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  • java求最大公约数(分解质因数)

    下面是四种用java语言编程实现的求最大公约数的方法:

    package gcd;
    
    import java.util.ArrayList;
    import java.util.List;
    
    public class gcd {
    	public static void main(String[] args) {
    		long startTime;
    		long endTime;
    		long durationTime;
    		
    		int[] testArray1 = new int[]{784, 988, 460, 732, 548, 998, 672, 1024, 888, 512};
    		int[] testArray2 = new int[]{1024, 82, 92, 128, 58, 2014, 512, 88, 582, 788};
    		
    		for (int i = 0; i < 10; i++) {
    			startTime = System.nanoTime();
    			System.out.println("欧几里得方法:" + Euclid(testArray1[i],testArray2[i]));
    			endTime = System.nanoTime();
    			durationTime = endTime - startTime;
    			System.out.println("欧几里得算法耗时:" + durationTime + "
    ");
    			
    			startTime = System.nanoTime();
    			System.out.println("连续整数检測法:" + consecutiveIntegersTest(testArray1[i], testArray2[i]));
    			endTime = System.nanoTime();
    			durationTime = endTime - startTime;
    			System.out.println("连续整数检測算法耗时:" + durationTime + "
    ");
    			
    			startTime = System.nanoTime();
    			System.out.println("辗转相减法:" + consecutiveSubstract(testArray1[i], testArray2[i]));
    			endTime = System.nanoTime();
    			durationTime = endTime - startTime;
    			System.out.println("辗转相减算法耗时:" + durationTime + "
    ");
    			
    			startTime = System.nanoTime();
    			System.out.println("分解质因数法:" + primeFactors(testArray1[i], testArray2[i]));
    			endTime = System.nanoTime();
    			durationTime = endTime - startTime;
    			System.out.println("分解质因数算法耗时:" + durationTime);
    		}
    		
    	}
    
    	/**
    	 * 欧几里得算法求最大公约数
    	 * @param no1
    	 * @param no2
    	 * @return
    	 */
    	public static int Euclid(int no1, int no2) {
    		int remainder;
    		remainder = no1%no2;
    		while(remainder != 0) {
    			no1 = no2;
    			no2 = remainder;
    			remainder = no1%no2;
    		}
    		return no2;
    	}
    	
    	/**
    	 * 连续整数检測法
    	 * @param m
    	 * @param n
    	 * @return
    	 */
    	public static int consecutiveIntegersTest(int m, int n) {
    		int t;
    		if (m > n) 
    			t = n;
    		else
    			t = m;
    		while(true) {
    			if (m%t == 0 && n%t == 0)
    				break;
    			else
    				t = t - 1;
    		}
    		return t;
    	}
    	
    	/**
    	 * 辗转相减法
    	 * @param num1
    	 * @param num2
    	 * @return
    	 */
    	public static int consecutiveSubstract(int num1, int num2) {
    		while(true) {
    			if (num1 > num2)
    				num1 -= num2;
    			else if (num1 < num2)
    				num2 -= num1;
    			else
    				return num1;
    		}
    	}
    	
    	/**
    	 * 分解质因数法
    	 * @param primeNum1
    	 * @param primeNum2
    	 * @return
    	 */
    	public static int primeFactors(int primeNum1, int primeNum2) {
    		int prime_gcd = 1;
    		int compareListSize;
    		int temp1, temp2;
    		int pn1 = primeNum1, pn2 = primeNum2;
    		List<Integer> num1List = new ArrayList<Integer>();
    		List<Integer> num2List = new ArrayList<Integer>();
    		List<Integer> sameNumList = new ArrayList<Integer>();
    		//求出质因数
    		for (int i = 2; i < pn1/2;) {		//注意此处用的是pn1,而不是primeNum1,primeNum1的值在以下的运行过程会不断减小
    			if (primeNum1%i == 0) {		//求余数,假设能被整除,返回true
    				temp1 = primeNum1 / i;		//求商
    				primeNum1 = temp1;		//将商赋值给primeNum1。又一次推断余数是否为0
    				num1List.add(i);		//将质因数放入num1List
    			} else if (primeNum1%i != 0) {
    				i = i + 1;		//假设余数不等于0。除数i加1,继续求余数
    			}
    		}
    		
    		for (int i = 2; i < pn2/2;) {
    			if (primeNum2%i == 0) {
    				temp2 = primeNum2 / i;
    				primeNum2 = temp2;
    				num2List.add(i);
    			} else if (primeNum2%i != 0) {
    				i = i + 1;
    			}
    		}
    		int num1ListSize = num1List.size();
    		int num2ListSize = num2List.size();
    		if (num1ListSize < num2ListSize) {
    			for (int i = 0; i < num1List.size();) {
    				if (num2List.contains(num1List.get(i))) {
    					prime_gcd *= num1List.get(i);
    					num2List.remove(num2List.indexOf(num1List.get(i)));
    					num1List.remove(i);
    					if (num1List.size() == 0 || num2List.size() == 0)
    						break;
    				} else {
    					i = i + 1;
    				}
    			}
    		} else {
    			for (int i = 0; i < num2List.size(); ) {
    				if (num1List.contains(num2List.get(i))) {
    					prime_gcd *= num2List.get(i);
    					num1List.remove(num1List.indexOf(num2List.get(i)));
    					num2List.remove(i);
    					if (num1List.size() == 0 || num2List.size() == 0)
    						break;
    				} else {
    					i = i + 1;
    				}
    			}
    		}
    		return prime_gcd;
    	}
    }
    


     

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  • 原文地址:https://www.cnblogs.com/mengfanrong/p/5058608.html
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