a + n + 1, cmp);
for(int i = 1; i <= n; i++)
cout << a[i].fi << " ";
cout << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers <b>a</b> and <b>b</b>, your task is to calculate and print the following four values:-
a+b
a-b
a*b
a/bThe input contains two integers a and b separated by spaces.
<b>Constraints:</b>
1 ≤ b ≤ a ≤ 1000
<b> It is guaranteed that a will be divisible by b</b>Print the mentioned operations each in a new line.Sample Input:
15 3
Sample Output:
18
12
45
5
<b>Explanation:-</b>
First operation is a+b so 15+3 = 18
The second Operation is a-b so 15-3 = 12
Third Operation is a*b so 15*3 = 45
Fourth Operation is a/b so 15/3 = 5, I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args){
FastReader read = new FastReader();
int a = read.nextInt();
int b = read.nextInt();
System.out.println(a+b);
System.out.println(a-b);
System.out.println(a*b);
System.out.println(a/b);
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader(){
InputStreamReader inr = new InputStreamReader(System.in);
br = new BufferedReader(inr);
}
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());
}
double nextDouble(){
return Double.parseDouble(next());
}
long nextLong(){
return Long.parseLong(next());
}
String nextLine(){
String str = "";
try{
str = br.readLine();
}
catch(IOException e){
e.printStackTrace();
}
return str;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers <b>a</b> and <b>b</b>, your task is to calculate and print the following four values:-
a+b
a-b
a*b
a/bThe input contains two integers a and b separated by spaces.
<b>Constraints:</b>
1 ≤ b ≤ a ≤ 1000
<b> It is guaranteed that a will be divisible by b</b>Print the mentioned operations each in a new line.Sample Input:
15 3
Sample Output:
18
12
45
5
<b>Explanation:-</b>
First operation is a+b so 15+3 = 18
The second Operation is a-b so 15-3 = 12
Third Operation is a*b so 15*3 = 45
Fourth Operation is a/b so 15/3 = 5, I have written this Solution Code: a,b=map(int,input().split())
print(a+b)
print(a-b)
print(a*b)
print(a//b), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Maruti is a software engineer at Newton School. He knows his basic salary but doesn't know the Gross salary he gets from the company after getting HRA(Home Rent Allowance) and DA(Dearness allowance) as it does not credit directly to his bank account.
As a friend of Maruti, tell him the Gross salary paid by Newton School to him.
<b>Note: The percentage of HRA and DA gave on basic salary is below. </b>
Basic Salary <= 10000 : HRA = 20%, DA = 80%
Basic Salary <= 20000 : HRA = 25%, DA = 90%
Basic Salary > 20000 : HRA = 30%, DA = 95%
Given an integer n(Basic salary of Maruti). Return Gross salary paid to him.There is an integer n(Basic salary of maruti) is given as input.
<b>Constraint:</b>
1 <b>≤</b> n <b>≤</b> 10<sup>5</sup>Return Gross salary paid to Maruti in Integer.Sample Input:
10000
Sample Output:
20000
Explanation:
HRA:2000
DA:8000
Total Gross salary=10000+2000+8000=20000, I have written this Solution Code: import java.util.*;
public class Main
{
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int ans=0;
if(n<=10000){
ans=2*n;
}
else if(n>10000&&n<=20000){
ans=(int)(2.15*n);
}
else ans=(int)(2.25*n);
System.out.println(ans);
return;
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import static java.lang.System.out;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader reader = new FastReader();
int n = reader.nextInt();
String S = reader.next();
int ncount = 0;
int tcount = 0;
for (char c : S.toCharArray()) {
if (c == 'N') ncount++;
else tcount++;
}
if (ncount > tcount) {
out.print("Nutan\n");
} else {
out.print("Tusla\n");
}
out.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: n = int(input())
s = input()
a1 = s.count('N')
a2 = s.count('T')
if(a1 > a2):
print("Nutan")
else:
print('Tusla'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: //Author: Xzirium
//Time and Date: 02:18:28 24 March 2022
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//-----------------------------------------------------------------------------------------------------------//
READV(N);
string S;
cin>>S;
ll n=0,t=0;
FORI(i,0,N)
{
if(S[i]=='N')
{
n++;
}
else if(S[i]=='T')
{
t++;
}
}
if(n>t)
{
cout<<"Nutan"<<endl;
}
else
{
cout<<"Tusla"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: static void printDataTypes(int a, long b, float c, double d, char e)
{
System.out.println(a);
System.out.println(b);
System.out.printf("%.2f",c);
System.out.println();
System.out.printf("%.4f",d);
