Codeforces 979 D. Kuro and GCD and XOR and SUM(异或和,01字典树)

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Codeforces 979 D. Kuro and GCD and XOR and SUM
题目大意:有两种操作:①给一个数v,加入数组a中②给出三个数x,k,s;从当前数组a中找出一个数u满足 u与x的gcd可以被k整除,u不大于s-x,且与x的异或和最大。
思路:之前没有碰到过异或和最值的问题,所以是懵逼的。学习了01字典树后把这题补出来。
碰到操作①就上树,上树过程中注意不断维护每个节点往后路径中的最小值(具体见代码细节);
碰到操作②,如果k==1,那么从树上找数的同时注意限制条件最小值不超过s-x;如果k>1,那么直接枚举找最值。

#include<iostream>
#include<set>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<climits>
using namespace std;
const int maxn=1e5+10;
bool a[maxn]; 
struct Trie_01
{
    static const int N = 32*maxn , M = 2;
    int node[N][M],value[N],rt,L;
    void init()
    { 
        fill_n(node[N-1],M,0);
        fill_n(value,N,INT_MAX);
        L = 0;
        rt = newnode();
    }
    int newnode()
    {
        fill_n(node[L],M,0);
        return L++; 
    }
    void add(int x)
    {
        int p = rt;
        value[p]=min(value[p],x);
        for (int i=31;i>=0;--i)
        {
            int idx = (x>>i)&1;
            if (!node[p][idx])
            {
                node[p][idx] = newnode();
            }
            p = node[p][idx];
            value[p]=min(value[p],x);
        }
    }
    int query(int x,int bound)
    {
        int p = rt;
        if (value[p]>bound)
            return -1;
        for (int i=31;i>=0;--i)
        {
            int idx = (x>>i)&1;
            if (node[p][idx^1]&&value[node[p][idx^1]]<=bound)
                p = node[p][idx^1];
            else
                p = node[p][idx];
        }
        return value[p];
    }
};
Trie_01 tree;
int main()
{
    int i,n;
    cin>>n;
    tree.init();
    for (i=0;i<n;++i)
    {
        int ty,s,x,k,v;
        scanf("%d",&ty);
        if (ty==1)
        {
            scanf("%d",&v);
            tree.add(v);
            a[v]=1;         
        }
        else
        {
            scanf("%d%d%d",&x,&k,&s);
            if (x%k)
                cout<<"-1\n";
            else if (k==1)
            {
                cout<<tree.query(x,s-x)<<endl;
            }
            else
            {
                int mx=-1,ans=-1;
                for (int j=k;j<=s-x;j+=k)
                {
                    if (a[j]&&(j^x)>mx)
                    {
                        mx=j^x;
                        ans=j;
                    }
                }
                cout<<ans<<endl;
            }
        }
    }
    return 0;
}

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