POJ 2155 （二维树状数组）

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题目大意：矩阵里面开始全是0，‘C x1,y1,x2,y2’就是将左上角为x1,y1,右下角为x2,y2,的这个矩阵内的数字进行变换，0变1,1变0；'Q x1 y1',输出a[x1][y1]的值。

解题思路：经典的二维树状数组，重叠消元，利用0和1之间变换特性，答案对2取余即可。

/* ***********************************************┆  ┏┓　　　┏┓ ┆┆┏┛┻━━━┛┻┓ ┆┆┃　　　　　　　┃ ┆┆┃　　　━　　　┃ ┆┆┃　┳┛　┗┳　┃ ┆┆┃　　　　　　　┃ ┆┆┃　　　┻　　　┃ ┆┆┗━┓　马　┏━┛ ┆┆　　┃　勒　┃　　┆　　　　　　┆　　┃　戈　┗━━━┓ ┆┆　　┃　壁　　　　　┣┓┆┆　　┃　的草泥马　　┏┛┆┆　　┗┓┓┏━┳┓┏┛ ┆┆　　　┃┫┫　┃┫┫ ┆┆　　　┗┻┛　┗┻┛ ┆************************************************ */#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>using namespace std;#define rep(i,a,b) for (int i=(a),_ed=(b);i<=_ed;i++)#define per(i,a,b) for (int i=(b),_ed=(a);i>=_ed;i--)#define pb push_back#define mp make_pairconst int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define mod 1000000007#define LL long long#define ULL unsigned long long#define MS0(X) memset((X), 0, sizeof((X)))#define SelfType intSelfType Gcd(SelfType p,SelfType q){return q==0?p:Gcd(q,p%q);}SelfType Pow(SelfType p,SelfType q){SelfType ans=1;while(q){if(q&1)ans=ans*p;p=p*p;q>>=1;}return ans;}#define Sd(X) int (X); scanf("%d", &X)#define Sdd(X, Y) int X, Y; scanf("%d%d", &X, &Y)#define Sddd(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")int n,m;int c[1005][1005];int lowbit(int x){    return x & (-x);}void update(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 sum(int x,int y){    int ans = 0;    for(int i=x;i;i-=lowbit(i))    {        for(int j=y;j;j-=lowbit(j))        {            ans += c[i][j];        }    }    return ans;}int main(){//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);ios::sync_with_stdio(0);cin.tie(0);Sd(t);while(t--)    {        MS0(c);        n = read(),m = read();        while(m--)        {            char c;            scanf("%c",&c);            if(c=='C')            {                int x1,x2,y1,y2;                x1 = read(),y1 = read();                x2 = read(),y2 = read();                update(x1,y1,1);                update(x2+1,y1,1);                update(x1,y2+1,1);                update(x2+1,y2+1,1);            }            else            {                int x,y;                x = read(),y = read();                printf("%d\n",sum(x,y)%2);            }        }        printf("\n");    }return 0;}

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