hdu-4989 Summary(水题)

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

4

1 2 3 4

2

5 5

 

Sample Output

25

10

 

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 ;
}