POJ5429 GCD & LCM Inverse

GCD & LCM Inverse

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 9913 Accepted: 1841

Description

Given two positive integers a and b, we can easily calculate the greatest common divisor (GCD) and the least common multiple (LCM) of a and b. But what about the inverse? That is: given GCD and LCM, finding a and b.

Input

The input contains multiple test cases, each of which contains two positive integers, the GCD and the LCM. You can assume that these two numbers are both less than 2^63.

Output

For each test case, output a and b in ascending order. If there are multiple solutions, output the pair with smallest a + b.

Sample Input

3 60

Sample Output

12 15

Source

POJ Achilles

算是钻空子解法了。。算法没变，C++ TLE,java ac..

import java.util.Scanner;public class Main {static long gcd(long a, long b) {long c;while(b != 0) {c = a % b;a = b;b = c;}return a;}public static void main(String[] args) {Scanner cin = new Scanner(System.in);long a, b, c, i;while(cin.hasNext()) {a = cin.nextLong();b = cin.nextLong();c = b / a;for(i = (long)Math.sqrt(c); i > 0; --i)if(c % i == 0 && gcd(i, c / i) == 1) {System.out.println(i*a + " " + c/i*a);break;}}}}