POJ 2155 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].

---

【题目分析】
    还是一道区间查询的题目，看到之后很容易想到用二维的树状数组去求解，但是这道题询问的是关于1、0这两种情况，而且还要求01反转。很容易，%2求解就可以了。
   

---

【代码】

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>using namespace std;const int MAXN = 1010;int lowbit(int x){return x&(-x);}int c[MAXN][MAXN];int n;int sum(int x,int y){    int ret = 0;    for(int i = x;i > 0;i -= lowbit(i))        for(int j = y;j > 0;j -= lowbit(j))            ret += c[i][j];    return ret;}void 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))            c[i][j] += val;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        int q;        scanf("%d%d",&n,&q);        memset(c,0,sizeof(c));        char op[10];        int x1,y1,x2,y2;        while(q--)        {            scanf("%s",op);            if(op[0] == 'C')            {                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);                add(x1,y1,1);                add(x2+1,y1,1);                add(x1,y2+1,1);                add(x2+1,y2+1,1);            }            else            {                scanf("%d%d",&x1,&y1);                if(sum(x1,y1)%2 == 0)printf("0\n");                else printf("1\n");            }        }        if(T > 0)printf("\n");    }    return 0;}