将圆心移动到目标的最小转数

给定半径为r的圆并以point(x1，y1)为中心，并给出一个point(x2，y2)。任务是使用最少的步数将圆心从给定的中心(x1，y1)移至目标(x2，y2)。第一步，我们可以在任意点上将销钉放在圆的边界上，然后围绕该销钉旋转圆弧任意角度，最后取下销钉。

例子 ：

Input : r = 2 
        x1 = 0, y1 = 0
        x2 = 0, y2 = 4
Output :1

Input  : r = 1
         x1 = 1, y1 = 1,
         x2 = 4, y2 = 4 
Output : 3

让我们考虑两个中心之间的直线。显然，要使中心移动最大距离，我们需要将其绕直线旋转180度。因此，每次我们可以移动中心的最大距离为2 * r。让我们继续每次以2 * r的距离移动中心，直到两个圆相交。现在显然可以通过围绕两个圆的交点之一旋转中心，使其到达最终位置，从而使中心移至最终位置。

每次使圆移动2 * r次(最后一次移动除外)，它都会小于<2 * r。设两点之间的初始距离为d。该问题的解决方案将是ceil(d / 2 * r)。

C++

// C++ program to find minimum number of
// revolutions to reach a target center
#include
using namespace std;
  
// Minimum revolutions to move center from
// (x1, y1) to (x2, y2)
int minRevolutions(double r, int x1, int y1,
                             int x2, int y2)
{
    double d = sqrt((x1 - x2)*(x1 - x2) +
                    (y1 - y2)*(y1 - y2));
    return ceil(d/(2*r));
}
  
// Driver code
int main()
{
    int r = 2, x1 = 0, y1 = 0, x2 = 0, y2 = 4;
    cout << minRevolutions(r, x1, y1, x2, y2);
    return 0;
}

Java

// Java program to find minimum number of
// revolutions to reach a target center
class GFG {
      
    // Minimum revolutions to move center 
    // from (x1, y1) to (x2, y2)
    static double minRevolutions(double r, 
         int x1, int y1, int x2, int y2) 
    {
          
        double d = Math.sqrt((x1 - x2) 
                   * (x1 - x2) + (y1 - y2) 
                   * (y1 - y2));
                     
        return Math.ceil(d / (2 * r));
    }
      
    // Driver Program to test above function
    public static void main(String arg[]) {
          
        int r = 2, x1 = 0, y1 = 0;
        int x2 = 0, y2 = 4;
          
        System.out.print((int)minRevolutions(r,
                              x1, y1, x2, y2));
    }
}
  
// This code is contributed by Anant Agarwal.

Python3

# Python program to find
# minimum number of
# revolutions to reach
# a target center
import math
  
# Minimum revolutions to move center from
# (x1, y1) to (x2, y2)
def minRevolutions(r,x1,y1,x2,y2):
  
    d = math.sqrt((x1 -x2)*(x1 - x2) +
         (y1 - y2)*(y1 - y2))
    return math.ceil(d//(2*r))
  
# Driver code
  
r = 2
x1 = 0
y1 = 0
x2 = 0
y2 = 4
  
print(minRevolutions(r, x1, y1, x2, y2))
  
# This code is contributed
# by Anant Agarwal.

C#

// C# program to find minimum number of
// revolutions to reach a target center
using System;
  
class GFG {
      
    // Minimum revolutions to move center 
    // from (x1, y1) to (x2, y2)
    static double minRevolutions(double r, 
            int x1, int y1, int x2, int y2) 
    {
          
        double d = Math.Sqrt((x1 - x2) 
                * (x1 - x2) + (y1 - y2) 
                * (y1 - y2));
                      
        return Math.Ceiling(d / (2 * r));
    }
      
    // Driver Program to test above function
    public static void Main() 
    {
        int r = 2, x1 = 0, y1 = 0;
        int x2 = 0, y2 = 4;
          
        Console.Write((int)minRevolutions(r,
                            x1, y1, x2, y2));
    }
}
  
// This code is contributed by nitin mittal.

输出 ：

1

时间复杂度： O(1)