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 ;

}

巴特西

hdu 4989(水题)

Summary

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