HDU-4989-Summary(STL之set）_hdu 4989-优快云博客

本文链接：https://blog.youkuaiyun.com/Li_Hongcheng/article/details/79116282

本文介绍了一个有趣的数学游戏算法：给定一组整数，通过计算所有可能的两两之和并去除重复项后求和。文章详细解释了游戏规则及输入输出格式，并提供了一个C++实现示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Summary

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.

#include<bits/stdc++.h>
using namespace std;

int main() {
	int n,a[105];
	set<int>s;
	while(~scanf("%d",&n)) {
		long long sum=0;
		s.clear();//清空！
		for(int i = 0; i < n; i++) {
			scanf("%d",&a[i]);
		}
		for(int i=0; i < n; i++) {
			for(int j=i+1; j<n; j++) {
				s.insert(a[i]+a[j]);
			}
		}
		set<int>::iterator it;//定义前向迭代器 
		for(it=s.begin(); it!=s.end(); it++)
			sum +=*it;
		printf("%lld\n",sum);
	}
}