AIZU 0005

GCD and LCM

Time Limit : 1 sec, Memory Limit : 65536 KB
Japanese version is here

GCD and LCM

Write a program which computes the greatest common divisor (GCD) and the least common multiple (LCM) of given a and b (0 < a, b ≤ 2,000,000,000). You can supporse that LCM(a, b) ≤ 2,000,000,000.

Input

Input consists of several data sets. Each data set contains a and b separated by a single space in a line. The input terminates with EOF.

Output

For each data set, print GCD and LCM separated by a single space in a line.

Sample Input

8 6
50000000 30000000

Output for the Sample Input

2 24
10000000 150000000

水题不解释
 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <iostream>
 5 
 6 using namespace std;
 7 
 8 typedef long long ll;
 9 
10 int gcd(int x,int y) {
11         return y > 0 ? gcd(y,x % y) : x;
12 }
13 int main() {
14 
15         int a,b;
16         while(~scanf("%d%d",&a,&b)) {
17             printf("%d ",gcd(a,b));
18             printf("%lld
",(ll)a / gcd(a,b) * b);
19         }
20 
21 }
View Code
原文地址:https://www.cnblogs.com/hyxsolitude/p/3590029.html