joj2510

2510: Product

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Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	1s	65536K	691	205	Standard

There are multiple test cases, the first line of each test case is a integer n. (2<=n<=20) The next line have n integers (every integer in the range [1,1000]). We can divide these n integers into two parts and get the sum of the integers in these two parts, define them as sumA and sumB. Please calc the max number of sumA*sumB and output it in a single line.

Sample Input

3
5 3 7
2
4 9

Sample Output

56
36

Hint

(5,3),(7)  sumA=8  sumB=7  sumA*sumB=56
(5),(3,7)  sumA=5  sumB=10 sumA*sumB=50
(3),(5,7)  sumA=3  sumB=12 sumA*sumB=36
(),(3,5,7) sumA=0  sumB=15 sumA*sumB=0
The max is 56.

Problem Source: xwbsw

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This problem is used for contest: 120 

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Problem Set with Online Judge System Version 3.12
Jilin University

Developed by skywind, SIYEE

 

 

 

 

 

#include<cstdio>
int n;
int a[22];
int sum=0;
int max=0;
void dfs(int i ,int ss)
{
    if(i==n-1)
    {
        ss=ss*(sum-ss);
        max=max>ss?max:ss;
        return ;
    }
    dfs(i+1,ss+a[i]);
    dfs(i+1,ss);
}
int main()
{
    while(scanf("%d",&n)==1)
    {
        sum=0;
        max=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        dfs(0,0);
        printf("%d\n",max);
    }
    return 0;
}