System.out.println();
System.out.println(e);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: void printDataTypes(int a, long long b, float c, double d, char e){
cout<<a<<endl;
cout<<b<<endl;
cout <<fixed<< std::setprecision(2) << c << '\n';
cout <<fixed<< std::setprecision(4) << d << '\n';
cout<<e<<endl;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: a=int(input())
b=int(input())
x=float(input())
g = "{:.2f}".format(x)
d=float(input())
e = "{:.4f}".format(d)
u=input()
print(a)
print(b)
print(g)
print(e)
print(u), 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: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){cout<<"FizzBuzz"<<" ";}
else if(i%5==0){cout<<"Buzz ";}
else if(i%3==0){cout<<"Fizz ";}
else{cout<<i<<" ";}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int x= sc.nextInt();
fizzbuzz(x);
}
static void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("FizzBuzz ");}
else if(i%5==0){System.out.print("Buzz ");}
else if(i%3==0){System.out.print("Fizz ");}
else{System.out.print(i+" ");}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: def fizzbuzz(n):
for i in range (1,n+1):
if (i%3==0 and i%5==0):
print("FizzBuzz",end=' ')
elif i%3==0:
print("Fizz",end=' ')
elif i%5==0:
print("Buzz",end=' ')
else:
print(i,end=' '), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){printf("FizzBuzz ");}
else if(i%5==0){printf("Buzz ");}
else if(i%3==0){printf("Fizz ");}
else{printf("%d ",i);}
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, multiples of 100 which are not multiples of 400 are not leap years.The input contains a single integer N
<b>Constraint:</b>
1 <= n <= 10<sup>4</sup>Print "YES" if the year is a leap year else print "NO".Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code:
function LeapYear(year){
// write code here
// return the output using return keyword
// do not use console.log here
if ((0 != year % 4) || ((0 == year % 100) && (0 != year % 400))) {
return 0;
} else {
return 1
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, multiples of 100 which are not multiples of 400 are not leap years.The input contains a single integer N
<b>Constraint:</b>
1 <= n <= 10<sup>4</sup>Print "YES" if the year is a leap year else print "NO".Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code: year = int(input())
if year % 4 == 0 and not year % 100 == 0 or year % 400 == 0:
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, multiples of 100 which are not multiples of 400 are not leap years.The input contains a single integer N
<b>Constraint:</b>
1 <= n <= 10<sup>4</sup>Print "YES" if the year is a leap year else print "NO".Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code: import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int side = scanner.nextInt();
int area = LeapYear(side);
if(area==1){
System.out.println("YES");}
else{
System.out.println("NO");}
}
static int LeapYear(int year){
if(year%400==0){return 1;}
if(year%100 != 0 && year%4==0){return 1;}
else {
return 0;}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, the task is to find the nearest greater element G[i] for every element of A[i] such that the element has index greater than i, If no such element exists, output -1.
More formally:
G[i] for an element A[i] = an element A[j] such that
j is minimum possible AND
j > i AND
A[j] > A[i]Each test case consists of two lines. The first line contains an integer N denoting the size of the array. The Second line of each test case contains N space separated positive integers denoting the values/elements in the array A.
Constraints:
1 <= N <= 10^5
1 <= Ai <= 10^9For each test case, print the next greater element for each array element separated by space in order.Sample Input
4
1 3 2 4
Sample Output
3 4 4 -1
Sample Input
4
4 3 2 1
Sample Output
-1 -1 -1 -1, I have written this Solution Code: def rightlarg(arr):
n = len(arr)
A =[None]*n
A[-1] = -1
Indx = [-1]*n
stack = []
s = 1
stack.append([arr[-1],n-1])
for i in range(n-2,-1,-1):
while s :
if stack[0][0]>arr[i] :
A[i] = stack[0][0]
Indx[i] = stack[0][1]
break
else:
stack.pop(0)
s -= 1
else:
A[i] = -1
stack.insert(0,[arr[i],i])
s += 1
for x in Indx:
#print("kjhx ",x," jhgk")
if x==-1:
print(x,end=" ")
else:
print(arr[x],end=" ")
n=int(input())
B=[int(x) for x in input().split()]
rightlarg(B), 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, the task is to find the nearest greater element G[i] for every element of A[i] such that the element has index greater than i, If no such element exists, output -1.
