Count the rectangles
TimeLimit: 2 Second MemoryLimit: 64 Megabyte
Totalsubmit: 64 Accepted: 36
Description
You are given numbers of rectangles made of '1's and '0's.Calculate how many small rectangles which have four '1's on the four corners.Input
There are several test cases, each test case starts with a line containing two positive integers n and m. n and m is the size of the rectangle (1<=n<=100, 1<=m<=100). Next follow a rectangle which contains only number '0' and '1'. The input will finish with the end of file.Output
The number of rectangles which meet the requirements.Sample Input
3 4
1 0 1 0
0 1 1 0
1 1 1 0
Sample Output
2
Source
zzx
思路：
代码：
#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> using namespace std; int map[110][110]; int the_last_sum; int the_long_x; int the_long_y; int n,m; int main() { while(scanf("%d%d",&n,&m) != EOF) { memset(map,0,sizeof(map)); for(int i = 1;i <= n;i ++) for(int j = 1;j <= m;j ++) scanf("%d",&map[i][j]); the_last_sum = 0; for(int i = 1;i <= n;i ++) for(int j = 1;j <= m;j ++) { the_long_x = 0; the_long_y = 0; if(map[i][j] == 1 && j + 1 <= m && i + 1 <= n) { for(int k = j + 1;k <= m;k ++) { if(map[i][k] == 1) { the_long_y = k - j; for(int l = i + 1;l <= n;l ++) { if(map[l][j] == 1) { the_long_x = l - i; if(the_long_x != 0 && the_long_y != 0) { if(i + the_long_x <= n && j + the_long_y <= m && map[i + the_long_x][j + the_long_y] == 1) the_last_sum ++; } } } } } } } printf("%d ",the_last_sum); } return 0; }