Marc's Cakewalk

                            Solution in C :

In  C  :





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


void swap(int *a,int *b)
{
    int temp;
    temp = *a;
    *a = *b;
    *b=temp;
}



int partition(int *x,int start,int end)
{
    int pivot,pindex,i;
    pivot = x[end];
    pindex = start;
    for(i=start;i<end;i++)
    {
        if(x[i]>=pivot)
          {
            swap(&x[i],&x[pindex]);
            pindex = pindex + 1;
          }
    }
    swap(&x[pindex],&x[end]);
    return pindex;
}
void quicksort(int *x,int start,int end)
{

    if(start<end)
   {
     int i = partition(x,start,end);
      quicksort(x,start,i-1);
      quicksort(x,i+1,end);


    }
}

int main(){
    int n; 
    scanf("%d",&n);
    int *calories = malloc(sizeof(int) * n);
    for(int calories_i = 0; calories_i < n; calories_i++)
    {
       scanf("%d",&calories[calories_i]);
    }
    quicksort(calories,0,n-1);
    int i;
    long long int sum = 0;
    for(i=0;i<n;i++)
    {
        sum += calories[i]*((long long int)pow(2,i));
    }    
    printf("%lld",sum);
    // your code goes here
    return 0;
}
                        

                        Solution in C++ :

In  C ++  :






#include "bits/stdc++.h"
using namespace std;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
static const int INF = 0x3f3f3f3f; static const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
typedef vector<int> vi; typedef pair<int, int> pii; typedef vector<pair<int, int> > vpii; typedef long long ll;
template<typename T, typename U> static void amin(T &x, U y) { if (y < x) x = y; }
template<typename T, typename U> static void amax(T &x, U y) { if (x < y) x = y; }

int main() {
	int n;
	while (~scanf("%d", &n)) {
		vector<int> c(n);
		for (int i = 0; i < n; ++ i)
			scanf("%d", &c[i]);
		sort(c.begin(), c.end());
		ll ans = 0;
		rep(i, n)
			ans += (ll)c[i] << (n - 1 - i);
		printf("%lld\n", ans);
	}
	return 0;
}
                    

                        Solution in Java :

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();
        int[] calories = new int[n];
        for(int calories_i=0; calories_i < n; calories_i++){
            calories[calories_i] = in.nextInt();
        }
        Arrays.sort(calories);
        long ans = 0;
        for (int i = 0; i < n; i++) {
            ans += ((long)calories[n-i-1])<<i;
        }
        System.out.println(ans);
    }
}
                    

                        Solution in Python : 
                            
In  Python3 :






#!/bin/python3

import sys


n = int(input().strip())
calories = list(map(int, input().strip().split(' ')))
calories.sort()

toplam = 0
count = 0
for kalori in calories[::-1]:
    toplam += kalori * (2 ** count)
    count += 1
print(toplam)