Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 16797		Accepted: 6312

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

12 10C 2 1 2 2Q 2 2C 2 1 2 1Q 1 1C 1 1 2 1C 1 2 1 2C 1 1 2 2Q 1 1C 1 1 2 1Q 2 1

Sample Output

1001

Source

,Lou Tiancheng

1 //4100K    438MS    C++    1052B    2014-04-01 21:00:26 2 /* 3  4     题意： 5         给出一个n*n的 二维区间，有小矩形取反操作和对某元素求值操作，对每个操作 6     做出处理。 7      8     树状数组： 9         这题和一般的树状数组不一样，一般的是对某个值更改，对某段求值，现在是刚好10     反过来，利用反向思维，讲原先更新的操作用来求值， 求值的操作用来更新即可。 11 12 */13 #include
     
      14 #include
      
       15 #define N 100516 int c[N][N];17 int lowbit(int i)18 {19     return i&(-i);20 }21 int update(int x,int y)22 {23     int s=0;24     for(int i=x;i
       
        0;i-=lowbit(i))33         for(int j=y;j>0;j-=lowbit(j))34             c[i][j]^=1;35 }36 int main(void)37 {38     int t,n,m;39     char op;40     int x1,x2,y1,y2;41     scanf("%d",&t);42     while(t--)43     {44         scanf("%d%d%*c",&n,&m);45         memset(c,0,sizeof(c));46         while(m--){47             scanf("%c",&op);48             if(op=='Q'){49                 scanf("%d%d%*c",&x1,&y1);50                 printf("%d\n",update(x1,y1));51             }else{52                 scanf("%d%d%d%d%*c",&x1,&y1,&x2,&y2);53                 getsum(x1-1,y1-1);54                 getsum(x1-1,y2);55                 getsum(x2,y1-1);56                 getsum(x2,y2);57             }58         }59         printf("\n");60     }61     return 0;62 }