2015年9月15号360笔试算法2

Problem Description 

 

当你学一些可视化程序设计语言时，老师经常让你设计并编程做出一个计算机，这时也许你会仿照windows系统自带的计算器外观和功能区设计，但是现在考试要你多做出一个有新功能的计算器，实现当输入一个数是，能够将这个数分解成一个或多个素因子乘积的形式，并按素因子的大小排列显示出来。大家对计算器中数的表示应该是很清楚的，下面显示出了0-9这是个数组的表示形式，每个数字占据5*3大小的字符区域。

 你能实现这个心能能吗？试试看吧！

 

输入

能输入多组测试数据，每组包括一个正整数n(n<=1000000)

 

 

输出

 

对于每个数，将它分解成若干个素数乘积的形式，应按从小到大的顺序输出，素因子之间用“*”的形式链接

 

 

样例输入

10
2

 

 

样例输出

 -     -   
  |   |    
 -  *  -   
|       |  
 -     -   
 
 -   
  |  
 -   
|    
 -   

 

 

解析：

此题主要有两个难点，第一是将一个整数分解为素数的乘积，第二是将各个数显示出来，

 

代码

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

public class WindowsNumber {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		
		Scanner scanner = new Scanner(System.in);
		while(scanner.hasNextInt()) {
			compute(scanner.nextInt());
		}
		scanner.close();
	}
	
	public static void compute(int number) {
		String[] number0 = new String[]{" - ", "| |", "   ", "| |", " - "};
		String[] number1 = new String[]{"   ", "  |", "   ", "  |", "   "};
		String[] number2 = new String[]{" - ", "  |", " - ", "|  ", " - "};
		String[] number3 = new String[]{" - ", "  |", " - ", "  |", " - "};
		String[] number4 = new String[]{"   ", "| |", " - ", "  |", "   "};
		String[] number5 = new String[]{" - ", "|  ", " - ", "  |", " - "};
		String[] number6 = new String[]{" - ", "|  ", " - ", "| |", " - "};
		String[] number7 = new String[]{" - ", "  |", "   ", "  |", "   "};
		String[] number8 = new String[]{" - ", "| |", " - ", "| |", " - "};
		String[] number9 = new String[]{" - ", "| |", " - ", "  |", " - "};
		
		List<List<String[]>> resultNumbers = new ArrayList<List<String[]>>();
		//保存被分解的数
		int[] result = new int[10000];
		int resultLength = 0;
		resultLength = decodeInt(number, result);
		for(int i=0; i<resultLength; i++) {
			List<String[]> resultNumber = new ArrayList<String[]>();
			int temp = result[i];
			while((temp%10 != 0) || (temp/10>0)) {
				int mod = temp % 10;
				temp = temp / 10;
				switch (mod) {
				case 0:
					resultNumber.add(0, number0);
					break;
				case 1:
					resultNumber.add(0, number1);
					break;
				case 2:
					resultNumber.add(0, number2);
					break;
				case 3:
					resultNumber.add(0, number3);
					break;
				case 4:
					resultNumber.add(0, number4);
					break;
				case 5:
					resultNumber.add(0, number5);
					break;
				case 6:
					resultNumber.add(0, number6);
					break;
				case 7:
					resultNumber.add(0, number7);
					break;
				case 8:
					resultNumber.add(0, number8);
					break;
				case 9:
					resultNumber.add(0, number9);
					break;
				}
			}
			resultNumbers.add(resultNumber);
			
		}
		
		//将最终结果输出
		for(int i=0; i<5; i++) {
			for(int j=0; j<resultLength; j++) {
				List<String[]> resultNumber = resultNumbers.get(j);
				for(int k=0; k<resultNumber.size(); k++) {
					System.out.print(resultNumber.get(k)[i]);
				}
				if((i==2) && (j!=resultLength-1))
					System.out.print(" * ");
				else
					System.out.print("   ");
			}
			System.out.println();
		}
	}
	
        //此处代码借用别人的
	public static int decodeInt(int n, int[] facArr) {
		int fac;
		int count;
		if (n < 4) {
			facArr[0] = n;
			return 1;
		}
		count = 0;
		// 下面的while循环为2试出n，直到2不是n的因子为止
		while ((n & 1) == 0) // 这里判断偶数用 （n &1）==0，这比(n % 2)==0更快
		{
			facArr[count++] = 2;
			n /= 2;
		}

		fac = 3; // 用3到sqrt(n)之间的奇数试除
		while (fac * fac <= n) // fac*fac <= n
		{
			while (n % fac == 0) {
				facArr[count++] = fac;
				n /= fac;
			}
			fac += 2;
		}
		if (n == 1)
			return count;

		facArr[count++] = n;
		return count;
	}

}