HDU 1108 最小公倍數（stein演算法）

最小公倍數

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 27030 Accepted Submission(s): 14945

Problem Description給定兩個正整數，計算這兩個數的最小公倍數。

Input輸入包含多組測試資料，每組只有一行，包括兩個不大於1000的正整數.

Output對於每個測試案例，給出這兩個數的最小公倍數，每個執行個體輸出一行。

Sample Input

10 14

Sample Output

70

 

stein演算法介紹：http://blog.csdn.net/deng_hui_long/article/details/9842269

stein演算法

import java.io.*;import java.util.*;public class Main {public static void main(String[] args) {new Main().work();}public void work(){Scanner sc=new Scanner(new BufferedInputStream(System.in));while(sc.hasNext()){int n=sc.nextInt();int m=sc.nextInt();int k=gcdStein(n,m);int num=(n*m)/k;System.out.println(num);}}public int gcdStein(int m,int n){if(n==0)return m;if(m==0)return n;if(m%2==0&&n%2==0)return 2*gcdStein(m>>1,n>>1);if(m%2==0)return gcdStein(m>>1,n);if(n%2==0)return gcdStein(m,n>>1);elsereturn gcdStein(Math.abs(m-n),Math.min(m, n));}}

輾轉相除法

import java.io.*;import java.util.*;public class Main {public static void main(String[] args) {new Main().work();}public void work(){Scanner sc=new Scanner(new BufferedInputStream(System.in));while(sc.hasNext()){int n=sc.nextInt();int m=sc.nextInt();int num=(n*m)/gcd(n,m);System.out.println(num);}}public int gcd(int n,int m){int a=Math.max(n, m);int b=Math.min(n, m);if(b==0)return a;return gcd(b,a%b);}}