319. Bulb Switcher

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.

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题意：第n轮拨动编号为n*k的灯泡,k=1,2,3,,4.....

1是1的倍数。

2是1,2的倍数。

3,是1,3的倍数

4是1,2,4,的倍数

因为除了平方因子，其他的都是成对出现，所以题目可以转化为求少于n的平方数的个数。

class Solution {public:    int bulbSwitch(int n) {        int ans=0;        for(int i=1;i*i<=n;i++){            ans++;        }        return ans;    }};