Acwing 200 Hankson的趣味题

题意：

给定$a, b, c, d$，求$x$满足$gcd(a, x) = b$且$lcm(c, x) = d$

思路：

枚举d的所有约数，判断每个数是否满足条件。（直接枚举会超时）

优化：枚举质数（相当于d质因子分解），$dfs$枚举所有约数，判断是否满足条件

Code:

#include <map>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;

ll a, b, c, d, x;
vector <pair<int, int>> v;
map <ll, bool> mp;

void dfs(int cnt, ll res){
    if(!mp[res] && __gcd(a, res)== b && c / __gcd(c, res) * res == d){
        x ++; 
        mp[res] = true;
    }
    if(cnt == (int)v.size()) return;
    int p = v[cnt].first, k = v[cnt].second;
    ll mid = 1;
    for(int i = 0; i <= k; i ++){
        dfs(cnt + 1, res * mid);
        mid *= p;
    }
}

int main(){
    
    int n;
    scanf("%d", &n);
    
    while(n --){
        x = 0;
        scanf("%lld%lld%lld%lld", &a, &b, &c, &d);
        
        v.clear();
        mp.clear();
        ll k = d;
        for(int i = 2; i <= k / i; i ++){
            if(k % i == 0){
                int s = 0;
                while(k % i == 0){
                    s ++;
                    k /= i;
                }
                v.push_back({i, s});
            }
        }
        if(k > 1) v.push_back({k, 1});
        
        dfs(0, 1);
        cout << x << endl;
    }
    
    return 0;
}