More formally:
G[i] for an element A[i] = an element A[j] such that
j is minimum possible AND
j > i AND
A[j] > A[i]Each test case consists of two lines. The first line contains an integer N denoting the size of the array. The Second line of each test case contains N space separated positive integers denoting the values/elements in the array A.
Constraints:
1 <= N <= 10^5
1 <= Ai <= 10^9For each test case, print the next greater element for each array element separated by space in order.Sample Input
4
1 3 2 4
Sample Output
3 4 4 -1
Sample Input
4
4 3 2 1
Sample Output
-1 -1 -1 -1, I have written this Solution Code: import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
// don't change the name of this class
// you can add inner classes if needed
class Main {
public static void main (String[] args) {
// Your code here
Scanner sc = new Scanner(System.in);
int arrSize = sc.nextInt();
int arr[] = new int[arrSize];
for(int i = 0; i < arrSize; i++)
arr[i] = sc.nextInt();
nextLarger(arr, arrSize);
}
static void nextLarger(int arr[], int arrSize)
{
Stack<Integer> s = new Stack<>();
int a = 0;
for(int i=arrSize-1;i>=0;i--)
{
a= arr[i];
while(!(s.empty() == true)){
if(s.peek()>a){break;}
s.pop();
}
if(s.empty() == true){arr[i]=-1;}
else{arr[i]=s.peek();}
s.push(a);
}
for(int i=0;i<arrSize;i++)
{
System.out.print(arr[i]+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, the task is to find the nearest greater element G[i] for every element of A[i] such that the element has index greater than i, If no such element exists, output -1.
More formally:
G[i] for an element A[i] = an element A[j] such that
j is minimum possible AND
j > i AND
A[j] > A[i]Each test case consists of two lines. The first line contains an integer N denoting the size of the array. The Second line of each test case contains N space separated positive integers denoting the values/elements in the array A.
Constraints:
1 <= N <= 10^5
1 <= Ai <= 10^9For each test case, print the next greater element for each array element separated by space in order.Sample Input
4
1 3 2 4
Sample Output
3 4 4 -1
Sample Input
4
4 3 2 1
Sample Output
-1 -1 -1 -1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
stack <long long > s;
long long a,b[n];
for(int i=0;i<n;i++){
cin>>b[i];}
for(int i=n-1;i>=0;i--){
a=b[i];
while(!(s.empty())){
if(s.top()>a){break;}
s.pop();
}
if(s.empty()){b[i]=-1;}
else{b[i]=s.top();}
s.push(a);
}
for(int i=0;i<n;i++){
cout<<b[i]<<" ";
}
}
, 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 out the integer N.You don't have to worry about the input, you just have to complete the function <b>printIntger()</b>
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>9</sup>Print the integer N.Sample Input:-
3
Sample Output:
3
Sample Input:
56
Sample Output:
56, I have written this Solution Code: static void printInteger(int N){
System.out.println(N);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to print out the integer N.You don't have to worry about the input, you just have to complete the function <b>printIntger()</b>
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>9</sup>Print the integer N.Sample Input:-
3
Sample Output:
3
Sample Input:
56
Sample Output:
56, I have written this Solution Code: void printInteger(int x){
printf("%d", x);
}, 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 out the integer N.You don't have to worry about the input, you just have to complete the function <b>printIntger()</b>
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>9</sup>Print the integer N.Sample Input:-
3
Sample Output:
3
Sample Input:
56
Sample Output:
56, I have written this Solution Code: void printIntger(int n)
{
cout<<n;
}, 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 out the integer N.You don't have to worry about the input, you just have to complete the function <b>printIntger()</b>
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>9</sup>Print the integer N.Sample Input:-
3
Sample Output:
3
Sample Input:
56
Sample Output:
56, I have written this Solution Code: n=int(input())
print (n), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer, print whether that integer is a prime number or not.First line of input contains an integer T, showing the number of test cases. Every test case is a single integer A.
