树状数组扩展（异或求和）

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题目：逃票的chanming

题意：操作0：读入p,q,v，并且a[p] ^= v, a[p + 1] ^= v, .. ,a[q] ^= v;操作1: 读入p,q,输出s = a[p] ^ a[p + 1] ^ a[p + 2]..^a[q]的结果;

思路：

代码：

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <math.h>#include <iostream>#include <algorithm>#include <queue>#include <stack>#include <vector>#include <set>#include <string>using namespace std;#define For(i,a) for(i=0;i<a;i++)#define Foru(i,a,b) for(i=a;i<=b;i++)#define Ford(i,a,b) for(i=a;i>=b;i--)#define clr(ar,vel) memset(ar,vel,sizeof(ar))#define PB push_back#define maxint 0x7fffffff#define maxn 500010template <int SZ>class BIT{    long long c[SZ+10];public :    void clear(){clr(c,0);}    long long getsum(int x){        long long s=0;        while(x>0)            s^=c[x], x-=x&-x;        return s;    }    void update(int x, int n){        while(x<=SZ)            c[x]^=n,x+=x&-x;    }};BIT <maxn> sum, sumi;int main(){int n, m;int act, s, t, v;//for(int i = 0; i < 10; i ++) cout << (i&1) << endl;scanf("%d%d",&n,&m);for(int i = 0; i < m; i ++) {scanf("%d%d%d",&act,&s,&t);if( !act ) {scanf("%d",&v);sum.update(s,v);sumi.update(s,v*(s&1));sum.update(t+1,v);sumi.update(t+1,v*((t+1)&1));}else{//cout << sum.getsum(t) << ' ' <<  sumi.getsum(t) << endl; int w = ((t+1)&1)*sum.getsum(t)^sumi.getsum(t);//cout << w << endl;w ^= (s&1)*sum.getsum(s-1)^sumi.getsum(s-1);printf("%d\n",w);}}return 0;}

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