POJ 1218 THE DRUNK JAILER

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THE DRUNK JAILER

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 27502 Accepted: 17079

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

[S

水题，只要理解题意，模拟一下就可以，题意大致就是：一开始监狱大门全都是关着的，然后n=1是，操作（对打开的门关上它，对关上的门打开它）所有1的倍数的门，n=2时操作所有2的倍数的门以此类推

ac代码：

#include <stdio.h>int main(){int n,a[101],t,cnt;int i,j;scanf("%d",&n);while(n--){scanf("%d",&t);cnt=0;for(i=1;i<101;i++){a[i]=1;}for(i=1;i<=t;i++){for(j=i;j<=t;j+=i){a[j]=a[j]*(-1);}}for(i=1;i<=t;i++){if(a[i]<0){cnt++;}}printf("%d\n",cnt);}return 0;}