Codeforces Round #828 (Div. 3) E2. Divisible Numbers （分解质因子，dfs判断x,y）

题目链接

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题目大意

　给定a,b,c,d四个数，其中a<ｃ,ｂ＜ｃ,现在让你寻找一对数(x,y),满足一下条件：

　1. a<x<c,b<y<d

　2. (x*y)%(a*b)==0

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题目思路

　因为(x*y)%(a*b)==0\(\rightarrow\)x*y\(~\)=\(~\)k*a*b\(~\)=\(~\)k*k1*k2

　所以我们要找的就是a,b的因子来保证a%k1\(~\)||\(~\)a%k2\(~\)||\(~\)b%k1\(~\)||\(~\)b%k2为0

　而每个数都能被拆成多个质因子的k次的乘积，也就是说：

　a\(~\)=\(~\)P\(_{1}\)\(^{k1}\)\(~\)*\(~\)P\(_{2}\)\(^{k2}\)\(~\)*\(~\)P\(_{3}\)\(^{k3}\)\(~\)*....*\(~\)P\(_{n}\)\(^{kn}\)

　b\(~\)=\(~\)S\(_{1}\)\(^{k1}\)\(~\)*\(~\)S\(_{2}\)\(^{k2}\)\(~\)*\(~\)S\(_{3}\)\(^{k3}\)\(~\)*....*\(~\)S\(_{n}\)\(^{kn}\)

\(~~~\)而且在质因子较少，便于搜索，所以我们直接拆出a,b的质因子去爆搜看能不能找到对应的x,y

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代码详情

# include<iostream>
# include<bits/stdc++.h>
using namespace std;
# define int long long
# define endl "\n"
const int N = 5e5 + 10;
map<int, int> mp;
vector<pair<int, int>> fac;
int a, b, c, d;
bool ok = false;
void fj(int n) //分解质因数
{
	for (int i = 2; i * i <= n; ++i) {
		if (n % i == 0) {
			int cnt = 0;
			while (n % i == 0) {
				n /= i;
				cnt++;
			}
			mp[i] += cnt;//保存质因数的次数k
		}
	}
	if (n > 1) mp[n]++;
}
int ans_x, ans_y;
void dfs(int id, int k1, int k2) {
	if (id == fac.size()) {
		int x = ((a / k1) + 1) * k1;
		int y = ((b / k2) + 1) * k2;
		if (x <= c && y <= d) {
			ok = true;
			ans_x = x;
			ans_y = y;
		}
		return;
	}
	int p = fac[id].first, cnt = fac[id].second;
	int pow = 1;
	for (int i = 1; i <= cnt; ++i) pow *= p;
	int res = 1;
	for (int i = 0; i <= cnt; ++i) {
		dfs(id + 1, k1 * res, k2 * (pow / res));
		res *= p;
	}
}
void solve() {
	mp.clear(), fac.clear();
	ok = false;
	cin >> a >> b >> c >> d;
	fj(a);
	fj(b);
	for (auto p : mp) fac.push_back(p);
	dfs(0, 1, 1);
	if (ok) cout << ans_x << " " << ans_y << endl;
	else cout << -1 << " " << -1 << endl;
}
int tt;
signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	tt = 1;

	cin >> tt;
	while (tt--)solve();
	return 0;
}