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(%d\n,max1);
}
return 0;
}
（poj 2479）宝，都看到这里了你确定不收藏一下？？