題目描述
There are n lockers in a hallway numbered sequentially from 1 to n. Initially, all the locker doors are closed. You make n passes by the lockers, each time starting with locker #1. On the ith pass, i = 1, 2, …, n, you toggle the door of every ith locker: if the door is closed, you open it, if it is open, you close it. For example, after the first pass every door is open; on the second pass you only toggle the even-numbered lockers (#2, #4, …) so that after the second pass the even doors are closed and the odd ones are opened; the third time through you close the door of locker #3 (opened from the first pass), open the door of locker #6 (closed from the second pass), and so on. After the last pass, which locker doors are open and which are closed? How many of them are open? Your task is write a program to output How many doors are open after the last pass? Assumptions all doors are closed at first.
輸入
a positive numbers n, total doors. n<=100000
輸出
a positive numbers ，the total of doors opened after the last pass.
樣例輸入
10
樣例輸出
3
代碼
#include<iostream>
#include<cmath>
using namespace std;
int main(){
int n;
cin>>n;
int m =sqrt(n);
cout<<m<<endl;
return 0;
}
超級簡單有沒有？是的，就這樣。爲什麼呢？下面有詳細解說。
思路
大多數情況下，每個數的因子是成對出現的，例如數15的因子有1和15，以及3和5，數12的因子有1和12，2和6，3和4，所以有偶數個因子。只有當這個數是完全平方數時，例如1,4,9,16,25,36等等，它的因子除成對出現的以外，還有它的整數平方根作爲單獨的因子，這些完全平方數的因子數爲奇數。
最後一次經過後，只有編號爲1,4,9,16,25,36,49,81…的門會開着，其餘編號的門都關着。打開的門總數爲(sqrt(n))
