P6225 [eJOI2019]异或橙子

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

(A_L)^(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("%lld
",find(r,1)^find(l-1,1)) ;
				 }else{
				 printf("%lld
",find(r,0)^find(l-1,0)) ;
				 }
			}else{
				cout<<0<<endl;
			} 
		} 
	} 
	return 0;
}