hdu 4989(水题)

Summary

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 835    Accepted Submission(s): 424

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

 

开long long 就好。

#include<iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
const int N = ;
long long a[N],b[N*N];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF){
        for(int i=;i<=n;i++){
            scanf("%lld",&a[i]);
        }
        int id = ;
        for(int i=;i<=n;i++){
            for(int j=i+;j<=n;j++){
                b[id++] = a[i]+a[j];
            }
        }
        sort(b,b+id);
        long long sum = b[];
        for(int i=;i<id;i++){
            if(b[i]==b[i-]) continue;
            sum=sum+b[i];
        }
        printf("%lld\n",sum);
    }
    return ;
}