CodeForces 979 D Kuro and GCD and XOR and SUM

Kuro and GCD and XOR and SUM

题解：

对于每个值先找到所有的因子。

然后每次add的时候，在他的所有因子都加入这个数。

然后询问的时候询问这个数的因子所构成的字典树。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int INF = 0x3f3f3f3f;
const LL mod = 1e9+7;
const int N = 1e5+10;
vector<int> son[N];
struct Node{
    int Min;
    Node * p[2];
    Node(){
        Min = INF;
        p[0] = p[1] = nullptr;
    }
}*rt[N];
bool vis[N];
void init(){
    for(int i = 1; i < N; i++){
        rt[i] = new Node();
        for(int j = i; j < N; j+=i){
            son[j].pb(i);
        }
    }
}
void Add(int k, int u){
    Node *tmp = rt[k];
    tmp -> Min = min(tmp -> Min, u);
    for(int i = 18; i >= 0; i--){
        int id = u>>i & 1;
        if(tmp -> p[id] == nullptr)
            tmp -> p[id] = new Node();
        tmp = tmp -> p[id];
        tmp -> Min = min(tmp -> Min, u);
    }
}
int Query(int x, int k, int s){
    Node *tmp = rt[k];
    if(x%k != 0 || tmp->Min+x > s)
        return -1;
    int ret = 0;
    for(int i = 18; i >= 0; i--){
        int id = x >> i & 1;
        if(tmp -> p[id^1] != nullptr && tmp -> p[id^1] -> Min + x <= s){
            tmp = tmp -> p[id^1];
            ret += (id^1) << i;
        }
        else {
            tmp = tmp -> p[id];
            ret += id << i;
        }
    }
    return ret;
}
int main(){
    init();
    int q, t, u, x, s, k;
    scanf("%d", &q);
    while(q--){
        scanf("%d", &t);
        if(t == 1){
            scanf("%d", &u);
            if(!vis[u]){
                vis[u] = 1;
                for(int k : son[u]){
                    Add(k, u);
                }
            }
        }
        else{
            scanf("%d%d%d", &x, &k, &s);
            printf("%d
", Query(x,k,s));
        }
    }
}
    return 0;