• Aizu:0005-GCD and LCM


    GCD and LCM

    Time limit 1000 ms
    Memory limit 131072 kB

    Problem Description

    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


    解题心得:

    1. 就是给你两个数,要求输出GCD和LCM。
    2. 其实就是求个GCD,LCM = a*b/GCD

    #include <algorithm>
    #include <stdio.h>
    #include <cstring>
    using namespace std;
    typedef long long ll;
    
    ll __gcd(ll a,ll b) {
        if(b == 0)
            return a;
        return __gcd(b,a%b);
    }
    
    int main() {
        ll a,b;
        while(scanf("%lld%lld",&a,&b) != EOF){
            ll gcd = __gcd(a, b);
            ll lcm = a * b / gcd;
            printf("%lld %lld
    ", gcd, lcm);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/GoldenFingers/p/9107136.html
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