poj 2479

poj 2479

Maximum sum

Time Limit: 1000MS

Memory Limit: 65536K

Total Submissions: 45591

Accepted: 14088

Description

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).

Input

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.

Output

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

Sample Input

1

10

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

Sample Output

13

Hint

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

Huge input,scanf is recommended.

这个题目也是一个最大和问题但是又不太一样，这个是从两边开始的，求的是两边最大和的最大值，所以说要在中间找一个点作为分割点（一个个试），再求两边的最大和，但是用之前的那个方法一个个算会超时，所以说我们建立两个数组，来记录每个点两边的最大和，即可一步完成

#includestdio.h

#includealgorithm

using namespace std;

#define MIN -0x3f3f3f3f

int main()

{

int a[50010],left[50010],right[50010];

int n;

//freopen(F:/y.txt,r,stdin);

int t;

scanf(%d,t);

while(t--)

{

scanf(%d,n);

for(int i=0;in;i++)

{

scanf(%d,a[i]);

}

left[0]=a[0];

for(int i=1;in;i++)

{

if(left[i-1]0)

left[i]=left[i-1]+a[i];

else

left[i]=a[i];

}

for(int i=1;in;i++)

{

if(left[i]left[i-1])

left[i]=left[i-1];

}

right[n-1]=a[n-1];

for(int i=n-2;i=0;i--)

{

if(right[i+1]0)

right[i]=right[i+1]+a[i];

else

right[i]=a[i];

}

for(int i=n-2;i=0;i--)

if(right[i]right[i+1])

right[i]=right[i+1];

int max1=MIN;

for(int i=0;in-1;i++)

{

max1=max(max1,left[i]+right[i+1]);

}

printf(%dn,max1);

}

return 0;

}

（poj 2479）宝，都看到这里了你确定不收藏一下？？