LeetCode[319] Bulb Switcher
2016-07-28 08:12
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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 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.
一般的数都会是灭的,因为它们有偶数个因子,只有完全平方数有奇数个因子,于是此题相当于在1~n中找完全平方数。(包括1)
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.
一般的数都会是灭的,因为它们有偶数个因子,只有完全平方数有奇数个因子,于是此题相当于在1~n中找完全平方数。(包括1)
class Solution { public: int bulbSwitch(int n) { if (n < 1) return 0; int i; for (i = 1; i * i <= n; i++); return i - 1; } };
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