Python程序，用于给定长度的跳转次数，以从二维平面的原点到达形式点（d，0）-yiteyi-C++库

给定三个正整数 a、 b 和 D .您当前位于无限二维坐标平面上的原点（0,0）。你可以跳到二维平面上的任意点，欧几里德距离等于 A. 或 B 从你现在的位置。任务是找到从（0，0）到达（d，0）所需的最小跳跃次数。

null

例如：

Input : a = 2, b = 3, d = 1 
Output : 2
First jump of length a = 2, (0, 0) -> (1/2, √15/2)
Second jump of length a = 2, (1/2, √15/2) -> (1, 0)
Thus, only two jump are required to reach 
(1, 0) from (0, 0).

Input : a = 3, b = 4, d = 11 
Output : 3
(0, 0) -> (4, 0) using length b = 4
(4, 0) -> (8, 0) using length b = 4
(8, 0) -> (11, 0) using length a = 3

                     # Python code to find the minimum number                   
                     # of jump required to reach                   
                     # (d, 0) from (0, 0)                   
                             
                     def                               minJumps(a, b, d):                   
                             
                                         temp                               =                               a                   
                                         a                               =                               min                               (a, b)                   
                                         b                               =                               max                               (temp, b)                   
                             
                                         if                               (d >                               =                               b):                   
                                         return                               (d                               +                               b                               -                               1                               )                               /                               b                   
                             
                                         # if d is 0                   
                                         if                               (d                               =                               =                               0                               ):                   
                                         return                               0                   
                             
                                         # if d is equal to a.                   
                                         if                               (d                               =                               =                               a):                   
                                         return                               1                   
                             
                                         # else make triangle, and only 2                   
                                         # steps required.                   
                                         return                               2                   
                             
                     # main()                   
                     a                               =                               3                   
                     b                               =                               4                   
                     d                               =                               11                   
                     print                               (                               int                               (minJumps(a, b, d)))                   
                             
                     # Contributed by _omg                   

输出

请参阅完整的文章从二维平面原点到达形状点（d，0）所需的给定长度的跳跃次数更多细节！

文章版权归作者所有，未经允许请勿转载。

THE END

技术文章