Poj1218_THE DRUNK JAILER(水题)

一、Description

A certain prison contains a long hall of n cells, each right next to each other. Each cell has a prisoner in it, and each cell is locked.

One night, the jailer gets bored and decides to play a game. For round 1 of the game, he takes a drink of whiskey,and then runs down the hall unlocking each cell. For round 2, he takes a drink of whiskey, and then runs down the

hall locking every other cell (cells 2, 4, 6, ?). For round 3, he takes a drink of whiskey, and then runs down the hall. He visits every third cell (cells 3, 6, 9, ?). If the cell is locked, he unlocks it; if it is unlocked, he locks it. He

repeats this for n rounds, takes a final drink, and passes out.

Some number of prisoners, possibly zero, realizes that their cells are unlocked and the jailer is incapacitated. They immediately escape.

Given the number of cells, determine how many prisoners escape jail.

Input

The first line of input contains a single positive integer. This is the number of lines that follow. Each of the following lines contains a single integer between 5 and 100, inclusive, which is the number of cells n.

Output

For each line, you must print out the number of prisoners that escape when the prison has n cells.

二、问题分析

此题属于小弟的消暑计划之一，通俗点就是水题。题意通俗易懂，解题没什么难度，要的就是这样的效果啊。哈哈哈！

三、Java代码

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner cin=new Scanner(System.in);
		int times=cin.nextInt();
		for(int i=0;i<times;i++){
			int n=cin.nextInt();
			boolean[] b=new boolean[n+1];
			int num=0;
			for(int l=1;l<=n;l++){
				b[l]=true;
			}
			for(int m=1;m<=n;m++){
				for(int j=2;j<=n;j++){
					if(b[m] && m%j==0){
						b[m]=false;
					}else if( !b[m] && m%j==0){
						b[m]=true;
					}
				}
			}
			for(int j=1;j<=n;j++){
				if(b[j])
					num++;
			}
			System.out.println(num);
		}
	}
}

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