• codeforces 706D Vasiliy's Multiset


    D. Vasiliy's Multiset

     

    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
    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 .

    #include<cstdio>
    #include<cmath>
    #include<cstring>
    using namespace std;
    struct Trie
    {
        Trie *next[2];
        int cnt;
        Trie()
        {
            next[0]=next[1]=0;
            cnt=0;
        }
    };
    Trie *rt=new Trie();
    void _insert(int x)
    {
        Trie *u=rt;
        for(int i=30;i>=0;i--)
        {
            int v=x&(1<<i)?1:0;
            if(u->next[v])
                u->next[v]->cnt++;
            else
            {
                Trie *p=new Trie();
                u->next[v]=p;
                p->cnt++;
            }
            u=u->next[v];
        }
    }
    void _delete(int x)
    {
        Trie *u=rt;
        for(int i=30;i>=0;i--)
        {
            int v=x&(1<<i)?1:0;
            u->next[v]->cnt--;
            u=u->next[v];
        }
    }
    int query(int x)
    {
        int res=0;
        Trie *u=rt;
        for(int i=30;i>=0;i--)
        {
            int v=x&(1<<i)?0:1;
            if(u->next[v]&&u->next[v]->cnt>0)
            {
                res+=pow(2.0,i);
                u=u->next[v];
            }
            else
                u=u->next[!v];
        }
        return res;
    }
    int main()
    {
        _insert(0);
        int q;
        scanf("%d",&q);
        while(q--)
        {
            char c;
            int num;
            getchar();
            scanf("%c %d",&c,&num);
            if(c=='+')
                _insert(num);
            else if(c=='-')
                _delete(num);
            else
                printf("%d
    ",query(num));
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/homura/p/5764312.html
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