【树状数组】[CodeForces - 341D]Iahub and Xors

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CodeForces - 341D
分析：异或有一个性质：a^b=c,a^c=b,b^c=a;
然后就可以，直接用二维树状数组，具体还可以参见上帝造题的七分钟。

代码：

#include<cstdio>#define MAXN 1000int n,m,c[MAXN+10][MAXN+10],ci[MAXN+10][MAXN+10],cj[MAXN+10][MAXN+10],cij[MAXN+10][MAXN+10];void Read(int &x){    char c;    while(c=getchar(),c!=EOF)        if(c>='0'&&c<='9'){            x=c-'0';            while(c=getchar(),c>='0'&&c<='9')                x=x*10+c-'0';            ungetc(c,stdin);            return;        }}inline int lowbit(int x){    return x&-x;}void update(int c[][MAXN+10],int x,int y,int d){    int j;    for(;x<=n;x+=lowbit(x))        for(j=y;j<=n;j+=lowbit(j))            c[x][j]^=d;}int getsum(int c[][MAXN+10],int x,int y){    int j,ret=0;    for(;x;x-=lowbit(x))        for(j=y;j;j-=lowbit(j))            ret^=c[x][j];    return ret;}int main(){    Read(n),Read(m);    int p,x1,y1,x0,y0,v,a1,a2,a3,a4;    while(m--){        Read(p);        if(p==2){            Read(x0),Read(y0),Read(x1),Read(y1),Read(v);            update(c,x0,y0,v);            update(ci,x0,y0,x0&1?v:0);            update(cj,x0,y0,y0&1?v:0);            update(cij,x0,y0,(x0*y0)&1?v:0);            update(c,x0,y1+1,v);            update(ci,x0,y1+1,x0&1?v:0);            update(cj,x0,y1+1,(y1+1)&1?v:0);            update(cij,x0,y1+1,(x0*(y1+1))&1?v:0);            update(c,x1+1,y0,v);            update(ci,x1+1,y0,(x1+1)&1?v:0);            update(cj,x1+1,y0,y0&1?v:0);            update(cij,x1+1,y0,((x1+1)*y0)&1?v:0);            update(c,x1+1,y1+1,v);            update(ci,x1+1,y1+1,(x1+1)&1?v:0);            update(cj,x1+1,y1+1,(y1+1)&1?v:0);            update(cij,x1+1,y1+1,((x1+1)*(y1+1))&1?v:0);        }        else{            Read(x0),Read(y0),Read(x1),Read(y1);            a1=(((x1+1)*(y1+1))&1?getsum(c,x1,y1):0)^((y1+1)&1?getsum(ci,x1,y1):0)^((x1+1)&1?getsum(cj,x1,y1):0)^getsum(cij,x1,y1);            a2=((x0*(y1+1))&1?getsum(c,x0-1,y1):0)^((y1+1)&1?getsum(ci,x0-1,y1):0)^(x0&1?getsum(cj,x0-1,y1):0)^getsum(cij,x0-1,y1);            a3=(((x1+1)*y0)&1?getsum(c,x1,y0-1):0)^(y0&1?getsum(ci,x1,y0-1):0)^((x1+1)&1?getsum(cj,x1,y0-1):0)^getsum(cij,x1,y0-1);            a4=((x0*y0)&1?getsum(c,x0-1,y0-1):0)^(y0&1?getsum(ci,x0-1,y0-1):0)^(x0&1?getsum(cj,x0-1,y0-1):0)^getsum(cij,x0-1,y0-1);            printf("%d\n",a1^a2^a3^a4);        }    }}

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