GCD & LCM Inverse
Time Limit: 2000MS | Memory Limit: 65536K | |
Total 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
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; } } } }
时间: 2024-10-19 16:48:31