Codeforces Round #367 (Div. 2) D. Vasiliy's Multiset （tire 树）

D. Vasiliy's Multiset

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Author has gone out of the stories about Vasiliy, so here is just a formal task description.

You are given q queries and a multiset A, initially containing only integer 0. There are three types of queries:

1. "+ x" — add integer x to multiset A.
2. "- x" — erase one occurrence of integer x from multiset A. It's guaranteed that at least one x is present in the multiset A before this query.
3. "? x" — you are given integer x and need to compute the value , i.e. the maximum value of bitwise exclusive OR (also know as XOR) of integer x and some integer y from the multiset A.

Multiset is a set, where equal elements are allowed.

Input

The first line of the input contains a single integer q (1 ≤ q ≤ 200 000) — the number of queries Vasiliy has to perform.

Each of the following q lines of the input contains one of three characters '+', '-' or '?' and an integer xi (1 ≤ xi ≤ 109). It's guaranteed that there is at least one query of the third type.

Note, that the integer 0 will always be present in the set A.

Output

For each query of the type '?' print one integer — the maximum value of bitwise exclusive OR (XOR) of integer xi and some integer from the multiset A.

Example

input

10+ 8+ 9+ 11+ 6+ 1? 3- 8? 3? 8? 11

output

11101413

Note

After first five operations multiset A contains integers 0, 8, 9, 11, 6 and 1.

The answer for the sixth query is integer  — maximum among integers , , ,  and .

题解: 裸的tire树。题好像没什么可以说的。。。

AC代码：

#pragma comment(linker, "/STACK:102400000,102400000")//#include<bits/stdc++.h>#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>#include<cstring>#include<vector>#include<map>#include<cmath>#include<queue>#include<set>#include<stack>using namespace std;typedef long long ll;typedef unsigned long long ull;#define mst(a) memset(a, 0, sizeof(a))#define M_P(x,y) make_pair(x,y)  #define rep(i,j,k) for (int i = j; i <= k; i++)  #define per(i,j,k) for (int i = j; i >= k; i--)  #define lson x << 1, l, mid  #define rson x << 1 | 1, mid + 1, r  const int lowbit(int x) { return x&-x; }  const double eps = 1e-8;  const int INF = 1e9+7; const ll inf =(1LL<<62) ;const int MOD = 1e9 + 7;  const ll mod = (1LL<<32);const int N = 101010; template <class T1, class T2>inline void getmax(T1 &a, T2 b) { if (b>a)a = b; }  template <class T1, class T2>inline void getmin(T1 &a, T2 b) { if (b<a)a = b; }int read(){int v = 0, f = 1;char c =getchar();while( c < 48 || 57 < c ){if(c=='-') f = -1;c = getchar();}while(48 <= c && c <= 57) v = v*10+c-48, c = getchar();return v*f;}map<int,int>mp[33];void add(int x,int k){    int sum = 0;    for(int i=29;i>=0;--i)    {        if((x>>i) & 1)            sum+= 1<<i;        mp[29-i][sum] += k;    }}int query(int x){    int ans = 0;    for(int i= 0;i<30; i++)    {        int r =29 - i;        if((x>>r) & 1)        {            if(!mp[i][ans])                ans += 1<<r;        }        else {            if(mp[i][ans + (1<<r)])                ans += 1<<r;        }    }    return ans;}int main() {#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);    #endif    int q;    q=read();    while(q--)    {        char c;        int x;        scanf(" %c %d",&c,&x);        if(c=='+') add(x,1);        else if(c=='-') add(x,-1);        else printf("%d\n",max(x,query(x)^x));    }    return 0;   }