Miku

画一画图就能知道规律

倘若询问区间是偶数

画图可知，答案为零

为奇数，则为

$AL$^$A{L+2}$·······$A_R$

然后奇数偶数分别开一个 树状数组

就行了

#include<iostream> 
#include<cstdio>
#include<algorithm>
using namespace std;
long long n,q;
long long k;
long long ans;
long long  l,r;
long long  tree[200001][2];
long long a[200001];
long long  lowbit(long long  x){
    return x&-x;
}
long long find(long long x,long long f){
    long long ans=0;
    while(x){
        ans^=tree[x][f];
        x-=lowbit(x); 
    }
    return ans; 
}
void up(long long x,long long  y,long long f){
    while(x<=n){
        tree[x][f]^=y;
        x+=lowbit(x);
    }
    return ;
}
long long x;
int main(){
    scanf("%lld%lld",&n,&q); 
    for(long long i=1;i<=n;++i){
        scanf("%lld",&a[i]);
        if(i%2){
            up(i,a[i],1);
        }else{
            up(i,a[i],0);
        }       
    }
    for(long long i=1;i<=q;++i){
        scanf("%lld%lld%lld",&k,&l,&r); 
        if(k==1)
        if(l%2){
            up(l,a[l]^r,1);// 因为是异或
            //所以说这样可以抵消原来的数 a[l] 
            a[l]=r;
        }else{
            up(l,a[l]^r,0);
            a[l]=r;
        }
        if(k==2){
            if((r-l+1)%2){
                if(l%2){
                 printf("%lldn",find(r,1)^find(l-1,1)) ;
                 }else{
                 printf("%lldn",find(r,0)^find(l-1,0)) ;
                 }
            }else{
                cout<<0<<endl;
            } 
        } 
    } 
    return 0;
}