Constraints
1 <= T <= 100
1 <= A <= 10^8If the given integer is prime, print 'Yes', else print 'No'.Sample Input
3
5
9
13
Output
Yes
No
Yes, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
try{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int testcase = Integer.parseInt(br.readLine());
for(int t=0;t<testcase;t++){
int num = Integer.parseInt(br.readLine().trim());
if(num==1)
System.out.println("No");
else if(num<=3)
System.out.println("Yes");
else{
if((num%2==0)||(num%3==0))
System.out.println("No");
else{
int flag=0;
for(int i=5;i*i<=num;i+=6){
if(((num%i)==0)||(num%(i+2)==0)){
System.out.println("No");
flag=1;
break;
}
}
if(flag==0)
System.out.println("Yes");
}
}
}
}catch (Exception e) {
System.out.println("I caught: " + e);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer, print whether that integer is a prime number or not.First line of input contains an integer T, showing the number of test cases. Every test case is a single integer A.
Constraints
1 <= T <= 100
1 <= A <= 10^8If the given integer is prime, print 'Yes', else print 'No'.Sample Input
3
5
9
13
Output
Yes
No
Yes, I have written this Solution Code: t=int(input())
for i in range(t):
number = int(input())
if number > 1:
i=2
while i*i<=number:
if (number % i) == 0:
print("No")
break
i+=1
else:
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer, print whether that integer is a prime number or not.First line of input contains an integer T, showing the number of test cases. Every test case is a single integer A.
Constraints
1 <= T <= 100
1 <= A <= 10^8If the given integer is prime, print 'Yes', else print 'No'.Sample Input
3
5
9
13
Output
Yes
No
Yes, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
long long n,k;
cin>>n;
long x=sqrt(n);
int cnt=0;
vector<int> v;
for(long long i=2;i<=x;i++){
if(n%i==0){
cout<<"No"<<endl;
goto f;
}}
cout<<"Yes"<<endl;
f:;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have a numbers printed on them. The numbers are in the range of 1 to 6, like any ordinary dice.
Given a number on one face of the dice , you need to print the number on the opposite face .
NOTE : In a normal dice sum of numbers on opposite faces is 7 .The first line of the input contains a single integer T, denoting the number of test cases. Then T test case follows. Each test case contains a single line of the input containing a positive integer N.
Constraints:
1 <= T <= 100
1 <= N <= 6For each testcase, print the number that is on the opposite side of the given face.Input:
2
6
2
Output:
1
5
Explanation:
Testcase 1: For dice facing number 6 opposite face will have the number 1., I have written this Solution Code: def get_opposite_face(n):
return 7-n
t = int(input())
for n in range(t):
print(get_opposite_face(int(input()))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have a numbers printed on them. The numbers are in the range of 1 to 6, like any ordinary dice.
Given a number on one face of the dice , you need to print the number on the opposite face .
NOTE : In a normal dice sum of numbers on opposite faces is 7 .The first line of the input contains a single integer T, denoting the number of test cases. Then T test case follows. Each test case contains a single line of the input containing a positive integer N.
Constraints:
1 <= T <= 100
1 <= N <= 6For each testcase, print the number that is on the opposite side of the given face.Input:
2
6
2
Output:
1
5
Explanation:
Testcase 1: For dice facing number 6 opposite face will have the number 1., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
cout << 7-n << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have a numbers printed on them. The numbers are in the range of 1 to 6, like any ordinary dice.
Given a number on one face of the dice , you need to print the number on the opposite face .
NOTE : In a normal dice sum of numbers on opposite faces is 7 .The first line of the input contains a single integer T, denoting the number of test cases. Then T test case follows. Each test case contains a single line of the input containing a positive integer N.
Constraints:
1 <= T <= 100
1 <= N <= 6For each testcase, print the number that is on the opposite side of the given face.Input:
2
6
2
Output:
1
5
Explanation:
Testcase 1: For dice facing number 6 opposite face will have the number 1., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
while(t-->0){
int n=Integer.parseInt(br.readLine());
System.out.println(6-n+1);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, find its Nth smallest non negative integer that is divisible by N.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000Output a single integer, the Nth smallest non negative multiple of N.Sample Input
5
Sample Output
20
Explanation: The 5th smallest integer divisible by 5 is 20. The smaller integers divisible by 5 are 0, 5, 10, and 15.
Sample Input
2
Sample Output
2, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static boolean checkdigit(int n, int k)
{
while(n!=0)
{
int rem=n%10;
if(rem==k){
return true;
}
n=n/10;
}
return false;
}
public static int findNthNumber(int n)
{
return 0;
}
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line = br.readLine();
int n=Integer.parseInt(line);
for(int i=n,count=1;count<n;i++)
{
if (checkdigit(i,n) || (i%n==0)){
count++;
}
if (count == n){
System.out.println(i);
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, find its Nth smallest non negative integer that is divisible by N.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000Output a single integer, the Nth smallest non negative multiple of N.Sample Input
5
Sample Output
20
Explanation: The 5th smallest integer divisible by 5 is 20. The smaller integers divisible by 5 are 0, 5, 10, and 15.
