To the Max

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.

Input

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

Output

15

代码如下：

 # include<stdio.h>

 # include<string.h>

 # define N

 int main(){

     int a[N][N],b[N];

     int n,i,j,k;

     while(scanf("%d",&n)!=EOF)

     {

         for(i=;i<n;i++)

             for(j=;j<n;j++)

                 scanf("%d",&a[i][j]);

             int max= -;

             for(i=;i<n;i++)    //第i行到第j行的最大子矩阵和

             {

                 memset(b,,sizeof(b));

                 for(j=i;j<n;j++)

                 {

                     int sum=;

                     for(k=;k<n;k++)

                     {

                         b[k] += a[j][k];

                         sum += b[k];

                         if(sum<)  sum=b[k];

                         if(sum>max)  max=sum;

                     }

                 }

             }

             printf("%d\n",max);

     }

     return ;

 }