给定一个长度为l和宽度为b的矩形,我们必须找到可以内接在该矩形中的最大圆。
例子:
Input : l = 4, b = 8
Output : 12.56
Input : l = 16 b = 6
Output : 28.26
从图中我们可以看出,矩形内可内切的最大圆的半径总是等于矩形短边的一半。所以从图中,
radius, r = b/2 &
Area, A = π * (r^2)
C++
// C++ Program to find the biggest circle
// which can be inscribed within the rectangle
#include
using namespace std;
// Function to find the area
// of the biggest circle
float circlearea(float l, float b)
{
// the length and breadth cannot be negative
if (l < 0 || b < 0)
return -1;
// area of the circle
if (l < b)
return 3.14 * pow(l / 2, 2);
else
return 3.14 * pow(b / 2, 2);
}
// Driver code
int main()
{
float l = 4, b = 8;
cout << circlearea(l, b) << endl;
return 0;
}
Java
// Java Program to find the
// biggest circle which can be
// inscribed within the rectangle
class GFG
{
// Function to find the area
// of the biggest circle
static float circlearea(float l,
float b)
{
// the length and breadth
// cannot be negative
if (l < 0 || b < 0)
return -1;
// area of the circle
if (l < b)
return (float)(3.14 * Math.pow(l / 2, 2));
else
return (float)(3.14 * Math.pow(b / 2, 2));
}
// Driver code
public static void main(String[] args)
{
float l = 4, b = 8;
System.out.println(circlearea(l, b));
}
}
// This code is contributed
// by ChitraNayal
Python 3
# Python 3 Program to find the
# biggest circle which can be
# inscribed within the rectangle
# Function to find the area
# of the biggest circle
def circlearea(l, b):
# the length and breadth
# cannot be negative
if (l < 0 or b < 0):
return -1
# area of the circle
if (l < b):
return 3.14 * pow(l // 2, 2)
else:
return 3.14 * pow(b // 2, 2)
# Driver code
if __name__ == "__main__":
l = 4
b = 8
print(circlearea(l, b))
# This code is contributed
# by ChitraNayal
C#
// C# Program to find the
// biggest circle which can be
// inscribed within the rectangle
using System;
class GFG
{
// Function to find the area
// of the biggest circle
static float circlearea(float l,
float b)
{
// the length and breadth
// cannot be negative
if (l < 0 || b < 0)
return -1;
// area of the circle
if (l < b)
return (float)(3.14 * Math.Pow(l / 2, 2));
else
return (float)(3.14 * Math.Pow(b / 2, 2));
}
// Driver code
public static void Main()
{
float l = 4, b = 8;
Console.Write(circlearea(l, b));
}
}
// This code is contributed
// by ChitraNayal
PHP
Javascript
输出:
12.56
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