HackerRank Beautiful Days at the Movies problem solution

In this HackerRank Beautiful Days at the Movies problem you have Given a range of numbered days, [i…j]  and a number k, to determine the number of days in the range that are beautiful. Beautiful numbers are defined as numbers where |i-reverse(i)|  is evenly divisible by k. If a day’s value is a beautiful number, it is a beautiful day. Return the number of beautiful days in the range.

Problem solution in Python programming.

a, b, k = map(int, raw_input().split())
ans = 0
for i in range(a, b+1):
    ans = ans + abs(not (i - int(str(i)[::-1]))%k)
print ans

Problem solution in Java Programming.

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner in = new Scanner(System.in);
        int i = in.nextInt();
        int j = in.nextInt();
        int k = in.nextInt();
        int count = 0;
        
        for (int a=i;j>a; a++){
            StringBuilder temp = new StringBuilder();
            temp.append(a);
            temp=temp.reverse();
            String temp1 = temp.toString();
            int aRev = Integer.parseInt(temp1);
            if(Math.abs((a-aRev)%k)==0){
                count++;
            }
        }
        
        System.out.println(count);
    }
}

Problem solution in C++ programming.

#include <iostream>
  
using namespace std;

bool isOk(int x, int mod) {
    int n = x;
    int m = 0;
    while (x > 0) {
        m = m * 10 + x % 10;
        x /= 10;
    }
    int delta = abs(n - m);
    delta %= mod;
    return (delta == 0);
}

int main() {
    int l, r, k;
    cin >> l >> r >> k;
    int ans = 0;
    for (int i = l; i <= r; i++) {
        if (isOk(i, k)) {
            ++ans;
        }
    }
    cout << ans << endl;
    return 0;
}

Problem solution in C programming.

#include<stdio.h>
int main()
{
int i=0,j=0,k=0,x=0,rem=0,sum=0,count=0;
scanf("%d%d%d",&i,&j,&k);
for(i;i<=j;i++)
{
x=i;
while(x!=0)
{
rem=x%10;
sum=(sum*10)+rem;
x=x/10;
}
if(abs(i-sum)%k==0)
count=count+1;
sum=0;
}
printf("%d",count);
return 0;
}