北大oj--1050

/*To the MaxTime Limit: 1000MSMemory Limit: 10000KTotal Submissions: 40480Accepted: 21437DescriptionGiven 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. InputThe 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].OutputOutput the sum of the maximal sub-rectangle.Sample Input40 -2 -7 0 9 2 -6 2-4 1 -4  1 -18  0 -2Sample Output15题意：程序要求输出由矩形圈起来的所有值的和为最大，为负数时，输出0 采用压缩法，将多行压缩成一行，进行判断，即，将二维数组转换为一维数组来进行判断 */ #include<stdio.h>#include<string.h>int a[101][101],n,temp[101];int solve(){    int i,sum=0,max=0;    for(i=0;i<n;i++)    {        sum+=temp[i];        if(sum<0)sum=0;        if(sum>max)max=sum;    }    return max;}int main(){    int i,j,k,max;    while(scanf("%d",&n)!=EOF)    {        for(i=0;i<n;i++)            for(j=0;j<n;j++)                scanf("%d",&a[i][j]);        max=0;        for(i=0;i<n;i++)        {            memset(temp,0,sizeof(temp));            for(j=i;j<n;j++)            {                for(k=0;k<n;k++)                {                    temp[k]+=a[j][k];                }                if(max<solve())max=solve();            }        }        printf("%d\n",max);    }    return 0;}