HackerRank Chocolate Feast problem solution

In this HackerRank Chocolate Feast problem, you need to complete the chocolateFeast function that has three integer variables as a parameter and need to return the number of chocolate bobby can eat after taking full advantage of the promotion.

Problem solution in Python programming.

def chocolates(N, C, M):
    
    total = int(N/C);
    
    empty_wrapper = total
    
    while empty_wrapper >= M:
        
        temp = int(empty_wrapper/M);
        
        total = total + temp;
        
        empty_wrapper = empty_wrapper - (temp*M) + temp;
    
    return total;


T = int(input())

result = [];

for i in range(T):
    
    (N, C, M) = map(int, input().split());
    
    result = result + [chocolates(N,C,M)];


print('\n'.join(map(str, result)))

Problem solution in Java Programming.

import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        
        scanner.nextLine();
        while (scanner.hasNext()) {
            int money = scanner.nextInt();
            int price = scanner.nextInt();
            int bonus = scanner.nextInt();
            
            int count = money / price;
            int wrappers = count;
            
            while (wrappers >= bonus) {
                int freebies = wrappers / bonus;
                count += freebies;
                wrappers = freebies + wrappers % bonus;
            }
            
            System.out.println(count);
        }
    }
}

Problem solution in C++ programming.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */ 
    ios_base::sync_with_stdio(0);
    int T, N, C, M, w, res;
    cin >> T;
    while (T--) {
        cin >> N >> C >> M;
        w = res = N/C;
        while (w >= M) {
            res += w/M;
            w = (w%M) + w/M;
        }
        cout << res << "\n";
    }
    return 0;
}

Problem solution in C programming.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
         int testcases=0;
        int N=0,M=0,C=0;
        int i=0, now=0;

        scanf("%d",&testcases);
        int temp = 0;
        int answer[testcases];

        for(i=0;i<testcases;i++)
        {

                scanf("%d",&N);
                scanf("%d",&C);
                scanf("%d",&M);
                temp = N/C;
                now=temp;
                answer[i]=temp;
                while (now>=M)
                {  
                answer[i] = answer[i] + (now/M);
                now=now - (now/M)*M + now/M;
                }

                //printf("first = %d Second = %d\n",N/C,((N/C)/M));

        }

        for(i=0;i<testcases;i++)
                printf("%d\n",answer[i]);
   return 0;
}

Problem solution in JavaScript programming.

function parseInts(line) {
    return line.split(' ').map(function(x) { return parseInt(x, 10) });
}

function solve(n, c, m) {
    var chocolates = Math.floor(n / c);
    return chocolates + Math.floor((chocolates - 1) / (m - 1));
}

function processData(input) {
    var lines = input.split('\n');
    for (var t = parseInts(lines.shift())[0]; t > 0; t -= 1) {
        var ints = parseInts(lines.shift());
        console.log(solve.apply(null, ints));
    }
}

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});