319. Bulb Switcher

标签：ret   man   less   other   lbs   code   use   except   cep   

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 nrounds.

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.

分析： so all number have even number of factors except square number(e.g: factor of 9:1,3,9).
square number must turn on because of odd number of factors(9: turn on at 1st, off at 3rd, on at 9th)
other number must turn off(6: turn on at 1st, off at 2nd, on at 3rd, off at 6th)
so we only need to compute the number of square number less equal than n

public class Solution {
    public int bulbSwitch(int n) {
        if(n == 0 || n==1) return n;
        return (int)Math.sqrt(n);
    }
}

319. Bulb Switcher

标签：ret   man   less   other   lbs   code   use   except   cep   

原文地址：http://www.cnblogs.com/joannacode/p/6132591.html