Sample Input
2
Sample Output
2, I have written this Solution Code: n = int(input())
ans = n*(n-1)
print(ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, find its Nth smallest non negative integer that is divisible by N.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000Output a single integer, the Nth smallest non negative multiple of N.Sample Input
5
Sample Output
20
Explanation: The 5th smallest integer divisible by 5 is 20. The smaller integers divisible by 5 are 0, 5, 10, and 15.
Sample Input
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;
cout<<(n*(n-1));
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
// cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array containing N integers and an integer K. Your task is to find the length of the longest Sub-Array with sum of the elements equal to the given value K.The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case consists of two lines. The first line of each test case contains two integers N and K and the second line contains N space-separated elements of the array.
Constraints:-
1<=T<=500
1<=N,K<=10^5
-10^5<=A[i]<=10^5
Sum of N over all test cases does not exceed 10^5For each test case, print the required length of the longest Sub-Array in a new line. If no such sub-array can be formed print 0.Sample Input:
3
6 15
10 5 2 7 1 9
6 -5
-5 8 -14 2 4 12
3 6
-1 2 3
Sample Output:
4
5
0, I have written this Solution Code: def lenOfLongSubarr(arr, N, K):
mydict = dict()
sum = 0
maxLen = 0
for i in range(N):
sum += arr[i]
if (sum == K):
maxLen = i + 1
elif (sum - K) in mydict:
maxLen = max(maxLen, i - mydict[sum - K])
if sum not in mydict:
mydict[sum] = i
return maxLen
if __name__ == '__main__':
T = int(input())
#N,K=list(map(int,input().split()))
for i in range(T):
N,k= [int(N)for N in input("").split()]
arr=list(map(int,input().split()))
N = len(arr)
print(lenOfLongSubarr(arr, N, k)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array containing N integers and an integer K. Your task is to find the length of the longest Sub-Array with sum of the elements equal to the given value K.The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case consists of two lines. The first line of each test case contains two integers N and K and the second line contains N space-separated elements of the array.
Constraints:-
1<=T<=500
1<=N,K<=10^5
-10^5<=A[i]<=10^5
Sum of N over all test cases does not exceed 10^5For each test case, print the required length of the longest Sub-Array in a new line. If no such sub-array can be formed print 0.Sample Input:
3
6 15
10 5 2 7 1 9
6 -5
-5 8 -14 2 4 12
3 6
-1 2 3
Sample Output:
4
5
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
unordered_map<long long,int> um;
int n,k;
cin>>n>>k;
long arr[n];
int maxLen=0;
for(int i=0;i<n;i++){cin>>arr[i];}
long long sum=0;
for(int i=0;i<n;i++){
sum += arr[i];
// when subarray starts from index '0'
if (sum == k)
maxLen = i + 1;
// make an entry for 'sum' if it is
// not present in 'um'
if (um.find(sum) == um.end())
um[sum] = i;
// check if 'sum-k' is present in 'um'
// or not
if (um.find(sum - k) != um.end()) {
// update maxLength
if (maxLen < (i - um[sum - k]))
maxLen = i - um[sum - k];
}
}
cout<<maxLen<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array containing N integers and an integer K. Your task is to find the length of the longest Sub-Array with sum of the elements equal to the given value K.The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case consists of two lines. The first line of each test case contains two integers N and K and the second line contains N space-separated elements of the array.
