bzoj 3261: Maximum XOR sum

3261: Maximum XOR and

Time Limit: 10 Sec  Memory Limit: 512 MB
Submit: 2637  Solved: 1078
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Description

Given a sequence of non-negative integers {a}, initial length N.

There are M operations, and there are two types of operations:

1. Ax: Add operation, which means adding a number x at the end of the sequence, and the length of the sequence is N+1.

2. Qlrx: Inquiry operation, you need to find a position p, satisfying l<=p<=r, so that:

a[p] xor a[p+1] xor ... xor a[N] xor x is the largest, and the output is the largest.

Input

The first line contains two integers N, M with the meanings shown in the problem description.  The second line contains N non-negative integers representing the initial sequence A . Next M lines, each line describes an operation, the format is as described in the title.    
 

Output

Assuming that there are T query operations, the output should have T lines, each with an integer representing the answer to the query.

Sample Input

5 5
2 6 4 3 6
A 1
Q 3 5 4
A 4
Q 5 7 0
Q 3 6 6
For test points 1-2, N,M<=5.
For test points 3-7, N,M<=80000.
For test points 8-10, N,M<=300000.
Among them, test points 1, 3, 5, 7, 9 guarantee no modification operation.
0<=a[i]<=10^7.

Sample Output

4
5
6

 

    Just change the suffix query to the total XOR prefix, and then write a persistent trie.

 

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=300005;
struct node{
	int Sum
	node *ch[2];
}nil[maxn*73],*rot[maxn*4],*cnt;
int now,Xor,N,L,R,X,M,ci[33],ans;
char C;

node *update(node *u,int x){
	node * ret = ++ cnt;
	* ret = * u, ret-> Sum ++;
	if(x<0) return ret;
	
	if(Xor&ci[x]) ret->ch[1]=update(ret->ch[1],x-1);
	else ret->ch[0]=update(ret->ch[0],x-1);
	
	return ret;
}

void query(node *l,node *r,int x){
	if(x<0) return;
	int u=(ci[x]&X)?0:1;
	if(r->ch[u]->Sum-l->ch[u]->Sum) ans+=ci[x],query(l->ch[u],r->ch[u],x-1);
	else query(l->ch[u^1],r->ch[u^1],x-1);
}

inline void solve(){
	while(M--){
		C=getchar();
		while(C!='A'&&C!='Q') C=getchar();
		if(C=='A') N++,scanf("%d",&now),Xor^=now,rot[N]=update(rot[N-1],25);
		else{
			scanf("%d%d%d",&L,&R,&X);
			X ^ = Xor, ans = 0;
			query(rot[L-1],rot[R],25);
			printf("%d\n",ans);
		}
	}
}

int main(){
	ci [0] = 1;
	for(int i=1;i<=25;i++) ci[i]=ci[i-1]<<1;
	
	scanf("%d%d",&N,&M);
	nil->Sum=0,N++;
	nil->ch[0]=nil->ch[1]=cnt=rot[0]=nil;
	rot[1]=update(rot[0],25);
	
	for(int i=2;i<=N;i++){
		scanf("%d",&now),Xor^=now;
		rot[i]=update(rot[i-1],25);
	}
	
	solve();
	
	return 0;
}