Summary

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

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.

Input

Multi test cases, every case occupies two lines, the first line contain n, then second line contain n numbers a₁, a₂, ……a_n
separated by exact one space. Process to the end of file.

[Technical Specification]

2 <= n <= 100

-1000000000 <= a_i <= 1000000000

Output

For each case, output the final sum.

Sample Input

4
1 2 3 4
2
5 5

Sample Output

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.


 

Source

BestCoder Round #8

题目的意思看了挺久的。，给你一个N，叫你输入N个数。然后从N个数中任意取两个数相加，得到一个新的数，将所有新的数求和就是答案。注意一点就好，相同的元素只算一个。所以去掉重复的就好。上代码

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int main()
{
	int n,i,j;
	while(~scanf("%d",&n))
	{
		int a[105];
		int b[10005];
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		__int64 sum=0;
		int l=0;
		for(i=0;i<n;i++)
			scanf("%d",&a[i]);
		for(i=0;i<n;i++)
			for(j=i+1;j<n;j++)
			{
				b[l++]=a[i]+a[j];
			}
			sort(b,b+l);
			int n1=unique(b,b+l)-b;  //去掉相同的。此时新数组个数n1.
			for(i=0;i<n1;i++)
				sum+=b[i];
			printf("%I64d\n",sum);
	}
	return 0;
}