用于从二维平面中的原点到达形状点 (d, 0) 所需的给定长度的跳跃次数的Python程序

Python程序，用于从二维平面中的原点到达形状点（d，0）所需的给定长度的跳跃次数

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)。您可以在 2D 平面中的任意点上以欧几里德距离跳跃，距离当前位置等于a或b 。任务是找到从 (0, 0) 到达 (d, 0) 所需的最小跳跃次数。

例子：

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

3

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