leetcode 672. Bulb Switcher II 轮流开关灯泡 + 数学公式

旧城等待， 2022-06-03 07:37 186阅读 0赞

There is a room with n lights which are turned on initially and 4 buttons on the wall. After performing exactly m unknown operations towards buttons, you need to return how many different kinds of status of the n lights could be.

Suppose n lights are labeled as number \[1, 2, 3 …, n\], function of these 4 buttons are given below:

Flip all the lights.  
Flip lights with even numbers.  
Flip lights with odd numbers.  
Flip lights with (3k + 1) numbers, k = 0, 1, 2, …  
Example 1:  
Input: n = 1, m = 1.  
Output: 2  
Explanation: Status can be: \[on\], \[off\]  
Example 2:  
Input: n = 2, m = 1.  
Output: 3  
Explanation: Status can be: \[on, off\], \[off, on\], \[off, off\]  
Example 3:  
Input: n = 3, m = 1.  
Output: 4  
Explanation: Status can be: \[off, on, off\], \[on, off, on\], \[off, off, off\], \[off, on, on\].  
Note: n and m both fit in range \[0, 1000\].

看到有些答案，再一次感觉到智商被碾压了

建议和[leetcode 319. Bulb Switcher 轮流开关灯泡 + 数学公式][leetcode 319. Bulb Switcher _ _] 一起学习

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    
    using namespace std;
    
    
    class Solution
    {
    public:
        int flipLights(int n, int m) 
        {
            n = min(n, 3);
            return min(1 << n, 1 + m * n);
        }
    };

[leetcode 319. Bulb Switcher _ _]: http://blog.csdn.net/jackzhang_123/article/details/78150363