半径为 1,2,3,4……….N 的替代同心圆的面积之和

鉴于绘制了半径为1,2,3,4…….N cm 的同心圆。最小圆圈的内部被涂成白色，而有角的区域被交替涂成绿色和白色，因此没有两个相邻区域的颜色相同。任务是找出以平方厘米为单位的绿色区域的总面积。

例子：

Input: N=100
Output: 15865.042900628456

Input: N=10
Output: 172.78759594743863

方法：第一个绿色区域的面积将是-

[π(R2²- R1²)]

同样，所有替代绿色圆圈的面积将是-

[π(R4²- R3²)], [π(R6²- R5²)]…..

绿色区域的总面积是绿色区域的面积之和是-

A  =  [ π { (R2² – R1²) + (R4² – R3²) + (R6² – R5²) …………. (R(N)² –  R(N-1)²}]

A =   [ π { ((R2 – R1)(R2 + R1)) + ((R4 – R3)(R4 + R3)) + ((R6 – R5)(R6 + R5)) …………. ((RN –  R(N – 1)(RN + R(N – 1)) } ]

由于 2 个同心圆的半径之差仅为 1 cm，因此R(N) – R(N – 1) = 1。因此，

A  =   [ π {R1 + R2 + R3 + R4 + R5 + R6 ………….R(N-1) + R(N) } ]

(R1 + R2 + R3 + R4 + R5 + R6 ………….R(N-1) + R(N) 形成一个等差数列，所以我们必须找到以起始半径为 1 和最后一个半径的等差数列之和为 N，公差为 1。

AP的总和是-

S_N = N * (2 * a +(N – 1) * d) / 2  

or 

S_N = N * (a + l) / 2

因此，绿色区域的面积是-

A = Sn * π 

下面是上述方法的实现：

C++

// C++ code for the above approach
#include 
using namespace std;
 
// Function to calculate area
// of concentric circles
double area(int N, int a, int l)
{
 
    // Formula to find sum of
    // Arithmetic Progression
    double S = N * (a + l) / 2;
 
    // return area
    return S * M_PI;
}
 
// Driver code
int main()
{
   
    // N is number of concentric circles
    int N = 100;
 
    // Radius of interior circle
    int a = 1;
 
    // l radius of exterior circle
    int l = N;
 
    // r is result
    double r = area(N, a, l);
 
    // Print result
    cout << fixed << setprecision(12) << r;
}
 
// This code is contributed by Samim Hossain Mondal.

Java

// Java code for the above approach
class GFG
{
   
// Function to calculate area
// of concentric circles
static double area(int N, int a, int l)
{
 
    // Formula to find sum of
    // Arithmetic Progression
    double S = N * (a + l) / 2;
 
    // return area
    return S * Math.PI;
}
 
// Driver code
public static void main(String args[])
{
   
    // N is number of concentric circles
    int N = 100;
 
    // Radius of interior circle
    int a = 1;
 
    // l radius of exterior circle
    int l = N;
 
    // r is result
    double r = area(N, a, l);
 
    // Print result
    System.out.println(r);
}
}
 
// This code is contributed by gfgking

Python3

# import pi from math module
from math import pi
 
# Function to calculate area
# of concentric circles
def area(N,a):
       
    # Formula to find sum of
    # Arithmetic Progression
    S = N * (a + l) / 2
     
    # return area
    return S * pi
 
# N is number of concentric circles
N = 100
 
# Radius of interior circle
a = 1
 
# l radius of exterior circle
l = N
 
# r is result
r = area(N, a)
 
# Print result
print(r)

C#

// C# program for the above approach
using System;
class GFG
{
 
  // Function to calculate area
  // of concentric circles
  static double area(int N, int a, int l)
  {
 
    // Formula to find sum of
    // Arithmetic Progression
    double S = N * (a + l) / 2;
 
    // return area
    return S * Math.PI;
  }
 
  // Driver Code:
  public static void Main()
  {
    // N is number of concentric circles
    int N = 100;
 
    // Radius of interior circle
    int a = 1;
 
    // l radius of exterior circle
    int l = N;
 
    // r is result
    double r = area(N, a, l);
 
    // Print result
    Console.WriteLine(r);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

       // JavaScript code for the above approach
 
       // Function to calculate area
       // of concentric circles
       function area(N, a) {
 
           // Formula to find sum of
           // Arithmetic Progression
           S = N * (a + l) / 2
 
           // return area
           return S * Math.PI
       }
        
       // N is number of concentric circles
       N = 100
 
       // Radius of interior circle
       a = 1
 
       // l radius of exterior circle
       l = N
 
       // r is result
       r = area(N, a)
 
       // Print result
       document.write(r)
        
 // This code is contributed by Potta Lokesh
   

输出

15865.042900628456

时间复杂度： O(1)
辅助空间： O(1)