Flowers - Hacker Rank Program

Problem Statement

You and your K-1 friends want to buy N flowers. Flower number i has cost c_i. Unfortunately the seller does not want just one customer to buy a lot of flowers, so he tries to change the price of flowers for customers who have already bought some flowers. More precisely, if a customer has already bought x flowers, he should pay (x+1)*c_i dollars to buy flower number i.
You and your K-1 friends want to buy all N flowers in such a way that you spend the least amount of money. You can buy the flowers in any order.
Input:
The first line of input contains two integers N and K (K <= N). The next line contains N space separated positive integers c₁,c₂,...,c_N.
Output:
Print the minimum amount of money you (and your friends) have to pay in order to buy all N flowers.
Constraints
1≤N,K≤100
Any ci is not more than 106
Result is guaranteed to be less than 231
Sample input #00

3 3
2 5 6

Sample output #00

13

Sample input #01

3 2
2 5 6

Sample output #01

15

Explanation :
Sample Case #00: In this example, all of you should buy one flower each. Hence, you'll have to pay 13 dollars.
Sample Case #01: Here one of the friend buys first two flowers in decreasing order of their price. So he will pay (0+1)*5 + (1+1)*2 = 9. And other friend will buy the costliest flower of cost 6. So total money need is 9+6=15.

MY SOLUTION:

#include<stdio.h>
int sort(const void *a,const void *b)
    {
    return(*(int *)a-*(int *)b);
}
int main(){
    int n,k,i;
    scanf("%d %d",&n,&k);
    int a[n];
    for(i=0;i<n;i++)
        scanf("%d",&a[i]);
    qsort(a,n,sizeof(int),sort);
    int result=0,flower=0;
    for(i=0;i<n;i++)
    {
        flower=i/k;
        result+=(flower+1)*a[(n-1)-i];
    }
    printf("%d",result);
    return 0;
}