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
这道题确实很经典，尤其在这个二进制的计算方面
详细的可以参考《浅谈信息学竞赛中的“0”和“1”》此论文，网上很多说的并不详细，大多只介绍了翻转，并没有介绍为何sum(x,y)%2能得到结果
论文里很详细的证明了
#include <iostream>#include <stdio.h>#include <string.h>#include <string>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <math.h>#include <bitset>#include <list>#include <algorithm>#include <climits>using namespace std;#define lson 2*i#define rson 2*i+1#define LS l,mid,lson#define RS mid+1,r,rson#define UP(i,x,y) for(i=x;i<=y;i++)#define DOWN(i,x,y) for(i=x;i>=y;i–)#define MEM(a,x) memset(a,x,sizeof(a))#define W(a) while(a)#define gcd(a,b) __gcd(a,b)#define LL long long#define N 1005#define INF 0x3f3f3f3f#define EXP 1e-8#define lowbit(x) (x&-x)const int mod = 1e9+7;int c[N][N],n,m,cnt,s,t;int a[N][N];int sum(int x,int y){int ret = 0;int i,j;for(i = x;i>=1;i-=lowbit(i)){for(j = y;j>=1;j-=lowbit(j)){ret+=c[i][j];}}return ret;}void add(int x,int y){int i,j;for(i = x;i<=n;i+=lowbit(i)){for(j = y;j<=n;j+=lowbit(j)){c[i][j]++;}}}int main(){int i,j,x,y,ans,t;int x1,x2,y1,y2;char op[10];scanf("%d",&t);while(t–){scanf("%d%d",&n,&m);MEM(c,0);MEM(a,0);while(m–){scanf("%s",op);if(op[0]=='C'){scanf("%d%d%d%d",&x1,&y1,&x2,&y2);x1++,y1++,x2++,y2++;add(x2,y2);add(x1-1,y1-1);add(x2,y1-1);add(x1-1,y2);}else{scanf("%d%d",&x1,&y1);x2 = x1,y2 = y1;x1++,y1++,x2++,y2++;printf("%d\n",sum(x1,y1));}}printf("\n");}return 0;}
，大海，别为森林的渺小而沮丧，
