HDU 4989 Summary（数学题暴力）

题目链接：http://acm.hdu.edu.cn/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 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

41 2 3 425 5

 

Sample Output

2510
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.
 

 

Source

BestCoder Round #8

代码如下：

//#pragma warning (disable:4786)#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <cstdlib>#include <climits>#include <ctype.h>#include <queue>#include <stack>#include <vector>#include <utility>#include <deque>#include <set>#include <map>#include <iostream>#include <algorithm>using namespace std;#define INF 1e18//typedef long long LL;typedef __int64 LL;int main(){    int n;    LL a[7000], b[7000];    while(~scanf("%d",&n))    {        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        for(int i = 0; i < n; i++)        {            scanf("%I64d",&a[i]);        }        int l = 0;        for(int i = 0; i < n-1; i++)        {            for(int j = i+1; j < n; j++)            {                b[l++] = a[i]+a[j];            }        }        sort(b,b+l);        for(int i = 1; i < l; i++)        {            if(b[i] == b[i-1])                b[i-1] = 0;        }        LL sum = 0;        for(int i = 0; i < l; i++)        {            sum+=b[i];        }        printf("%I64d\n",sum);    }    return 0;}