hdu 4989 Summary（set）

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题目链接：http://hdu.hustoj.com/showproblem.php?pid=4989

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 a ₁, a ₂, ……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 
  
    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：求所有sum（i，j）（1<=i<j<=n）的和,要求不同对的i，j的和相同元素只加一次

思路：因为不保存相同元素，所以用set来保存一下

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<map>
#include<set>

using namespace std;

#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)

const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9+7;
const int N = 1e6+5;

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    std::ios::sync_with_stdio(false);
    int n;
    ll x,a[105];
    set<ll>s;
    while(~scanf("%d",&n))
    {
    	s.clear();
    	for(int i=1;i<=n;i++)
    		scanf("%lld",&a[i]);
    	for(int i=1;i<=n;i++)
    	{
    		for(int j=i+1;j<=n;j++)
    		{
    			s.insert(a[i]+a[j]);
			}
		}
		ll ans=0;
		set<ll>::iterator iter;
		for(iter=s.begin();iter!=s.end();iter++)
			ans+=*iter;
		printf("%lld\n",ans);
	}
    return 0;
}