poj2155

二维树状数组，对于一个矩阵，迅速修改其一个点的值，迅速求其左上角连续子矩阵的和。对于每个翻转修改矩阵的四个角，对于询问求出该点的左上角矩阵和看奇偶性。

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

#define maxn 1005

int c[maxn][maxn];
int Row, Col;

inline int Lowbit(const int &x)
{
	return x & (-x);
}

int Sum(int i, int j)
{
	int tempj, sum = 0;
	while (i > 0)
	{
		tempj = j;
		while (tempj > 0)
		{
			sum += c[i][tempj];
			tempj -= Lowbit(tempj);
		}
		i -= Lowbit(i);
	}
	return sum;
}

void Update(int i, int j, int num)
{
	int tempj;
	while (i <= Row)
	{
		tempj = j;
		while (tempj <= Col)
		{
			c[i][tempj] += num;
			tempj += Lowbit(tempj);
		}
		i += Lowbit(i);
	}
}

int main()
{
//	freopen("t.txt", "r", stdin);
	int x, n, t;
	scanf("%d", &x);
	while (x--)
	{
		memset(c, 0, sizeof(c));
		scanf("%d%d", &n, &t);
		getchar();
		Row = n;
		Col = n;
		for (int i = 0; i < t; i++)
		{
			char ch;
			int x1, x2, y2, y1;
			scanf("%c", &ch);
			if (ch == 'C')
			{
				scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
				x2++;
				y2++;
				Update(x1, y1, 1);
				Update(x1, y2, -1);
				Update(x2, y1, -1);
				Update(x2, y2, 1);
			}
			else
			{
				scanf("%d%d", &x1, &y1);
				printf("%d\n", (1 & Sum(x1, y1)));
			}
			getchar();
		}
		printf("\n");
	}
	return 0;
}