题目链接：

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

1 2 3 4

5 5

Sample Output

题意：

求生成的数的和,重复的去掉;

思路：

水题;

AC代码：

#include <bits/stdc++.h>

using namespace std;

const int N=3e5+;

typedef long long ll;

int n,a[];

queue<int>qu;

map<int,int>mp;

int main()

{

    while(scanf("%d",&n)!=EOF)

    {

        ll ans=;

        for(int i=;i<=n;i++)

        {

            scanf("%d",&a[i]);

        }

        for(int i=;i<=n;i++)

        {

            for(int j=i+;j<=n;j++)

            {

                mp[a[i]+a[j]]=;

                qu.push(a[i]+a[j]);

            }

        }

        while(!qu.empty())

        {

            int fr=qu.front();

            qu.pop();

            if(mp[fr]==)ans+=fr,mp[fr]=;

        }

        printf("%lld\n",ans);

    }

    return ;

}

巴特西

hdu-4989 Summary(水题)

Summary

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