HDU 2028 Lowest Common Multiple Plus (Stein演算法 求最大公約數)

Lowest Common Multiple Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26267 Accepted Submission(s): 10625

Problem Description求n個數的最小公倍數。

Input輸入包含多個測試執行個體，每個測試執行個體的開始是一個正整數n，然後是n個正整數。

Output為每組測試資料輸出它們的最小公倍數，每個測試執行個體的輸出佔一行。你可以假設最後的輸出是一個32位的整數。

Sample Input

2 4 63 2 5 7

Sample Output

1270

 最小共倍數= 兩個數的乘積/最大公約數Stein演算法

import java.io.*;import java.util.*;/* * @author denghuilong  *   * 2013-8-8下午7:05:42 **/public class Main {public static void main(String[] args) {Scanner sc=new Scanner(new BufferedInputStream(System.in));while(sc.hasNextInt()){int n=sc.nextInt();long sum=1;for(int i=0;i<n;i++){long t=sc.nextLong();sum*=t/GcdStein(sum,t);}System.out.println(sum);}}//Stein演算法  求最大公約數public static long GcdStein(long sum,long t){if(sum==0)return t;if(t==0)return sum;if(sum%2==0&&t%2==0)return 2*GcdStein(sum>>1,t>>1);else if(sum%2==0)return GcdStein(sum>>1,t);else if(t%2==0)return GcdStein(sum,t>>1);elsereturn GcdStein(Math.abs(sum-t),Math.min(sum, t));}}

輾轉相除法 求最大公約數 

import java.io.*;import java.math.BigInteger;import java.util.*;/* * @author denghuilong  *   * 2013-8-8下午7:05:37 **/public class Main {public static void main(String[] args) {Scanner sc=new Scanner(new BufferedInputStream(System.in));while(sc.hasNextInt()){int n=sc.nextInt();BigInteger sum=BigInteger.ONE;for(int i=0;i<n;i++){BigInteger big=sc.nextBigInteger();sum=sum.multiply(big.divide(Gcd(sum,big)));}System.out.println(sum);}}//  輾轉相除法 求最大公約數public static BigInteger Gcd(BigInteger sum,BigInteger big){if(big.compareTo(BigInteger.ZERO)==0)return sum;return Gcd(big,sum.mod(big));}}