hdu 4989 Summary（BestCoder Round #8 1001)-优快云博客

本文链接：https://blog.youkuaiyun.com/caduca/article/details/39124135

本文介绍了一个简单的总结游戏算法，玩家需要从一组数字中挑选两两配对并求和，然后去除重复的和值，最终计算剩余数值之和。文章通过一个具体示例详细解释了算法流程，并提供了一段使用C++实现的高效代码。

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Summary

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 76 Accepted Submission(s): 53

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
HintFirstly 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.

水题，数据太小，set判重

代码：

//15ms
#include <iostream>
#include <cstdio>
#include <cstring>
#include <set>
using namespace std;
long long a[100];
set<long long> s;
set<long long>::iterator it;
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%I64d",&a[i]);
        }
        s.clear();
        for(int j=0;j<n-1;j++)
        {
            for(int k=j+1;k<n;k++)
            {
                s.insert(a[j]+a[k]);
            }
        }
        long long ans=0;
        for(it=s.begin();it!=s.end();it++)
        {
            ans=ans+*it;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}