题目：
A problem that is simple to solve in one dimension is often much more difficult to solve in more than one dimension. Consider satisfying a boolean expression in conjunctive normal form in which each conjunct consists of exactly 3 disjuncts. This problem (3-SAT) is NP-complete. The problem 2-SAT is solved quite efficiently, however. In contrast, some problems belong to the same complexity class regardless of the dimensionality of the problem.
Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the subrectangle with the largest sum is referred to as the maximal sub-rectangle. A sub-rectangle is any contiguous sub-array of size 1 × 1 or greater located within the whole array. 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-hand corner:
9 2
−4 1
−1 8
and has the 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 N2 integers separated by white-space (newlines and spaces). These N2 integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on 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
The output is the sum of the maximal sub-rectangle.
Sample Input
4 0 -2 -7
0 9 2 -6
2 -4 1 -4
1 -1 8 0 -2
Sample Output
15
题意：
给你一个矩阵，让你求出矩阵中和最大的值。
代码如下：
#include<stdio.h>
#include<string.h>
int n,a[200][200];//储存矩阵;
int sum[200];
int main()
{
while(~scanf("%d",&n))
{
int i,j,k,h;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
scanf("%d",&a[i][j]);
}
}
int min=-99999999;//极小化；
for(i=0;i<n;i++)
{
memset(sum,0,sizeof sum);
for(j=i;j<n;j++)
{
h=0;
for(k=0;k<n;k++)
{
sum[k]+=a[j][k];
if(h>=0)//如果和小于0就不用接着算了；
h+=sum[k];//很像数列求和的方法；
else
h=sum[k];
if(h>min)//储存最大的值；
min=h;
}
}
}
printf("%d\n",min);
}
return 0;
}