codeforces 703D

题意：求区间出现偶数次的数的异或值。序列长度是1000000，操作是1000000.

题解：求出区间不同的数的异或值，然后区间异或值^不同的数的异或值 = 出现偶数次的数的异或值。不同数的异或值用离线的方式处理，用树状数组维护。

#include<bits/stdc++.h>
using namespace std;
class FenWick{
private:
	int n;
	vector<int> d;
public:
	void init(int n){
		this->n = n+1;
		d = vector<int>(this->n,0);
	}
	int lowbit(int x){
		return x&-x;
	}
	void add(int p,int x){
		while(p<n){
			d[p]+=x;
			p+=lowbit(p);
		}
	}
	void xor_add(int p,int x){
//		cout<<p<<"----"<<x<<endl;
		while(p<n){
			d[p]^=x;
			p+=lowbit(p);
		}
	}
	int sum(int p){
		int ret = 0;
		while(p){
			ret+=d[p];
			p-=lowbit(p);
		}
		return ret;
	}
	int xor_sum(int p){
		int ret = 0;
		while(p){
			ret^=d[p];
			p-=lowbit(p);
		}
		return ret;
	}
};

struct Node{
	int x,y,id;
	int ret;
	Node(int x,int y,int id):x(x),y(y),id(id){}
	Node(){}
};
bool cmpY(const Node &a,const Node &b){
	return a.y<b.y;
}
bool cmpID(const Node &a,const Node &b){
	return a.id<b.id;
}
int main(){
	ios_base::sync_with_stdio(false);
	int n;
	while(scanf(" %d",&n)==1){
		vector<int> a(n+1,0);
		for(int i=1;i<=n;i++){
			//cin>>a[i];
			scanf(" %d",&a[i]);
		}

		int q;
		//cin>>q;
		scanf(" %d",&q);
		vector<Node> pd(q);
		for(int i=0;i<q;i++){
			//cin>>pd[i].x>>pd[i].y;
			scanf(" %d %d",&pd[i].x,&pd[i].y);
			pd[i].id=i;
		}
		sort(pd.begin(),pd.end(),cmpY);

		FenWick fw;
		fw.init(n);
		map<int,int> kk;

		for(int i=1,j=0;i<=n&&j<(int)pd.size();i++){
			if(kk.count(a[i])){
				fw.xor_add(kk[a[i]],a[i]);
			}
			kk[a[i]]=i;
			fw.xor_add(i,a[i]);
			a[i]^=a[i-1];
			//计算出不同数的xor  全部数的xor 异或它就是答案了
//			cout<<"XOR:"<<fw.xor_sum(i)<<endl;
			while(j<(int)pd.size()&&pd[j].y==i){
//				cout<<pd[j].x<<","<<pd[j].y<<endl;
//				cout<<a[i]<<","<<a[pd[j].x-1]<<","<<fw.xor_sum(i)<<","<<fw.xor_sum(pd[j].x-1)<<endl;
				pd[j].ret = a[i]^a[pd[j].x-1]^fw.xor_sum(i)^fw.xor_sum(pd[j].x-1);
				//cout<<(12^0^12^0)<<endl;
				//cout<<(a[i]^a[pd[j].x-1]^fw.xor_sum(i)^fw.xor_sum(pd[j].x-1))<<endl;
				j++;
			}
		}
		sort(pd.begin(),pd.end(),cmpID);
		for(int i=0;i<q;i++){
			//cout<<pd[i].ret<<endl;
			printf("%d\n",pd[i].ret);
		}
	}
	return 0;
}