hdu 4989 Summary（水题）

Summary

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 836    Accepted Submission(s): 425

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n个数两两之和，去重后加起来的和

思路：把两两之和求出来，然后排序，去重，最后加起来即可~

代码：

#include 
#include 
#include 
#include 
#include 
using namespace std;
#define N 110
int a[N],b[N*N];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        int cnt=0;
        for(int i=1; i<=n; i++)
            for(int j=i+1; j<=n; j++)
                b[cnt++]=a[i]+a[j];
        sort(b,b+cnt);
        int tot=0;
        b[tot++]=b[0];
        for(int i=1;i