有趣的Java编程题

如下是在网上看到的一道题目：

大厅里有100盏灯，每盏灯都编了号码，分别为1-100。每盏灯由一个开关来控制。（开关按一下，灯亮，再按一下灯灭。开关的编号与被控制的灯相同。）开始时，灯是全灭的。如今按照如下规则按动开关。
第一次，将全部的灯点亮。
第二次，将全部2的倍数的开关按一下。
第三次，将全部3的倍数的开关按一下。
以此类推。第N次，将全部N的倍数的开关按一下。
问第N次（N大于等于2，且小于等于100）按完之后，大厅里还有几盏灯是亮的。

 

请编程实现上面的逻辑，以N为参数

个人思路以下：

1，灯的状态变化是什么。灯的开关只有俩种状态，非开即关。因此用一个变量定义一下开关状态，我用数组的值A或B来定义灯的开关，其实我以为用一个布尔值会更好。

if(i%n==0){
	if(lamp[i]=="A"){
        lamp[i]="B";
}else {
	lamp[i]="A";
       }
}

2，灯的状态如何变。第一次，全变。第二次，2的倍数变一次。第三次，3的倍数变一次。以此类推能够得出：每一次，但凡是次数整数倍的都会变。因此我用取余为0来判断是否进行变换。那么，它的开关次数的逐次递增用一个for循环来解决。

for(int n=1;n<N+1;n++){
	for(int i=0;i<100;i++){
//the change of "A"and"B"means:lamp is open or close
	    if(i%n==0){
		   if(lamp[i]=="A"){
			lamp[i]="B";
		}else {
			lamp[i]="A";
		      }
		}
	}
}

3，在灯的变化结果已经出来时，就是统计最后开着的灯的数量。用判断，累计。

（PS：对于count有定义却没有初始化值，致使一直出错却找不到缘由，通常来讲定义都有默认初始化的值，可是总有不肯定隐患发生，因此仍是要谨慎***）

for(int i=0;i<100;i++){

		System.out.print(lamp[i]);
	if(lamp[i]=="A"){
//output the number of lamp how many is open
		count=count+1;
				
	}
			
}
System.out.println();
System.out.print("the number of open lamp is:"+count);

所有代码显示以下：

import java.util.Scanner;

class Demo12{
public static void main(String[] args){
	change();
}
	public static void change(){
		//I have forget to initialize the value count,that make a false and I spend half an hour to find it.
		int count=0;
		int N;
		String[] lamp={"B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B","B"};
		
		//define the volue of N as the number of times
		System.out.println("please input an number:");
		Scanner key=new Scanner(System.in);
		N=key.nextInt();
		for(int n=1;n<N+1;n++){
			for(int i=0;i<100;i++){
			//the change of "A"and"B"means:lamp is open or close
				if(i%n==0){
					if(lamp[i]=="A"){
						lamp[i]="B";
					}else {
						lamp[i]="A";
							}
						}
					}
				}
		//output the situation of lamp
		for(int i=0;i<100;i++){
			
				System.out.print(lamp[i]);
			if(lamp[i]=="A"){
				//output the number of lamp how many is open
				count=count+1;
				
			}
			
		}
		System.out.println();
		System.out.print("the number of open lamp is:"+count);
	}

}