PKU 2155

/*
Matrix
Time Limit: 3000MS  Memory Limit: 65536K
Total Submissions: 5565  Accepted: 2047

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

Source

*/

#include
#include
#define M 1005

int c[M][M], N;

int lowbit(int x)
{
 return x & (-x);
}

void modify(int x, int y, int delta) // 包含原数组节点a[x][y]的树状数组都要加上delta
{
 int i, j;
 for (i = x; i <= N; i += lowbit(i))
 {
  for (j = y; j <= N; j += lowbit(j))
  {
   c[i][j] += delta;
  }
 }
}

int sum(int x, int y)
{
 int res = 0, i, j;
 
 for (i = x; i > 0; i -= lowbit(i))
 {
  for (j = y; j > 0; j -= lowbit(j))
  {
   res += c[i][j];
  }
 }
 return res;
}

int main()
{
 int t, rec;
 
 scanf("%d", &t);
 rec = t;
 while (t--)
 {
  if (t != rec - 1)
  {
   putchar('/n');
  }
  int cmd;
  memset(c, 0, sizeof(c));
  scanf("%d %d", &N, &cmd);
  while (cmd--)
  {
   char optr[10];
   scanf("%s", optr);
   if (optr[0] == 'C')
   {
    int x1, y1, x2, y2;
    scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
    modify(x1, y1, 1);
    modify(x2 + 1, y2 + 1, 1);
    modify(x1, y2 + 1, 1);
    modify(x2 + 1, y1, 1);
   }
   else
   {
    int x, y;
    scanf("%d %d", &x, &y);
    printf("%d/n", sum(x, y) % 2);
   }
  }
 }
 return 0;
}