二维数组数组（区间更新+单点查询）

http://poj.org/problem?id=2155

题意：给出一个矩阵，初始为0，两种操作，1、C x1 y1 x2 y2 将该矩阵元素0变1，1变0. 2、Q x y 询问该点值。

解法：二维前缀和：

我们可以令差分数组表示与 的差。

二维差分：https://www.cnblogs.com/LMCC1108/p/10753451.html

//#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
typedef long long ll ;
#define int ll
#define mod 1000000007
#define gcd __gcd
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j)  for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//ll lcm(ll a , ll b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define SC scanf
#define INF  0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
using namespace std;
const int N = 1e6+100;
const int maxn = 1e3+9;
int a[maxn][maxn] , c[maxn][maxn];
char str;
int n , q ;
int cnt  ;
int lowerbit(int x){
    return x&(-x);
}

void add(int x , int y , int val){
    while(x <= n){
        int j = y ;
        while(j <= n){
            c[x][j] += val;
            j += lowerbit(j);
        }
        x += lowerbit(x);
    }
}

int getsum(int x , int y){
    int ans = 0 ;
    while(x){
        int j = y ;
        while(j){
            ans += c[x][j];
            j -= lowerbit(j);
        }
        x -= lowerbit(x);
    }
    return ans ;
}
void init(){
    ME(c , 0);
}
void solve(){
    if(cnt) cout << endl;
    cnt++;
    init();
    scanf("%lld%lld" , &n , &q);
    while(q--){
        cin >> str;
        if(str == 'C'){
            int x1 , y1 , x2 , y2 ;
            scanf("%lld%lld%lld%lld" , &x1 , &y1 , &x2 , &y2);
            add(x1 , y1 , 1);
            add(x1 , y2+1 , -1);
            add(x2+1 , y1 , -1);
            add(x2+1 , y2+1 , 1);
        }else{
            int x , y ;
            scanf("%lld%lld" , &x , &y);
            cout << (getsum(x , y)&1) << endl;
        }
    }
}

signed main()
{
    //ios::sync_with_stdio(false);
    //cin.tie(0); cout.tie(0);
    cnt = 0;
    int t ;
    scanf("%lld" , &t);
    while(t--){
        solve();
    }
}