aoj 0005 GCD and LCM

标签：turn   inpu   sam   number   eve   iostream   which   ever   space   

Write a program which computes the greatest common divisor (GCD) and the least common multiple (LCM) of given a and b.

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.

Constraints

• 0 < a, b ≤ 2,000,000,000
• LCM(a, b) ≤ 2,000,000,000
• The number of data sets ≤ 50

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
辗转相除法求最大公约数，然后最小公倍数就是a * b / 最大公约数。
代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#define Max 1001
using namespace std;
typedef long long LL;
LL gcd(LL a,LL b) {
    if(b == 0)return a;
    return gcd(b,a % b);
}
LL lcm(LL a,LL b) {
    return a / gcd(a,b) * b;
}
int main(){
    LL a,b;
    while(~scanf("%lld%lld",&a,&b)) {
        printf("%lld %lld\n",gcd(a,b),lcm(a,b));
    }
    return 0;
}

aoj 0005 GCD and LCM

标签：turn   inpu   sam   number   eve   iostream   which   ever   space   

原文地址：https://www.cnblogs.com/8023spz/p/9460935.html