Leetcode - Brainteaser - 319. Bulb Switcher（规律题）

1. Problem Description

There are n bulbs that are initially off. You first turn on all the bulbs. Then, you turn off every second bulb. On the third round, you toggle every third bulb (turning on if it's off or turning off if it's on). For the ith round, you toggle every i bulb. For the nth round, you only toggle the last bulb. Find how many bulbs are on after n rounds.

 

Example:

Given n = 3.

At first, the three bulbs are [off, off, off].

After first round, the three bulbs are [on, on, on].

After second round, the three bulbs are [on, off, on].

After third round, the three bulbs are [on, off, off].

So you should return 1, because there is only one bulb is on.

很常见的一道规律题。一开始所有灯泡都是关着的，第一次（从0开始）每隔一个改变灯泡的状态，第二次每隔两个，第三次每隔三个……以此类推，给出初始灯泡数目n，问n次后剩下几个亮的。

2. My solution

找规律，剩下的都是完全平方数。

class Solution
{
public:
    int bulbSwitch(int n)
    {
       int i;
       for(i=1;i*i<=n;i++) ;
       return i-1;
    }
};