可以刻在较大圆圈中的较小圆圈的数量

给定两个正整数R1和R2 ，其中R1和R2分别代表较大和较小圆的半径，任务是找出可以放置在较大圆内的较小圆的数量，以使较小的圆接触边界较大的圆圈。

例子：

Input: R1 = 3, R2 = 1
Output: 6
Explanation: The radii of the circles are 3 and 1. Therefore, the circle of radius 1 can be inscribed in the circle of radius 3.

From the above representation, the total number of smaller circles that can be inscribed with touching of the boundary of the larger circle is 6.

Input: R1 = 5, R2 = 4
Output: 1

方法：可以通过找到半径为R2的较小圆在半径为R1的圆的中心处形成的角度，然后将其除以360度来解决给定的问题。
请按照以下步骤解决给定的问题：

• 如果R1的值小于R2 ，则不可能内接一个圆。因此，打印0 。
• 如果R1的值小于2 * R2 ，即较小的圆的直径大于较大的圆的半径，则只能内切一个圆。因此，打印1 。
• 否则，使用以下公式找到较大圆圈中心的较小圆圈所成的角度，然后除以360度以获取可被刻写的圆圈总数并打印该值。

$\frac{1}{\pi}{|sin^{-1}(\frac{R2}{R1-R2}) * 180|}$

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count number of smaller
// circles that can be inscribed in
// the larger circle touching its boundary
int countInscribed(int R1, int R2)
{
    // If R2 is greater than R1
    if (R2 > R1)
        return 0;
 
    // Stores the angle made
    // by the smaller circle
    double angle;
 
    // Stores the ratio
    // of R2 / (R1 - R2)
    double ratio;
 
    // Stores the count of smaller
    // circles that can be inscribed
    int number_of_circles = 0;
 
    // Stores the ratio
    ratio = R2 / (double)(R1 - R2);
 
    // If the diameter of smaller
    // circle is greater than the
    // radius of the larger circle
    if (R1 < 2 * R2) {
        number_of_circles = 1;
    }
 
    // Otherwise
    else {
 
        // Find the angle using formula
        angle = abs(asin(ratio) * 180)
                / 3.14159265;
 
        // Divide 360 with angle
        // and take the floor value
        number_of_circles = 360
                            / (2
                               * floor(angle));
    }
 
    // Return the final result
    return number_of_circles;
}
 
// Driver Code
int main()
{
    int R1 = 3;
    int R2 = 1;
 
    cout << countInscribed(R1, R2);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
  
// Function to count number of smaller
// circles that can be inscribed in
// the larger circle touching its boundary
static int countInscribed(int R1, int R2)
{
     
    // If R2 is greater than R1
    if (R2 > R1)
        return 0;
 
    // Stores the angle made
    // by the smaller circle
    double angle;
 
    // Stores the ratio
    // of R2 / (R1 - R2)
    double ratio;
 
    // Stores the count of smaller
    // circles that can be inscribed
    int number_of_circles = 0;
 
    // Stores the ratio
    ratio = R2 / (double)(R1 - R2);
 
    // If the diameter of smaller
    // circle is greater than the
    // radius of the larger circle
    if (R1 < 2 * R2)
    {
        number_of_circles = 1;
    }
 
    // Otherwise
    else
    {
         
        // Find the angle using formula
        angle = Math.abs(Math.asin(ratio) * 180) /
                3.14159265;
 
        // Divide 360 with angle
        // and take the floor value
        number_of_circles = (int)(360 /
        (2 * Math.floor(angle)));
    }
     
    // Return the final result
    return number_of_circles;
}
 
// Driver Code
public static void main(String args[])
{
    int R1 = 3;
    int R2 = 1;
     
    System.out.println(countInscribed(R1, R2));
}
}
 
// This code is contributed by ipg2016107

Python3

# Python3 program for the above approach
import math
 
# Function to count number of smaller
# circles that can be inscribed in
# the larger circle touching its boundary
def countInscribed(R1, R2):
     
    # If R2 is greater than R1
    if (R2 > R1):
        return 0
 
    # Stores the angle made
    # by the smaller circle
    angle = 0
 
    # Stores the ratio
    # of R2 / (R1 - R2)
    ratio = 0
 
    # Stores the count of smaller
    # circles that can be inscribed
    number_of_circles = 0
 
    # Stores the ratio
    ratio = R2 / (R1 - R2)
 
    # If the diameter of smaller
    # circle is greater than the
    # radius of the larger circle
    if (R1 < 2 * R2):
        number_of_circles = 1
     
    # Otherwise
    else:
 
        # Find the angle using formula
        angle = (abs(math.asin(ratio) * 180) /
                 3.14159265)
        
        # Divide 360 with angle
        # and take the floor value
        number_of_circles = (360 / (2 *
               math.floor(angle)))
     
    # Return the final result
    return number_of_circles
 
# Driver Code
if __name__ == "__main__":
 
    R1 = 3
    R2 = 1
 
    print (int(countInscribed(R1, R2)))
 
# This code is contributed by ukasp

C#

// C# program for the above approach
using System;
 
class GFG{
  
// Function to count number of smaller
// circles that can be inscribed in
// the larger circle touching its boundary
static int countInscribed(int R1, int R2)
{
     
    // If R2 is greater than R1
    if (R2 > R1)
        return 0;
 
    // Stores the angle made
    // by the smaller circle
    double angle;
 
    // Stores the ratio
    // of R2 / (R1 - R2)
    double ratio;
 
    // Stores the count of smaller
    // circles that can be inscribed
    int number_of_circles = 0;
 
    // Stores the ratio
    ratio = R2 / (double)(R1 - R2);
 
    // If the diameter of smaller
    // circle is greater than the
    // radius of the larger circle
    if (R1 < 2 * R2)
    {
        number_of_circles = 1;
    }
 
    // Otherwise
    else
    {
         
        // Find the angle using formula
        angle = Math.Abs(Math.Asin(ratio) * 180) /
                3.14159265;
 
        // Divide 360 with angle
        // and take the floor value
        number_of_circles = (int)(360 /
        (2 * Math.Floor(angle)));
    }
     
    // Return the final result
    return number_of_circles;
}
 
// Driver Code
public static void Main()
{
    int R1 = 3;
    int R2 = 1;
     
    Console.WriteLine(countInscribed(R1, R2));
}
}
 
// This code is contributed by mohit kumar 29

输出：

6

时间复杂度： O(1)
辅助空间： O(1)