CodeForces 979 D.Kuro and GCD and XOR and SUM (01字典树*n)

D.Kuro and GCD and XOR and SUM

Kuro is currently playing an educational game about numbers. The game focuses on the greatest common divisor (GCD), the XOR value, and the sum of two numbers. Kuro loves the game so much that he solves levels by levels day by day.

Sadly, he’s going on a vacation for a day, and he isn’t able to continue his solving streak on his own. As Katie is a reliable person, Kuro kindly asked her to come to his house on this day to play the game for him.

Initally, there is an empty array a. The game consists of q tasks of two types. The first type asks Katie to add a number ui to a. The second type asks Katie to find a number v existing in a such that ki∣GCD(xi,v), xi+v≤si, and xi⊕v is maximized, where ⊕ denotes the bitwise XOR operation, GCD(c,d) denotes the greatest common divisor of integers c and d, and y∣x means x is divisible by y, or report -1 if no such numbers are found.

Since you are a programmer, Katie needs you to automatically and accurately perform the tasks in the game to satisfy her dear friend Kuro. Let’s help her!

Input

The first line contains one integer q (2≤q≤105) — the number of tasks the game wants you to perform.

q lines follow, each line begins with an integer ti — the type of the task:

If ti=1, an integer ui follow (1≤ui≤105) — you have to add ui to the array a.
If ti=2, three integers xi, ki, and si follow (1≤xi,ki,si≤105) — you must find a number v existing in the array a such that ki∣GCD(xi,v), xi+v≤si, and xi⊕v is maximized, where ⊕ denotes the XOR operation, or report -1 if no such numbers are found.
It is guaranteed that the type of the first task is type 1, and there exists at least one task of type 2.

Output

For each task of type 2, output on one line the desired number v, or -1 if no such numbers are found.

Examples

Input

5
1 1
1 2
2 1 1 3
2 1 1 2
2 1 1 1

Output

2
1
-1

Input

10
1 9
2 9 9 22
2 3 3 18
1 25
2 9 9 20
2 25 25 14
1 20
2 26 26 3
1 14
2 20 20 9

Output

9
9
9
-1
-1
-1

Note

In the first example, there are 5 tasks:

The first task requires you to add 1 into a. a is now {1}.
The second task requires you to add 2 into a. a is now {1,2}.
The third task asks you a question with x=1, k=1 and s=3. Taking both 1 and 2 as v satisfies 1∣GCD(1,v) and 1+v≤3. Because 2⊕1=3>1⊕1=0, 2 is the answer to this task.
The fourth task asks you a question with x=1, k=1 and s=2. Only v=1 satisfies 1∣GCD(1,v) and 1+v≤2, so 1 is the answer to this task.
The fifth task asks you a question with x=1, k=1 and s=1. There are no elements in a that satisfy the conditions, so we report -1 as the answer to this task.

题意：

q次操作：
(1,x)加入x
(2,x,k,s)查询

查询操作：
在已有的数中找到一个数v，满足k∣GCD(x,v),且x+v<=s，同时还要求x异或v的值最大

输出每次查询的结果v，如果找不到满足条件的数就输出-1

思路:

k∣GCD(x,v)意味着gcd(x,v)%k=0，则题目条件转化为x%k=0且v%k=0

如果单是异或最大值可以用01字典树解决。
处理x%k=0且v%k=0:
建立1e5棵字典树，对于每棵树k，只存k的倍数，每次加入x，把x的因子对应字典树都插入x
查询则到k树中找和x异或最大的v就行了，因为k树中只有k的倍数，所以找到的v一定满足v%k=0
注意还有一种情况是x%k!=0，这种情况直接输出-1不用找了。

因为还需要满足x+v<=s，查询操作中有的路径是不能走的，因为可能走之后找到的v一定不满足这个条件，因此插入的时候更新路径上x的最小值，查询的时候就可以判断走某条路径是否满足该条件。

code:

#include<bits/stdc++.h>
using namespace std;
const int maxm=1e5+5;
const int maxd=17;//1e5最多17位
struct Node{
    int a[maxm*maxd*15][2],tot;
    int rt[maxm];
    int val[maxm*maxd*15];
    void init(){
        tot=0;
        for(int i=1;i<maxm;i++){
            rt[i]=++tot;
        }
    }
    void add(int x,int k){//把x插入k树下
        int node=rt[k];
        for(int i=maxd;i>=0;i--){
            int v=(x>>i&1);
            if(!a[node][v]){
                a[node][v]=++tot;
                node=a[node][v];
                val[node]=x;
            }else{
                node=a[node][v];
                val[node]=min(val[node],x);//因为需要满足x+val<=s，路径取min
            }
        }
        val[node]=x;
    }
    int ask(int x,int k,int s){//查询
        if(x%k)return -1;//这个不加会wa84，题目不保证x%k==0
        int node=rt[k];
        for(int i=maxd;i>=0;i--){
            int v=(x>>i&1);
            if(a[node][v^1]&&x+val[a[node][v^1]]<=s){//走满足条件的
                node=a[node][v^1];
            }else{//
                node=a[node][v];
            }
        }
        if(x+val[node]>s||val[node]==0)return -1;
        return val[node];
    }
}t;
vector<int>fac[maxm];
signed main(){
    for(int i=1;i<maxm;i++){//预处理每个数的因子
        for(int j=i;j<maxm;j+=i){
            fac[j].push_back(i);
        }
    }
    t.init();//不要忘了初始化
    int q;
    cin>>q;
    while(q--){
        int d;
        cin>>d;
        if(d==1){//add
            int x;
            cin>>x;
            for(int v:fac[x]){//每个因子的字典树都添加一次
                t.add(x,v);
            }
        }else{//d==2,ask
            int x,k,s;
            cin>>x>>k>>s;
            int ans=t.ask(x,k,s);//在k树下找答案
            cout<<ans<<endl;
        }
    }
    return 0;
}