Matrix

Description

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

解析：

       二维树状数组。

       我们可以记录下每个点被修改的次数，最终答案即为总修改次数模2，于是就可以用二维树状数组完成。

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
using namespace std;

const int Max=1010;
int t,n,m,k;
int sum[Max][Max];

inline int get_int()
{
	int x=0,f=1;
	char c;
	for(c=getchar();(!isdigit(c))&&(c!='-');c=getchar());
	if(c=='-') {f=-1;c=getchar();}
	for(;isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+c-'0';
	return x*f;
}

inline void add(int x,int y,int num)
{
	for(int i=x;i<=n;i+=i&-i)
	  for(int j=y;j<=n;j+=j&-j)
	    sum[i][j] += num;
}

//注意这种写法是错误的！因为 y 的值会改变！！
/*
inline int Q(int x,int y)
{
	int ans=0;
	for(;x;x-=x&-x)
	  for(;y;y-=y&-y)
	    ans+=sum[x][y];
	return ans;
}
*/

inline int Q(int x,int y)
{
	int ans=0;
	for(int i=x;i;i-=i&-i)
	  for(int j=y;j;j-=j&-j)
	    ans+=sum[i][j];
	return ans;
}

int main()
{
	freopen("lx.in","r",stdin);

	t=get_int();
	while(t--)
	{
	  n=get_int(),m=get_int();
	  memset(sum,0,sizeof(sum));
	  while(m--)
	  {
		char ch=getchar();
		if(ch == 'C')
		{
		  int x1=get_int(),y1=get_int(),x2=get_int(),y2=get_int();
		  add(x1,y1,1),add(x1,y2+1,-1),add(x2+1,y1,-1),add(x2+1,y2+1,1);
		}
		else
		{
		  int x=get_int(),y=get_int();
		  cout<<Q(x,y)%2<<"\n";
		}
	  }
	  cout<<"\n";
	}

	return 0;
}