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: Accepted)

#include <stdio.h>

int main(void) {
    int p = 0;
    int pmin[1000] = { 0 };
    int pminmax = 0;
    int n = 1, k = 1;
    int i = 0, j = 0;
    int sum = 0;
    int swap = 0;

    scanf("%d %d", &n, &k);

    scanf("%d", &p);
    pmin[0] = p;
    pminmax = p;

    for (i = 1; i < k; i++) {
        scanf("%d", &p);
        pmin[i] = p;
        if (pminmax < p)pminmax = p;
    }

    for (; i < n; i++) {
        scanf("%d", &p);
        if (pminmax <= p)continue;
        for (j = 0; j < k; j++) {
            if (pmin[j] > p) {
                swap = pmin[j];
                pmin[j] = p;
                p = swap;
            }
        }
        pminmax = p;
    }

    for (j = 0; j < k; j++) {
        sum += pmin[j];
    }

    printf("%d\n", sum);
    return 0;
}