Constraints:-
1<=T<=500
1<=N,K<=10^5
-10^5<=A[i]<=10^5
Sum of N over all test cases does not exceed 10^5For each test case, print the required length of the longest Sub-Array in a new line. If no such sub-array can be formed print 0.Sample Input:
3
6 15
10 5 2 7 1 9
6 -5
-5 8 -14 2 4 12
3 6
-1 2 3
Sample Output:
4
5
0, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0){
if(t%10==0){
System.gc();
}
int arrsize=sc.nextInt();
int k=sc.nextInt();
int[] arr=new int[arrsize];
for(int i=0;i<arrsize;i++){
arr[i]=sc.nextInt();
}
int subsize=0;
int sum=0;
HashMap<Integer, Integer> hash=new HashMap<>();
for(int i=0;i<arrsize;i++){
sum+=arr[i];
if(sum==k){
subsize=i+1;
}
if(!hash.containsKey(sum)){
hash.put(sum,i);
}
if(hash.containsKey(sum-k)){
if(subsize<(i-hash.get(sum-k))){
subsize=i-hash.get(sum-k);
}
}
}
System.out.println(subsize);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: def compound_interest(principle, rate, time):
Amount = principle * (pow((1 + rate / 100), time))
CI = Amount - principle
print( '%.2f'%CI)
principle,rate,time=map(int, input().split())
compound_interest(principle,rate,time), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: function calculateCI(P, R, T)
{
let interest = P * (Math.pow(1.0 + R/100.0, T) - 1);
return interest.toFixed(2);
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int p,r,t;
cin>>p>>r>>t;
double rate= (float)r/100;
double amt = (float)p*(pow(1+rate,t));
cout << fixed << setprecision(2) << (amt - p);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, 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 Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] s= br.readLine().split(" ");
double[] darr = new double[s.length];
for(int i=0;i<s.length;i++){
darr[i] = Double.parseDouble(s[i]);
}
double ans = darr[0]*Math.pow(1+darr[1]/100,darr[2])-darr[0];
System.out.printf("%.2f",ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers your task is to calculate the maximum integer among the given integers.The input contains three integers a, b, and c
<b>Constraint:</b>
1<=integers<=10000Print the maximum integer among the given integers.Sample Input:-
2 6 3
Sample Output:-
6
Sample Input:-
48 100 100
Sample Output:
100, I have written this Solution Code: a,b,c=[int(a) for a in input().split()]
print(max(a,b,c)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers your task is to calculate the maximum integer among the given integers.The input contains three integers a, b, and c
<b>Constraint:</b>
1<=integers<=10000Print the maximum integer among the given integers.Sample Input:-
2 6 3
Sample Output:-
6
Sample Input:-
48 100 100
Sample Output:
100, I have written this Solution Code: import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int p = scanner.nextInt();
int tm = scanner.nextInt();
int r = scanner.nextInt();
int intrst = MaxInteger(p,tm,r);
System.out.println(intrst);
}
static int MaxInteger(int a ,int b, int c){
if(a>=b && a>=c){return a;}
if(b>=a && b>=c){return b;}
return c;}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of 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: You are given an array of integers. Consider absolute difference between all the pairs of the the elements. You need to find Kth smallest absolute difference. If the size of the array is N then value of K will be less than N and more than or equal to 1.The first line of input contains number of test cases T.
The first line of each test case contains a two integers N and K denoting the number of elements in the array A and difference you need to output. The second line of each test case contains N space separated integers denoting the elements of the array A
Constraints:
1<= T <= 10
2 <= N <= 100000
1 <= K < N < 100000
0 <= A[i] <= 100000For each test case, output Kth smallest absolute difference.Input :
1
6 2
1 3 4 1 3 8
Output :
0
Explanation :
Test case 1: First smallest difference is 0, between the pair (1, 1) and second smallest absolute difference difference is also 0 between the pairs (3, 3)., I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main{
public static void main(String[] args)throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine().trim());
while (t-- > 0) {
String str[] = read.readLine().trim().split(" ");
int n = Integer.parseInt(str[0]);
int k = Integer.parseInt(str[1]);
int arr[] = new int[n];
str = read.readLine().trim().split(" ");
for (int i = 0; i < n; i++)
arr[i] = Integer.parseInt(str[i]);
System.out.println(Math.abs(small(arr, k)));
}
}
public static int small(int arr[], int k) {
Arrays.sort(arr);
int l = 0, r = arr[arr.length - 1] - arr[0];
while (r > l) {
int mid = l + (r - l) / 2;
if (count(arr, mid) < k) {
l = mid + 1;
} else {
r = mid;
}
}
return r;
}
public static int count(int arr[], int mid) {
int ans = 0, j = 0;
for (int i = 1; i < arr.length; ++i) {
while (j < i && arr[i] - arr[j] > mid) {
++j;
}
ans += i - j;
}
return ans;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers <b>a</b> and <b>b</b>, your task is to calculate and print the following four values:-
a+b
a-b
a*b
a/bThe input contains two integers a and b separated by spaces.
