Java程序，用于计算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-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。

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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

                     // Java code to find the minimum number                   
                     // of jump required to reach                   
                     // (d, 0) from (0, 0).                   
                     import                               java.io.*;                   
                             
                     class                               GFG {                   
                             
                                         // Return the minimum jump of length either a or b                   
                                         // required to reach (d, 0) from (0, 0).                   
                                         static                               int                               minJumps(                               int                               a,                               int                               b,                               int                               d)                   
                                         {                   
                                         // Assigning maximum of a and b to b                   
                                         // and assigning minimum of a and b to a.                   
                                         int                               temp = a;                   
                                         a = Math.min(a, b);                   
                                         b = Math.max(temp, b);                   
                             
                                         // if d is greater than or equal to 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                               ;                   
                                         }                   
                             
                                         // Driver code                   
                                         public                               static                               void                               main(String[] args)                   
                                         {                   
                                         int                               a =                               3                               , b =                               4                               , d =                               11                               ;                   
                                         System.out.println(minJumps(a, b, d));                   
                                         }                   
                     }                   
                             
                     // This code is contributed by vt_m                   

输出：

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