319. Bulb Switcher

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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 i-th round, you toggle every i bulb. For the n-th round, you only toggle the last bulb. Find how many bulbs are on after n rounds.

Example:

Input: 3
Output: 1 
Explanation: 
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.

Approath #1: Math. [Java]

class Solution {
    public int bulbSwitch(int n) {
        return (int)Math.sqrt(n);
    }
}

Analysis:

A bulb ends up on if it is switched an odd number of times.

Call them bulb 1 to bulb n. Bulb i is switched in round d if and only if d divides i. So bulb i ends up on if and only if it has an odd number of divisors.

Divisors come in pairs, like i = 12 has divisors 1 and 12, 2 and 6, and 3 and 4. Except when i is a square, like 36 has divisors 1 and 36, 2 and 18, 3 and 12, 4 and 9, and double divisor 6. So bulb i ends up on if and only if i is a square.

So just count the square numbers.

Let R = int(sqrt(n)). That is the root of the largestsquare in the range [1, n]. And 1 is the smallest root. So you have the roots from 1 to R, that‘s R roots. Which correspond to the R squares. So int(sqrt(n)) is the answer. 

Reference:

https://leetcode.com/problems/bulb-switcher/discuss/77104/Math-solution..

319. Bulb Switcher

标签：only   reference   tps   nbsp   exce   NPU   nal   number   after   

原文地址：https://www.cnblogs.com/ruruozhenhao/p/10805330.html