Marc's Cakewalk

Problem Statement :

Marc loves cupcakes, but he also likes to stay fit. Each cupcake has a calorie count, and Marc can walk a distance to expend those calories. If Marc has eaten  cupcakes so far, after eating a cupcake with  calories he must walk at least  miles to maintain his weight.

consumption. In this case, our minimum miles is calculated as .

Given the individual calorie counts for each of the cupcakes, determine the minimum number of miles Marc must walk to maintain his weight. Note that he can eat the cupcakes in any order.

Function Description

Complete the marcsCakewalk function in the editor below.

marcsCakewalk has the following parameter(s):

int calorie[n]: the calorie counts for each cupcake


long: the minimum miles necessary

Input Format

The first line contains an integer , the number of cupcakes in .
The second line contains  space-separated integers, .

Solution :


                            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;

int partition(int *x,int start,int end)
    int pivot,pindex,i;
    pivot = x[end];
    pindex = start;
            pindex = pindex + 1;
    return pindex;
void quicksort(int *x,int start,int end)

     int i = partition(x,start,end);


int main(){
    int n; 
    int *calories = malloc(sizeof(int) * n);
    for(int calories_i = 0; calories_i < n; calories_i++)
    int i;
    long long int sum = 0;
        sum += calories[i]*((long long int)pow(2,i));
    // 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.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(;
        int n = in.nextInt();
        int[] calories = new int[n];
        for(int calories_i=0; calories_i < n; calories_i++){
            calories[calories_i] = in.nextInt();
        long ans = 0;
        for (int i = 0; i < n; i++) {
            ans += ((long)calories[n-i-1])<<i;

                        Solution in Python : 
In  Python3 :


import sys

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

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

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