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 Returns 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;
*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)
View More Similar Problems
Jenny's Subtrees
Jenny loves experimenting with trees. Her favorite tree has n nodes connected by n - 1 edges, and each edge is ` unit in length. She wants to cut a subtree (i.e., a connected part of the original tree) of radius r from this tree by performing the following two steps: 1. Choose a node, x , from the tree. 2. Cut a subtree consisting of all nodes which are not further than r units from node x .
View Solution →Tree Coordinates
We consider metric space to be a pair, , where is a set and such that the following conditions hold: where is the distance between points and . Let's define the product of two metric spaces, , to be such that: , where , . So, it follows logically that is also a metric space. We then define squared metric space, , to be the product of a metric space multiplied with itself: . For
View Solution →Array Pairs
Consider an array of n integers, A = [ a1, a2, . . . . an] . Find and print the total number of (i , j) pairs such that ai * aj <= max(ai, ai+1, . . . aj) where i < j. Input Format The first line contains an integer, n , denoting the number of elements in the array. The second line consists of n space-separated integers describing the respective values of a1, a2 , . . . an .
View Solution →Self Balancing Tree
An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ
View Solution →Array and simple queries
Given two numbers N and M. N indicates the number of elements in the array A[](1-indexed) and M indicates number of queries. You need to perform two types of queries on the array A[] . You are given queries. Queries can be of two types, type 1 and type 2. Type 1 queries are represented as 1 i j : Modify the given array by removing elements from i to j and adding them to the front. Ty
View Solution →Median Updates
The median M of numbers is defined as the middle number after sorting them in order if M is odd. Or it is the average of the middle two numbers if M is even. You start with an empty number list. Then, you can add numbers to the list, or remove existing numbers from it. After each add or remove operation, output the median. Input: The first line is an integer, N , that indicates the number o
View Solution →