81 lines
2.3 KiB
C++
81 lines
2.3 KiB
C++
// Source : https://leetcode.com/problems/mirror-reflection/description/
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// Author : Hao Chen
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// Date : 2018-06-27
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/***************************************************************************************
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*
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* There is a special square room with mirrors on each of the four walls. Except for
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* the southwest corner, there are receptors on each of the remaining corners, numbered
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* 0, 1, and 2.
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*
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* The square room has walls of length p, and a laser ray from the southwest corner
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* first meets the east wall at a distance q from the 0th receptor.
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*
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* Return the number of the receptor that the ray meets first. (It is guaranteed that
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* the ray will meet a receptor eventually.)
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*
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*
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*
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*
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* Example 1:
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*
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*
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* Input: p = 2, q = 1
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* Output: 2
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* Explanation: The ray meets receptor 2 the first time it gets reflected back to the
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* left wall.
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*
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*
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*
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*
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* Note:
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*
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*
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* 1 <= p <= 1000
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* 0 <= q <= p
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*
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***************************************************************************************/
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/*
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* Solution
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* --------
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*
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* We know the following things:
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* 1）every reflection will increase the step of `q`.
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* 2) when reach the top, the reflection would go down, when reach the bottom the reflection would go up.
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*
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* So, we can image if there have two walls, left one and right one, then the reflection can go up instanstly,
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*
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* - the reflection points on left wall would be even times of `q`.
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* - the reflection points on right wall would be odd times of `q`.
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*
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* And in the right wall, the receptors `#0` would be the times of `2p`.
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*
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* So, we need find the least common multiple of `p` and `q`, then we can have the answer.
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*/
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class Solution {
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private:
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//GCD - greatest common divisor 最大公因数
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int greatestCommonDivisor (int a, int b) {
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if(b) while((a %= b) && (b %= a));
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return a + b;
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}
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//LCM - least common multiple 最小公倍数
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int leastCommonMultiple(int a, int b) {
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return a * b / greatestCommonDivisor(a, b);
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}
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public:
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int mirrorReflection(int p, int q) {
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int lcm = leastCommonMultiple(p, q);
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if (lcm % (2*p) == 0 ) return 0;
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int nq = lcm / q;
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if (nq % 2 == 0 ) return 2;
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return 1;
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}
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};
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