<b>Constraints:</b>
1 ≤ b ≤ a ≤ 1000
<b> It is guaranteed that a will be divisible by b</b>Print the mentioned operations each in a new line.Sample Input:
15 3
Sample Output:
18
12
45
5
<b>Explanation:-</b>
First operation is a+b so 15+3 = 18
The second Operation is a-b so 15-3 = 12
Third Operation is a*b so 15*3 = 45
Fourth Operation is a/b so 15/3 = 5, I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args){
FastReader read = new FastReader();
int a = read.nextInt();
int b = read.nextInt();
System.out.println(a+b);
System.out.println(a-b);
System.out.println(a*b);
System.out.println(a/b);
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader(){
InputStreamReader inr = new InputStreamReader(System.in);
br = new BufferedReader(inr);
}
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());
}
double nextDouble(){
return Double.parseDouble(next());
}
long nextLong(){
return Long.parseLong(next());
}
String nextLine(){
String str = "";
try{
str = br.readLine();
}
catch(IOException e){
e.printStackTrace();
}
return str;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers <b>a</b> and <b>b</b>, your task is to calculate and print the following four values:-
a+b
a-b
a*b
a/bThe input contains two integers a and b separated by spaces.
<b>Constraints:</b>
1 ≤ b ≤ a ≤ 1000
<b> It is guaranteed that a will be divisible by b</b>Print the mentioned operations each in a new line.Sample Input:
15 3
Sample Output:
18
12
45
5
<b>Explanation:-</b>
First operation is a+b so 15+3 = 18
The second Operation is a-b so 15-3 = 12
Third Operation is a*b so 15*3 = 45
Fourth Operation is a/b so 15/3 = 5, I have written this Solution Code: a,b=map(int,input().split())
print(a+b)
print(a-b)
print(a*b)
print(a//b), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A (0 indexed) of n integers from 0 to n-1 (each integer exactly once), output an array such that A[i] becomes A[A[i]].
Make sure you don't use extra memory than the array itself.The first line of the input contains a number n, the number of integers in the array.
The second line of the input contains n numbers, the elements of A.
<b>Constraints:</b>
1 <= n <= 100000
0 <= A[i] <= n-1Output the required array of n integers.Sample Input 1:
5
4 2 3 0 1
Sample Output 1:
1 3 0 4 2
Sample Input 2:
10
9 5 1 4 7 8 0 6 3 2
Sample Output 2:
2 8 5 7 6 3 9 0 4 1
<b>Explanation 1:</b>
A[0] will be A[A[0]]=A[4]=1
A[1] will be A[A[1]]=A[2]=3
A[2] will be A[A[2]]=A[3]=0
A[3] will be A[A[3]]=A[0]=4
A[4] will be A[A[4]]=A[1]=2, I have written this Solution Code: function simpleArrangement(n, arr) {
// write code here
// do not console.log
// return the output as an array
const newArr = []
// for (let i = 0; i < n; i++) {
// arr[i] += (arr[arr[i]] % n) * n;
// }
// Second Step: Divide all values by n
for (let i = 0; i < n; i++) {
newArr.push(arr[arr[i]])
}
return newArr
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A (0 indexed) of n integers from 0 to n-1 (each integer exactly once), output an array such that A[i] becomes A[A[i]].
Make sure you don't use extra memory than the array itself.The first line of the input contains a number n, the number of integers in the array.
The second line of the input contains n numbers, the elements of A.
<b>Constraints:</b>
1 <= n <= 100000
0 <= A[i] <= n-1Output the required array of n integers.Sample Input 1:
5
4 2 3 0 1
Sample Output 1:
1 3 0 4 2
Sample Input 2:
10
9 5 1 4 7 8 0 6 3 2
Sample Output 2:
2 8 5 7 6 3 9 0 4 1
<b>Explanation 1:</b>
A[0] will be A[A[0]]=A[4]=1
A[1] will be A[A[1]]=A[2]=3
A[2] will be A[A[2]]=A[3]=0
A[3] will be A[A[3]]=A[0]=4
A[4] will be A[A[4]]=A[1]=2, I have written this Solution Code: n = int(input())
a = input().split()
print(" ".join(a[int(x)] for x in a)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A (0 indexed) of n integers from 0 to n-1 (each integer exactly once), output an array such that A[i] becomes A[A[i]].
Make sure you don't use extra memory than the array itself.The first line of the input contains a number n, the number of integers in the array.
The second line of the input contains n numbers, the elements of A.
