Our life is frittered away by detail ... Simplify, simplify.

by Henry Thoreau

（美國哲學家亨利·梭羅說，我們的生活被瑣碎的細節消磨殆盡，要簡化，要簡化！）

所以，如果能夠找到現成的解決方案，我們就沒必要自己埋頭研究演算法了（除非我們的目的在於研究演算法本身！）。

package exercises.ch6Methods;

import java.util.*;

//JHTP Exercise 6.27 (Greatest Common Divisor)
//by [email protected]
 

/**
 * 
 *6.27 (Greatest Common Divisor) The greatest common divisor (GCD) of two integers is the 
 *largest integer that evenly divides each of the two numbers. Write a method gcd that 
 *returns the greatest common divisor of two integers. [Hint: You might want to use Euclid’s algorithm. 
 *You can find information about it at en.wikipedia.org/wiki/Euclidean_algorithm.] Incorporate the method 
 *into an application that reads two values from the user and displays the result.
 *
 */

public class GreatedCommonDivisor {


	
	//此演算法基於本人2008年的求最大公約數的方法，https://blog.csdn.net/hpdlzu80100/article/details/2290499
	public static long gcd1(long n1, long n2) {
		
	long g = 0;
	
	for(long i=1;i<=Math.min(n1,n2);i++)
	{
	if (n1 % i == 0 && n2 % i == 0)
	   g = i;
	}

	return g;
}
	
	//此演算法https://blog.csdn.net/jiaolipe/article/details/1476068，採用了Euclid’s algorithm（歐幾里得演算法）
    private static long gcd2 (long num1, long num2){
        while (num1 != num2)
        {
           if (num1 > num2)
              num1 = num1 - num2;
           else
              num2 = num2 - num1;
        }

        return num1;
   }


	public static void main(String args[]) {
		long number1;
		long number2;
		long gcd1;
		long gcd2;
		
	    Scanner scan=new Scanner(System.in);
	    
	    do {
	    System.out.printf("請輸入第一個整數(輸入-1退出)：");
	    number1 = scan.nextLong();
		if(number1 == -1)
		{System.out.print("已退出程式。");
		    break;
	    }
		
	    System.out.printf("請輸入第二個整數(輸入-1退出)：");
	    number2 = scan.nextLong();
		if(number2 == -1)
		{System.out.print("已退出程式。");
		    break;
	    }
	    
		long beginTime=System.nanoTime();	
		gcd1 = gcd1(number1, number2);
		long endTime=System.nanoTime();
		double duration1=(double)(endTime-beginTime);
	    System.out.printf("方法1：%d和%d的最大公約數為：%d，共用時%f納秒%n", number1, number2, gcd1, duration1);
	    
		beginTime=System.nanoTime();	
		gcd2 = gcd2(number1, number2);
		endTime=System.nanoTime();
		double duration2=(double)(endTime-beginTime);
	    System.out.printf("方法2：%d和%d的最大公約數為：%d，共用時%f納秒%n", number1, number2, gcd2, duration2);
	    System.out.printf("方法2的用時是方法1的%.2f%%%n", duration2/duration1 * 100);
	    
	    System.out.println();
	    } while (number1 != -1);
	    
	    System.out.println("已退出程式。");
	    scan.close();
	}
	}

 

執行結果：