Codeforces 706D Trie树/multiset

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 x_i (1 ≤ x_i ≤ 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 x_i and some integer from the multiset A.

Example

Input

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

Output

11
10
14
13

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 .

题意：

+表示吧这个数加到集合中,-表示把这个数从集合中减去一次,？表示集合里面的一个y使的x^y最大;

分析：

Trie树。

#include<bits/stdc++.h>
using namespace std;
const int N=1e7+9;
int a[N][2],cal[N];
int tot;
void add(int x,int y) //插入或删除一个元素
{
    int cur=0;
    for(int i=31;i>=0;i--){
        int t=(x>>i)&1;
        if(!a[cur][t])a[cur][t]=tot++;
        cur=a[cur][t];
        cal[cur]+=y;
    }
}

int query(int x)
{
    int cur=0,ans=0;
    for(int i=31;i>=0;i--){
        int t=(x>>i)&1;
        if(cal[a[cur][t^1]]){
            ans+=(1<<i);
            cur=a[cur][t^1];
        }
        else cur=a[cur][t];
    }
    return ans;
}
int main()
{
    int n,x;
    char op;
    tot=1;
    add(0,1);
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        cin>>op>>x;
        if(op=='+'){
            add(x,1);
        }
        else if(op=='-'){
            add(x,-1);
        }
        else{
            printf("%d
",query(x));
        }
    }
    return 0;
}

或者可以直接用multiset模拟：

#include<bits/stdc++.h>
using namespace std;
const int N=1e7+9;

int main()
{
    int n,x;
    char op;
    multiset<int>s;
    s.insert(0);
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        cin>>op>>x;
        if(op=='+'){
            s.insert(x);
        }
        else if(op=='-'){
            s.erase(s.find(x));
        }
        else{
            int ans=0;
            for(int i=31;i>=0;i--){
                ans|=(~x&(1<<i));
                multiset<int>::iterator it=s.lower_bound(ans);
                if(it==s.end()||*it>=ans+(1<<i)){
                   // cout<<ans<<endl;
                    ans^=1<<i;
                }
            }
            printf("%d
",x^ans);
        }
    }
    return 0;
}