hdu 1337 The Drunk Jailer(规律)

The Drunk Jailer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 821    Accepted Submission(s): 677

Problem 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

 

 

Recommend

Ignatius.L

 

思路：逛同一个号码两次，和没走是一样。所以当某个数的因子数为奇数时这个囚房就没锁。

所以问题就相当给你一个数，问这个数以内有多少个方数。

#include<iostream>#include<cmath>using namespace std;long long s;int main(){   int cas;    while(cin>>cas)    {        while(cas--)        {            cin>>s;            s=(long long )sqrt((double)s);            cout<<s<<"\n";        }    }}