（规律）poj - 1218 牛客OI赛制测试赛2 D-星光晚餐

THE DRUNK JAILER

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 29037		Accepted: 17926

Description

A certain * contains a long hall of n cells, each right next to each other. Each cell has a *er 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 *ers, possibly zero, realizes that their cells are unlocked and the jailer is incapacitated. They immediately escape. 
Given the number of cells, determine how many *ers 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 *ers that escape when the * has n cells. 

Sample Input

2
5
100

Sample Output

2
10

Source

Greater New York 2002

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链接：https://www.nowcoder.com/acm/contest/185/D
来源：牛客网

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 131072K，其他语言262144K
64bit IO Format: %lld

题目描述

Johnson和Nancy要在星光下吃晚餐。这是一件很浪漫的事情。

为了增加星光晚餐那浪漫的氛围，他拿出了一个神奇的魔法棒，并且可以按照一定的规则，改变天上星星的亮暗。

Johnson想考考Nancy，在他挥动魔法棒后，会有多少颗星星依旧闪耀在天空。他知道，Nancy一定会一口说出答案。

Nancy当然知道怎么做啦，但她想考考你！

Johnson先将天上n个星星排成一排，起初它们都是暗的。

他告诉他的妹子，他将挥动n次魔法棒，第i次挥动会将编号为i的正整数倍的星星的亮暗反转，即亮的星星转暗，暗的星星转亮。

Johnson想问Nancy，最终会有多少个星星依旧闪亮在天空。

输入描述:

一个整数n，含义请见题目描述。

输出描述:

一个整数ans，即n次操作后会有多少个星星依旧闪亮。

示例1

输入

复制

3

输出

复制

1

示例2

输入

复制

7

输出

复制

2

备注:

对于60%的数据：n≤2×106
对于100%的数据：n≤1018

#include<stdio.h>
#include<math.h>
int main(){
    long long n;
    scanf("%lld",&n);
    printf("%lld",(long long)sqrt(n*1.0));
    return 0;
}