Bulb Switcher

最新推荐文章于 2018-10-21 15:30:48 发布

原创最新推荐文章于 2018-10-21 15:30:48 发布 · 1.4k 阅读

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本文探讨了一个有趣的灯泡开关问题：初始状态下所有灯泡关闭，经过n轮操作后有多少灯泡保持开启状态。每轮操作中，对于第i轮会切换每个i的倍数位置上的灯泡状态。通过分析得出，最终处于开启状态的灯泡数量等于不大于n的完全平方数的数量。

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