Problem:
Score : 200 points

Problem Statement

A shop sells N kinds of fruits, Fruit 1, \ldots, N, at prices of p_1, \ldots, p_N yen per item, respectively. (Yen is the currency of Japan.)

Here, we will choose K kinds of fruits and buy one of each chosen kind. Find the minimum possible total price of those fruits.

Constraints

1 \leq K \leq N \leq 1000

1 \leq p_i \leq 1000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
p_1 p_2 \ldots p_N

Output

Print an integer representing the minimum possible total price of fruits.

Sample Input 1

5 3
50 100 80 120 80

Sample Output 1

210

This shop sells Fruit 1, 2, 3, 4, and 5 for 50 yen, 100 yen, 80 yen, 120 yen, and 80 yen, respectively.

The minimum total price for three kinds of fruits is 50 + 80 + 80 = 210 yen when choosing Fruit 1, 3, and 5.

Sample Input 2

1 1
1000

Sample Output 2

1000

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#define SIZE 1001

int med(int x, int y, int z){
    if(x < y){
        if(y < z)return y; else if(z < x)return x; else return z;
    }
    else{
        if(x < z)return x; else if(z < y)return y; else return z;
    }
}

void quick_sort(int a[], int left, int right){
    if(left < right){
        int i = left, j = right, tmp, pivot = med(a[i], a[i+(j-i)/2], a[j]);
        while(1){
            while(a[i] < pivot)i++;
            while(pivot < a[j])j--;
            if(i >= j)break;
            tmp = a[i]; a[i] = a[j]; a[j] = tmp;
            i++; j--;
            quick_sort(a, left, i-1);
            quick_sort(a, j+1, right);
        }
    }
}

int main(void){
    int n, k, i, min = 0;
    int p[SIZE];
    scanf("%d %d", &n, &k);
    for(i = 0; i < n; i++)
        scanf("%d", &p[i]);
    
    quick_sort(p, 0, n-1);
    
    for(i = 0; i < k; i++)
        min += p[i];
    printf("%d", min);
    return 0;
}