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本文介绍了一个算法问题：给定一组数字，计算所有不同数字对的和，并去除重复项后的总和。文章提供了实现思路及C++代码示例。

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Summary

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 836 Accepted Submission(s): 425

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

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.

题意：求n个数两两之和，去重后加起来的和

思路：把两两之和求出来，然后排序，去重，最后加起来即可~

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 110
int a[N],b[N*N];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        int cnt=0;
        for(int i=1; i<=n; i++)
            for(int j=i+1; j<=n; j++)
                b[cnt++]=a[i]+a[j];
        sort(b,b+cnt);
        int tot=0;
        b[tot++]=b[0];
        for(int i=1;i<cnt;i++)
            if(b[i]!=b[i-1]) b[tot++]=b[i];
        long long sum=0;
        for(int i=0;i<tot;i++)
            sum+=b[i];
        printf("%lld\n",sum);
    }
    return 0;
}