题目：
Today, Osama gave Fadi an integer X, and Fadi was wondering about the minimum possible value of max(a,b) such that LCM(a,b) equals X. Both a and b should be positive integers.

LCM(a,b) is the smallest positive integer that is divisible by both a and b. For example, LCM(6,8)=24, LCM(4,12)=12, LCM(2,3)=6.

Of course, Fadi immediately knew the answer. Can you be just like Fadi and find any such pair?

Input
The first and only line contains an integer X (1≤X≤1012).

Output
Print two positive integers, a and b, such that the value of max(a,b) is minimum possible and LCM(a,b) equals X. If there are several possible such pairs, you can print any.

题意：给定一个数x，找出两个数a,b，使得LCM（a,b）= x；求满足条件并且使max(a,b)尽可能小的a和b。
这题乍一看有点懵，然后看了一下数据范围，发现可以暴力，所以写着试了一发，然后AC；
思路：
Xoxoxo首先判断x是否素数，如果是则输出1和它本身；如果不是，则进行枚举，枚举的复杂度是O(sqrt(n))。

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
bool is_prim(ll x) {
	ll m = sqrt(x);
	for (ll i = 2; i <= m; i++) {
		if (x % i == 0)
			return false;
	}
	return true;
}
ll gcd(ll a, ll b) {
	if (b == 0)
		return a;
	else
		return gcd(b, a % b);
}
int main()
{
	std::ios::sync_with_stdio(false);
	std::cin.tie(0);
	std::cout.tie(0);
	ll x; cin >> x;
	if (is_prim(x))
		cout << "1 " << x << endl;
	else {
		ll maxn = 0;
		for (ll i = 1; i <= sqrt(x); i++) {
			if (x % i == 0) {
				if (gcd(i, x / i) == 1)
					maxn = i;
			}
		}
		if (maxn)
			cout << min(x / maxn, maxn) << " " << max(x / maxn, maxn) << endl;
		else
			cout << "1 " << x << endl;
	}
	return 0;
}