Pku oj 1218 THE DRUNK JAILER(开灯问题）

THE DRUNK JAILER

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 26252 Accepted: 16258

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. 

Sample Input

25100

Sample Output

210

Source

Greater New York 2002

输入一个n，表示有n个回合n个房间，第一回合改变1的倍数的房间锁的状态，第二回合改变2的倍数房间锁的状态，以此类推，数不太，O(n^2)暴力就好

#include<cstdio>#include<cstring>#include<iostream>using namespace std;const int maxn = 105;int vis[maxn];int main(){    int n;    int ans;    scanf("%d",&n);    while(n--)    {        int cnt = 0;        scanf("%d",&ans);        memset(vis,0,sizeof(vis));        for(int i=1;i<=ans;i++)        {            for(int j=1;j<=ans;j++)            {                if(j%i == 0)                {                    if(!vis[j]) vis[j] = 1;                    else vis[j] = 0;                }            }        }        for(int i=1;i<=ans;i++)        {            if(vis[i]) cnt++;        }        printf("%d\n",cnt);    }    return 0;}