C. Fadi and LCM

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.

Examples inputCopy 2 outputCopy 1 2 inputCopy 6 outputCopy 2 3 inputCopy 4 outputCopy 1 4 inputCopy 1 outputCopy 1 1

题意描述：这道题就是给你一个数X，让你输出两个数，这两个数要满足：1.LCM(m,n)=x 也就是说，这两个数的最小公倍数是X。2.使得max(a,b)的值最小，举个例子，如果X为6的话，那么有两种情况：LCM(1,6)和LCM(2,3)，要使得max(a,b)的值最小，所以输出的就是2和3。刚看到题的时候，脑袋很迷糊，在想使得max(a,b)的值最小是什么意思，弄懂了这一点，就很简单了。要想max(a,b)的值最小，那么就从1枚举到sqrt(X)，求得最大的因子。

AC代码如下：

import java.util.Scanner;
public class Main {
	private static int[] ans=new int[1000001];
	public static void main(String[] args){
		Scanner cin=new Scanner(System.in);
		long x=cin.nextLong();
		long max=0;
		for(long i=1;i*i<=x;i++) {
			if(lcm(i,x/i)==x&&gcd(i,x/i)==1) {
				max=Math.max(max, i);
			}
		}
		System.out.println(max+" "+x/max);
	}
	public static long gcd(long m,long n) {
		return m%n==0?n:gcd(n,m%n);
	}
	public static long lcm(long m,long n) {
		return m*n/gcd(m,n);
	}
}

原题链接https://codeforces.com/contest/1285/problem/C