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 the "the elements in" that rectangle. The problem the sub-rectangle with the largest sum are 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:

7 |
-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 in a line by itself, indicating the size of the square two-dimensional a Rray. This was followed by N 2 integers separated by whitespace (spaces and newlines). These is the N 2 integers of the array, presented in Row-major order. That's, all numbers on the first row, left-to-right, then all numbers-second row, left-to-right, etc. N may be as large as 100. The numbers in the array would 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

1#include <stdio.h>2#include <string.h>3 #defineMAXN 1054 intMain ()5 {6     //freopen ("A.txt", "R", stdin);7     intI,j,k,n,t,sum,max;8     intA[MAXN][MAXN];9      while(SCANF ("%d", &n)! =EOF)Ten     { OneMemset (A,0,sizeof(a)); A          for(i=1; i<=n;++i) -         { -              for(j=1; j<=n;++j) the             { -scanf"%d",&t); -a[i][j]=a[i-1][j]+T; -             } +         } -max=0; +          for(i=1; i<=n;++i) A         { at              for(j=i;j<=n;++j) -             { -sum=0; -                  for(k=1; k<=n;++k) -                 { -t=a[j][k]-a[i-1][k]; insum+=T; -                     if(sum<0) sum=0; to                     if(Sum>max) max=sum; +                 } -             } the         } *printf"%d\n", max); $     }Panax Notoginseng     return 0; -}

AC

to the Max