CF198 div1 D

简单说就是左边x,y按照奇偶分为四种对于答案的影响都是不相关的

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 1e3+5;
int N,M;
ll tree[4][MAXN][MAXN];

int ju(int x, int y) {
    int tt = 0;
    if(x%2) tt ++;
    if(y%2) tt += 2;
    return tt;
}
void Update(int x,int y, ll v) {
    int ty = ju(x,y);
    for(int i = x; i <= N; i += i&-i)
        for(int j = y; j <= N; j += j&-j)
            tree[ty][i][j] ^= v;
} 
ll Sum(int x,int y) {
    int ty = ju(x,y);
    ll ans = 0;
    for(int i = x; i > 0; i -= i&-i)
        for(int j = y; j > 0; j -= j&-j)
            ans ^= tree[ty][i][j];
    return ans;
}

int main(){
    while(~scanf("%d %d",&N,&M)) {
        memset(tree ,0, sizeof(tree));

        for(int i = 1; i <= M; ++i) {
            int a,b,c,d; int ty; ll v;
            scanf("%d",&ty);
            if(ty == 2) {
                scanf("%d %d %d %d %lld",&a,&b,&c,&d,&v);
                Update(a,b,v); 
                Update(c+1,b,v);
                Update(a,d+1,v);
                Update(c+1,d+1,v);  
            }else {
                ll ans = 0;
                scanf("%d %d %d %d",&a,&b,&c,&d);
                ans ^= Sum(c,d);
                ans ^= Sum(a-1,d);
                ans ^= Sum(c,b-1);
                ans ^= Sum(a-1,b-1);
                printf("%lld
",ans);
            }
        }
    }
    return 0;   
}