THE DRUNK JAILER(POJ--1218

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.
题意:总共有n个监狱,the jailer闲着无聊做了个游戏,跑第一圈时把所有的监狱都打开,跑第二圈时把是2的倍数的监狱都锁上,跑第三圈时把是3的倍数的监狱原来打开的就锁上原来锁上的就打开,以此类推,总共跑n圈,求n圈后还有多少间监狱是开着的.(类似于开灯问题)

Sample Input

25100

Sample Output

210
<span style="font-size:18px;">#include <cstdio>#include <math.h>#include <algorithm>#include <cstring>#define pi 3.141592654using namespace std;int main(){    int T;    bool a[150];    scanf("%d",&T);    while(T--)    {        int n;        scanf("%d",&n);        memset(a,0,sizeof(a));        for(int i=2; i<=n; i++)            for(int j=1; j<=n; j++)                if(j%i==0)                    a[j]=!a[j];        int sum=0;        for(int i=1; i<=n; i++)            if(!a[i])                sum++;        printf("%d\n",sum);    }    return 0;}</span>