leecode_319 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 · 196 阅读

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本文探讨了在经过n轮操作后，初始全关状态的灯中亮灯的数量变化规律。每轮操作中，操作者会切换特定位置的灯的状态（开或关）。最终，文章提供了一个简洁的解决方案来计算最终亮灯的数量。

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.

class Solution {
public:
    int bulbSwitch(int n) {
        int res=sqrt(n);
        return res;
        
    }   
};