中心平方数

给定数字n ，任务是找到^第n^个中心平方数。

中心正方形数是一个中心的图形数，它在一个正方形中给出一个点的数量，其中一个点位于中心，而围绕中心点的所有其他点则位于连续的正方形层中。可以使用公式n ² +(n-1) ²来计算^第N^个中心平方数。

例子 ：

Input : n = 2
Output : 5

Input : n = 9
Output : 145

1. 查找第n个中心平方数
   如果仔细观察，我们会注意到第n个中心平方数可以看作是两个连续平方数的总和(1点，4点，9点，16点等)。

   我们可以使用以下公式找到第n个中心平方数。

   n-th Centered Square Number = n² + (n-1)²

   下面是实现：

   C++

   // C++ program to find nth
   // Centered square number.
   #include 
     
   using namespace std;
     
   // Function to calculate Centered
   // square number function
   int centered_square_num(int n)
   {
       // Formula to calculate nth
       // Centered square number
       return n * n + ((n - 1) * (n - 1));
   }
     
   // Driver Code
   int main()
   {
       int n = 7;
       cout << n << "th Centered square number: ";
       cout << centered_square_num(n);
       return 0;
   }

   Java

   // Java program to find nth Centered square
   // number
   import java.io.*;
     
   class GFG {
     
       // Function to calculate Centered
       // square number function
       static int centered_square_num(int n)
       {
           // Formula to calculate nth
           // Centered square number
           return n * n + ((n - 1) * (n - 1));
       }
         
       // Driver Code
       public static void main (String[] args) 
       {
           int n = 7;
           System.out.print( n + "th Centered"
                          + " square number: "
                    + centered_square_num(n));
       }
   }
     
   // This code is contributed by anuj_67.

   Python3

   # Python program to find nth
   # Centered square number.
     
     
   # Function to calculate Centered
   # square number function
   def centered_square_num(n):
     
       # Formula to calculate nth
       # Centered square number
       return n * n + ((n - 1) * (n - 1))
     
     
   # Driver Code
   n = 7
   print("%sth Centered square number: " %n,
                     centered_square_num(n))

   C#

   // C# program to find nth
   // Centered square number.
   using System;
     
   public class GFG {
     
       // Function to calculate Centered
       // square number function
       static int centered_square_num(int n)
       {
           // Formula to calculate nth
           // Centered square number
           return n * n + ((n - 1) * (n - 1));
       }
         
       // Driver Code
     
       static public void Main (){
       int n = 7;
       Console.WriteLine( n + "th Centered"
                       + " square number: "
                  + centered_square_num(n));
       }
   }
     
   // This code is contributed by anuj_67.

   ---

   输出 ：

   7th Centered square number: 85
   

2. 检查N是否为中心平方数：
   • 前几个居中平方数字是：

     1,5,13,25,41,61,85,113,145,181,…………

   • 由于第n个中心平方数为

     H(n) = n * n + ((n - 1) * (n - 1))

   • 该公式表明，第n个中心平方数的数量平方依赖于n。因此，尝试找到N = H(n)方程的正整数根。

     H(n) = nth centered-square-number number
     N = Given Number
     
     Solve for n:
     H(n) = N
     n * n + ((n - 1) * (n - 1)) = N
     
     Applying Shridharacharya Formula
     The positive root of equation (i)
     n = (9 + sqrt(36*N+45))/18; 
     

   • 获得n后，检查它是否为整数。如果n – floor(n)为0，则n为整数。

   下面是上述方法的实现：

   CPP

   #include 
   using namespace std;
     
   bool centeredSquare_number(int N) 
   {     
       float n = (9 + sqrt(36*N+45))/18;  
       return (n - (int) n) == 0; 
   } 
     
   int main() 
   { 
       int i = 13;
       cout<

   Java

   // Java Code implementation of the above approach
   class GFG {
         
       static int centeredSquare_number(int N) 
       {     
           float n = (9 + (float)Math.sqrt(36*N+45))/18;  
           if (n - (int) n == 0)
               return 1;
           else
               return 0;
       } 
         
       // Driver code
       public static void main (String[] args) 
       { 
           int i = 13;
           System.out.println(centeredSquare_number(i));
       } 
         
   }
     
   // This code is contributed by Yash_R
   

   Python3

   # Python3 implementation of the above approach
   from math import sqrt
     
   def centeredSquare_number(N) : 
     
       n = (9 + sqrt(36 * N + 45))/18; 
       if (n - int(n)) == 0 :
           return 1
       else :
           return 0
     
   # Driver Code
   if __name__ == "__main__" : 
     
       i = 13;
       print(centeredSquare_number(i));
     
   # This code is contributed by Yash_R
   

   C＃

   // C# Code implementation of the above approach
   using System;
     
   class GFG {
         
       static int centeredSquare_number(int N) 
       {     
           float n = (9 + (float)Math.Sqrt(36 * N + 45))/18;  
           if (n - (int) n == 0)
               return 1;
           else
               return 0;
       } 
         
       // Driver code
       public static void Main (String[] args) 
       { 
           int i = 13;
           Console.WriteLine(centeredSquare_number(i));
       } 
         
   }
     
   // This code is contributed by Yash_R
   

   输出：

   0
   

参考： https ： //en.wikipedia.org/wiki/Centered_square_number