poj 2429 GCD & LCM Inverse 【java】+【数学】

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GCD & LCM Inverse

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 9928 Accepted: 1843

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

题意：给出你最大公约数和最小公倍数，让你求出原来的两个数a,b；特别的如果有多组的话，输出和最小的哪一组。

分析：由GCD和LCM之间的关系可得a*b/GCD= LCM；没有特别的方法只好枚举，但是我们可以得出a*b = LCM/GCD；我们要找的是最后的和最小的，所以从sqrt（b/=a）开始到1枚举就好了。

注：如果用c/c++会超时的。最后经学长指导改用java就过了。。。又学了一招。

代码：

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

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