poj2479,

## Problem:

Maximum sum
 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 20936 Accepted: 6347
Description
Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below: 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.
Source
POJ Contest,Author:Mathematica@ZSU

## Source Code:

#include <stdio.h>
int main()
{
freopen("input.in", "r", stdin);
freopen("output.out", "w", stdout);
int t, n, i, max, sum;
int a, b, c;
scanf("%d", &t);
while (t--)
{
scanf("%d", &n);
max = 0;
sum = 0;
if (n == 1)
{
scanf("%d", &a);
printf("%d\n", a);
continue;
}
if (n == 2)
{
scanf("%d", &a);
scanf("%d", &a);
printf("%d\n", a + a);
continue;
}
//b[i] is the max sum from 0 to i.
for (i = 0; i < n; i++)
{
scanf("%d", &a[i]);
sum += a[i];
if (sum < 0)
sum = 0;
else if (sum > max)
max = sum;
b[i] = max;
}
//c[i] is the max sum from i to the end.
max = 0;
sum = 0;
for (i = n - 1; i >= 0; i--)
{
sum += a[i];
if (sum < 0)
sum = 0;
else if (sum > max)
max = sum;
c[i] = max;
}
max = 0;
for (i = 0; i < n - 1; i++)
if (b[i] + c[i + 1] > max)
max = b[i] + c[i + 1];
printf("%d\n", max);
}

return (0);
}