<b>Constraints:</b>
1 <= n <= 100000
0 <= A[i] <= n-1Output the required array of n integers.Sample Input 1:
5
4 2 3 0 1
Sample Output 1:
1 3 0 4 2
Sample Input 2:
10
9 5 1 4 7 8 0 6 3 2
Sample Output 2:
2 8 5 7 6 3 9 0 4 1
<b>Explanation 1:</b>
A[0] will be A[A[0]]=A[4]=1
A[1] will be A[A[1]]=A[2]=3
A[2] will be A[A[2]]=A[3]=0
A[3] will be A[A[3]]=A[0]=4
A[4] will be A[A[4]]=A[1]=2, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
String s = br.readLine();
String[] str = s.split(" ");
int arr[] = new int[n];
for(int i=0;i<n;i++){
arr[i]=Integer.parseInt(str[i]);
}
for(int i =0;i<n;i++){
System.out.print(arr[arr[i]]+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A (0 indexed) of n integers from 0 to n-1 (each integer exactly once), output an array such that A[i] becomes A[A[i]].
Make sure you don't use extra memory than the array itself.The first line of the input contains a number n, the number of integers in the array.
The second line of the input contains n numbers, the elements of A.
<b>Constraints:</b>
1 <= n <= 100000
0 <= A[i] <= n-1Output the required array of n integers.Sample Input 1:
5
4 2 3 0 1
Sample Output 1:
1 3 0 4 2
Sample Input 2:
10
9 5 1 4 7 8 0 6 3 2
Sample Output 2:
2 8 5 7 6 3 9 0 4 1
<b>Explanation 1:</b>
A[0] will be A[A[0]]=A[4]=1
A[1] will be A[A[1]]=A[2]=3
A[2] will be A[A[2]]=A[3]=0
A[3] will be A[A[3]]=A[0]=4
A[4] will be A[A[4]]=A[1]=2, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
long long n,k;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
for(int i=0;i<n;i++){
cout<<a[a[i]]<<" ";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import static java.lang.System.out;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader reader = new FastReader();
int n = reader.nextInt();
String S = reader.next();
int ncount = 0;
int tcount = 0;
for (char c : S.toCharArray()) {
if (c == 'N') ncount++;
else tcount++;
}
if (ncount > tcount) {
out.print("Nutan\n");
} else {
out.print("Tusla\n");
}
out.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: n = int(input())
s = input()
a1 = s.count('N')
a2 = s.count('T')
if(a1 > a2):
print("Nutan")
else:
print('Tusla'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: //Author: Xzirium
//Time and Date: 02:18:28 24 March 2022
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//-----------------------------------------------------------------------------------------------------------//
READV(N);
string S;
cin>>S;
ll n=0,t=0;
FORI(i,0,N)
{
if(S[i]=='N')
{
n++;
}
else if(S[i]=='T')
{
t++;
}
}
if(n>t)
{
cout<<"Nutan"<<endl;
}
else
{
cout<<"Tusla"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Mohit has three integers A, B, and C with him he wants to find the average of these three integers however he is weak in maths, so help him to find the average. You need to report the floor of the average value.<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>Average()</b> that takes integers A, B, and C as arguments.
Constraints:-
1 <= A, B, and C <= 10000Return the floor of average of these numbers.Sample Input:-
3 4 5
Sample Output:-
4
Sample Input:-
3 4 4
Sample Output:-
3, I have written this Solution Code: def Average(A,B,C):
return (A+B+C)//3
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Mohit has three integers A, B, and C with him he wants to find the average of these three integers however he is weak in maths, so help him to find the average. You need to report the floor of the average value.<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>Average()</b> that takes integers A, B, and C as arguments.
Constraints:-
1 <= A, B, and C <= 10000Return the floor of average of these numbers.Sample Input:-
3 4 5
Sample Output:-
4
Sample Input:-
3 4 4
Sample Output:-
3, I have written this Solution Code:
int Average(int A,int B, int C){
return (A+B+C)/3;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Mohit has three integers A, B, and C with him he wants to find the average of these three integers however he is weak in maths, so help him to find the average. You need to report the floor of the average value.<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>Average()</b> that takes integers A, B, and C as arguments.
Constraints:-
1 <= A, B, and C <= 10000Return the floor of average of these numbers.Sample Input:-
3 4 5
Sample Output:-
4
Sample Input:-
3 4 4
Sample Output:-
3, I have written this Solution Code:
static int Average(int A,int B, int C){
return (A+B+C)/3;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
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