可以在一个大圆内接的较小圆的数目
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，任务是找出可放置在较大圆内的较小圆的数量，使得较小圆接触较大圆的边界。
示例:
输入: R1 = 3，R2 = 1 输出: 6 说明:圆的半径为 3 和 1。因此，半径为 1 的圆可以内接在半径为 3 的圆上。
根据上面的表示，通过接触大圆的边界可以内切的小圆的总数是 6。
输入: R1 = 5，R2 = 4 T3】输出: 1
逼近:求半径较小的圆 R2 在半径为的圆的圆心处与 R1 所成的角度，再除以 360 度，即可求解给定问题。 按照以下步骤解决给定问题:
- 如果 R1 的值小于 R2 ，那么单圈是不可能落款的。因此，打印 0 。
- 如果 R1 的值小于 2 * R2 ，即如果较小圆的直径大于较大圆的半径，则只能内接一个圆。因此，打印 1 。
- 否则，用下面的公式求出大圆中心处的小圆所成的角，然后除以 360 度，得到可内切圆的总数，并打印出该值。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
java 描述语言
<script>
// Javascript program for the above approach
// Function to count number of smaller
// circles that can be inscribed in
// the larger circle touching its boundary
function countInscribed(R1, R2)
{
// If R2 is greater than R1
if (R2 > R1)
return 0;
// Stores the angle made
// by the smaller circle
let angle;
// Stores the ratio
// of R2 / (R1 - R2)
let ratio;
// Stores the count of smaller
// circles that can be inscribed
let 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 = Math.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
let R1 = 3;
let R2 = 1;
document.write(countInscribed(R1, R2))
// This code is contributed by Hritik
</script>
Output:
6
时间复杂度:O(1) T5辅助空间:** O(1)
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