学步园

Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below:

 

 

 

Your task is to calculate d(A).

The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input.
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case.

Print exactly one line for each test case. The line should contain the integer d(A).

1

10
1 -1 2 2 3 -3 4 -4 5 -5

13

In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer.

Huge input,scanf is recommended.

 

 

 

#include <iostream>

using namespace std;

int data[32];
int sumCal[100][100];
int n;

int calmax(int low,int high)//求连续子串最大和的函数;
{
	if(sumCal[low][high]!=0) return sumCal[low][high];
	
	int maxTohere=0, maxAll=0;
	for(int i=low; i<=high; i++)
	{
		maxTohere += data[i];
		if(maxTohere<0) maxTohere=0;
		if(maxTohere>maxAll) maxAll=maxTohere;
	}
	sumCal[low][high]=maxAll;
	return sumCal[low][high];
}

int divide( int k)
{
	int sum1=0,sum2=0,sum=0;
	sum1=calmax(0,k);
	sum2=calmax(k+1,n-1);
	sum=sum1+sum2;
	return sum;
}

int main()
{
	memset(data,-1,sizeof(data));
	
	cin>>n;
	for(int i=0; i<n; i++) cin>>data[i];
	int max=0, tmp=0;
	for (int j=0; j<n; j++)
	{
		tmp=divide(j);//从不同位置划分成两个子串
		if(tmp>max) max=tmp;
	}
	cout<<max;
}