Problem:
Score : 200 points

Problem Statement

Snuke is having a barbeque party.

At the party, he will make N servings of Skewer Meal.

Example of a serving of Skewer Meal

He has a stock of 2N skewers, all of which will be used in Skewer Meal. The length of the i-th skewer is L_i.
Also, he has an infinite supply of ingredients.

To make a serving of Skewer Meal, he picks 2 skewers and threads ingredients onto those skewers.
Let the length of the shorter skewer be x, then the serving can hold the maximum of x ingredients.

What is the maximum total number of ingredients that his N servings of Skewer Meal can hold, if he uses the skewers optimally?

Constraints

1≦N≦100

1≦L_i≦100

For each i, L_i is an integer.

Input

The input is given from Standard Input in the following format:

N
L_1 L_2 ... L_{2N}

Output

Print the maximum total number of ingredients that Snuke's N servings of Skewer Meal can hold.

Sample Input 1

2
1 3 1 2

Sample Output 1

3

If he makes a serving using the first and third skewers, and another using the second and fourth skewers, each serving will hold 1 and 2 ingredients, for the total of 3.

Sample Input 2

5
100 1 2 3 14 15 58 58 58 29

Sample Output 2

135

Solution in C (Clang 3.8.0) (judge verdict: Wrong Answer)


#include <stdio.h>

int main ()
{
    int n,sum=0,temp,i,j;
    int l[100]={};
    
    scanf("%d",&n);
    for(i=0; i<2*n; i++)
    {
        scanf("%d", &l[i]);
    }
 
    for(i=0; i<2*n; i++)
    {
        for(j=2*n-1; j>i; j--)
        {
            if(l[j-1] < l[j])
            {
                temp = l[j];
                l[j] = l[j-1];
                l[j-1] = temp;
            }
        }
    }
    
    for(i=0; i<2*n; i++)
    {
        if(i%2 != 0)
        {
            sum += l[i];
        }
    }
    
    printf("%d\n",sum);
    
    return 0;
}