POJ

Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1 <= i, j <= N).

We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a '0' then change it into '1' otherwise change it into '0'). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions.

1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2).
2. Q x y (1 <= x, y <= n) querys A[x, y].

Input

The first line of the input is an integer X (X <= 10) representing the number of test cases. The following X blocks each represents a test case.

The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2 y2", which has been described above.

Output

For each querying output one line, which has an integer representing A[x, y].

There is a blank line between every two continuous test cases.

Sample Input

1
2 10
C 2 1 2 2
Q 2 2
C 2 1 2 1
Q 1 1
C 1 1 2 1
C 1 2 1 2
C 1 1 2 2
Q 1 1
C 1 1 2 1
Q 2 1

Sample Output

1
0
0
1

代码：

#include <cstdio>
#include <cstring>

using namespace std;

const int MAXN = 1050;

int N,T;
int tree[2*MAXN][2*MAXN];

int lowbit(int x)
{
        return x&(-x);
}
int Add(int x,int y,int val)
{
        for (int i=x ; i<=N ; i+=lowbit(i))
                for (int j=y ; j<=N ; j+=lowbit(j))
                        tree[i][j] += val;
}
int Query(int x,int y)
{
        int ans=0;
        for (int i=x ; i>0 ; i-=lowbit(i))
                for (int j=y ; j>0 ; j-=lowbit(j))
                        ans += tree[i][j];
        return ans;
}

int main(){
	
	int X;
	scanf("%d",&X);
	char s[10];
	int a,b,c,d;
	while(X--){
		scanf("%d %d",&N,&T);
		memset(tree,0,sizeof tree);
		while(T--){
			scanf("%s",s);
			if(s[0] == 'C'){
				scanf("%d %d %d %d",&a,&b,&c,&d);
				Add(a,b,1);//这四步是关键，仔细理解。 
				Add(c+1,b,1);
				Add(a,d+1,1);
				Add(c+1,d+1,1);
			}
			else {
				scanf("%d %d",&a,&b);
				if(Query(a,b)%2 == 0)printf("0
");
				else printf("1
");
			}
		}
		if(X)printf("
");
	}
	
	return 0;
}