THE DRUNK JAILER POJ - 1218

THE DRUNK JAILER POJ - 1218

题目描述
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.

Sample Input
2
5
100

Sample Output
2
10

大致题意
有n个监狱,编号1到n，开始都是关闭的，进行n局游戏，第一局，把所有的监狱都打开，第i(i>=2)局，把编号为 i 的倍数的监狱的状态改变（打开变为关闭或关闭变为打开）。问n局游戏以后，有多少个监狱为打开状态。

思路
用一维数组来记录监狱门的开关状态，模拟进行n次游戏即可

代码如下

#include <iostream> 
#include <cstdio>
#include <cstdlib>
#include <fstream>
#include <math.h>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include<sstream>
#include<ctime>
using namespace std;

int main()
{
    int t,n,i,j;
    int f[105];
    cin>>t;
    while(t--)
    {
        cin>>n;
        for(i=1;i<=n;i++)
        {
           f[i]=1;
           for(j=2;j<=n;j++)
           if(i%j==0)
           f[i]=!f[i];
       }
       int sum=0;
       for(i=1;i<=n;i++)
       sum+=f[i];
       cout<<sum<<endl;
    }

}