最小公倍数 java

今天起床迟了，30多个小时的连续作战，睡个懒觉也无可厚非。。。

然后看到一个题目，一个很另类的记录整个求两个整数的最小公倍数的过程。于是小写了一下，发现过程中问题还不少，果然底层代码一段时间不写就生疏了。

先看两个简洁方法求最小公倍数

一最常用

两数之积 分别同时除以 两数，能同时除得整数的为公倍数，最小者为最小公倍数，当然，从最小数开始循环，第一个公倍数即为最小：

假定两个数分别为m,n
if(m

二：效率很高的写法

两数相乘除去两数中重复部分即可

int minMultiple(int a, int b) {
		int r = a, s = a, t = b;
		if (a < b) {
			r = a;
			a = b;
			b = r;
		}
		while (r != 0) {
			r = a % b;
			a = b;
			b = r;
		}
		return s * t / a;
	} 

三：模拟中学时代的最小公倍数求法

还记得吗，我们写两个数到一排，然后在左边分别写出两个数都有的部分，然后重复，最后把框框外的数相乘即为最小公倍数。

package calcumulation;

import java.util.ArrayList;
import java.util.Scanner;

public class LCM {
		static int a;
		static int b;
		
		static ArrayList primeNumber;
		public static void main(String []  args){
			Scanner in  = new Scanner(System.in);
			//note:I do not know what the size of your a and b ,
			//so I am just give you the prime number that less than 100,you can change it below
			LCM.findPrime();
	        System.out.println(primeNumber);
	        
			tag: while(in.hasNext()){
				
				a = in.nextInt();
				b = in.nextInt();
				//test 0,if zero,simple,do nothing and continue...
				if(a==0 ||b ==0) continue tag;
				int [][] aTable = new int[primeNumber.size()][2];
				int [][] bTable = new int[primeNumber.size()][2];;

		/***********table cal************/
			for(int i = 0;i < primeNumber.size() ; i++){
				for(int j = 0;j < primeNumber.size() ; j++){
					if( a%primeNumber.get(i) == 0 ) {
						aTable[i][0] = primeNumber.get(i);
						aTable[i][1]++;
						a = a/primeNumber.get(i);
					}
					if( b%primeNumber.get(i) == 0 ) {
						bTable[i][0] = primeNumber.get(i);
						bTable[i][1]++;
						b = b/primeNumber.get(i);
					}
				}
			}
		//end cal table
			
			/************print LCM******************/
			int i=0,j=0;
			while(aTable[i][0] != 0){
					System.out.println("This is show a prime number is :"+aTable[i][0] +" occurs "+aTable[i][1]);
					i++;
			}		
			while(bTable[j][0] != 0){
					System.out.println("This is show b prime number is :"+bTable[j][0] +" occurs "+bTable[j][1]);
					j++;
				}

			int result = 1;
			//get the same value of a and b
			int [] temp = new int[primeNumber.size()];
			int k = 0;
			for(int m = 0;m < primeNumber.size() ; m++){
				if(aTable[m][0] !=0 ){
				for(int n = 0;n < primeNumber.size() ; n++){
						if(aTable[m][0] == bTable[n][0]) { 
							temp[k] = aTable[m][0]; k++;
							aTable[m][1]--;
							bTable[n][1]--;
						}//inner if
				}//inner for
				}//if outer
			}
			int l = 0;
			while(temp[l] !=0) { result *=temp[l]; l++; }
			
			// previous value multiple the rest of not identical value
			for(int m = 0;m < primeNumber.size() ; m++){
				if(aTable[m][0] !=0 && aTable[m][1] >0) result *=(aTable[m][0] * aTable[m][1] );
			}
				for(int n = 0;n < primeNumber.size() ; n++){
					if(bTable[b][0] !=0 && bTable[n][1] >0) result *= (bTable[n][0] * bTable[n][1] );
				}

			
			//print
			System.out.println(" LCM is :" + result);
			
			
		}//end outer for
		}//end non-daemon
		
		//prime cal
		static  void findPrime(){
			//get prime number less than 100
	        primeNumber = new ArrayList();
	       label: for (int i = 2; i < 100; i++) {
	        		int sqrt = (int) Math.sqrt(i);
	        			if(i ==2 || i%2 != 0){
	        					for( int j=2; j <=sqrt ; j++){
	        						if( i%j == 0) continue label;
	        					}
	        				primeNumber.add(i);
	        			}
	                }
	            }	
}

输出测试：

[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
120
150
This is show a prime number is :2 occurs 3
This is show a prime number is :3 occurs 1
This is show a prime number is :5 occurs 1
This is show b prime number is :2 occurs 1
This is show b prime number is :3 occurs 1
This is show b prime number is :5 occurs 2
30
 LCM is :600
150
180
This is show a prime number is :2 occurs 1
This is show a prime number is :3 occurs 1
This is show a prime number is :5 occurs 2
This is show b prime number is :2 occurs 2
This is show b prime number is :3 occurs 2
This is show b prime number is :5 occurs 1
30
 LCM is :900