Summary

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Small W is playing a summary game. Firstly, He takes N numbers. Secondly he takes out every pair of them and add this two numbers, thus he can get N∗(N−1)2 new numbers. Thirdly he deletes the repeated number of the new numbers. Finally he gets the sum of the left numbers. Now small W want you to tell him what is the final sum.

Multi test cases, every case occupies two lines, the first line contain n, then second line contain n numbers a1,a2,…,an separated by exact one space. Process to the end of file.
[Technical Specification]
2 ≤n≤100
-1000000000≤ai≤1000000000

For each case, output the final sum.

4
1 2 3 4
2
5 5

25
10
Hint
Firstly small W takes any pair of 1 2 3 4 and add them, he will get 3 4 5 5 6 7. Then he deletes the repeated numbers, he will get 3 4 5 6 7, Finally he gets the sum=3+4+5+6+7=25.

/*题解：
很简单的题，但是需要注意测试数据的范围，用__int64输出结果
*/

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
__int64 a[110],b[10000],sum;
int main()
{
    int i,j,k,n;
    while(scanf("%d",&n)!=EOF)
	{
		memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(i=0; i<n; i++)
            scanf("%I64d",&a[i]);
        for(i=0,k=0; i<n-1; i++)
		{
            for(j=i+1; j<n; j++)
			{
                b[k++] = a[i]+a[j];
            }
        }
        sort(b,b+k);
        for(i=1,sum=b[0]; i<k; i++)
		{
            if(b[i]!=b[i-1])
                sum+=b[i];
            else
                continue;
        }
        printf("%I64d\n",sum);
    }
}