To the Max

         To the Max
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Time Limit: 1 Second      Memory Limit: 32768 KB
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Problem
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.
As an example, the maximal sub-rectangle of the array:

0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2

is in the lower left corner:

9 2
-4 1
-1 8

and has a sum of 15.
The input consists of an N x N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N 2 integers separated by whitespace (spaces and newlines). These are the N 2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].

Output
  Output the sum of the maximal sub-rectangle.

Example Input
4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2
Output
15

#include<stdio.h>
#include<string.h>
#define MAXN 105
int main()
{
    //freopen("a.txt","r",stdin);
    int i,j,k,n,t,sum,max;
    int a[MAXN][MAXN];
    while (scanf("%d",&n)!=EOF)
    {
        memset(a,0,sizeof(a));
        for (i=1;i<=n;++i)
        {
            for (j=1;j<=n;++j)
            {
                scanf("%d",&t);
                a[i][j]=a[i-1][j]+t;
            }
        }
        max=0;
        for (i=1;i<=n;++i)
        {
            for (j=i;j<=n;++j)
            {
                sum=0;
                for (k=1;k<=n;++k)
                {
                    t=a[j][k]-a[i-1][k];
                    sum+=t;
                    if (sum<0) sum=0;
                    if (sum>max) max=sum;
                }
            }
        }
        printf("%d
",max);
    }
    return 0;
}