TO THE MAX

Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*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. 

Input

The input consists of an N * 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.

Sample Input

40 -2 -7 0 9 2 -6 2-4 1 -4  1 -18  0 -2

Sample Output

15

#include<stdio.h>#include<algorithm>#include<iostream>#include<string.h>#include<stdlib.h>#include<queue>#include<vector>#include<math.h>#include<map>using namespace std;const int MAX = 100+10;int main(){   int a[MAX][MAX], n, b[MAX][MAX], t, maxx, sum;    scanf("%d", &n);    for(int i = 1; i<=n; ++i)for(int j = 1;j<=n; ++j){scanf("%d", &t);a[i][j] = a[i-1][j]+t;}maxx = sum = 0;for(int i=  1; i<=n;++i){for(int j = i; j<=n;++j){sum = 0;for(int k =1; k<=n;++k){sum+=a[j][k]-a[i][k];if(sum<0)sum = 0;if(sum>maxx)maxx = sum;}}}printf("%d", maxx);    return 0;}