HackerRank Marc’s Cakewalk problem solution

Problem solution in Python.

#!/bin/python3

import sys


n = int(input().strip())
cal = list(map(int, input().strip().split(' ')))
# your code goes here
cal.sort()
cal.reverse()

miles=0
for i in range(n):
    cup=cal[i]*(2**i)
    miles+=cup
    
print(miles)

Problem solution in Java.

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) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        Integer[] calories = new Integer[n];
        for(int calories_i=0; calories_i < n; calories_i++){
            calories[calories_i] = in.nextInt();
        }
        List<Integer> calList = Arrays.asList(calories);
        Collections.sort(calList, Collections.reverseOrder());

        long cals = 0;
        for (int i=0;i<calList.size();i++) {
            cals += Math.pow(2,i)*calList.get(i);
        }
        System.out.println(cals);
    }
}

Problem solution in C++.

#include <bits/stdc++.h>

using namespace std;

int main(){
    int n;
    cin >> n;
    vector<int> calories(n);
    for(int calories_i = 0; calories_i < n; calories_i++){
       cin >> calories[calories_i];
    }
    // your code goes here
    sort(calories.begin(),calories.end());
    reverse(calories.begin(),calories.end());
    long long temp=1,ans=0;
    for(int i=0;i<n;i++)
    {
        ans+=calories[i]*temp;
        temp*=2;
    }
    printf("%lldn",ans);
    return 0;
}

Problem solution in C.

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
    int n; 
    scanf("%d",&n);
    int a[n];
    int i,j,t;
    long int c=0;
    for(i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
        for(i=0;i<n;i++){
            for(j=n-1;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++){
            c=c+(a[i]*(pow(2,i)));
        }
    printf("%ld",c);
    return 0;
}