用数字1到9构成的所有N个数字回文数的总和可被9整除

给定数字N ，任务是找到可被9整除的所有N个数字回文数字(由1到9的数字组成)的总和。

例子：

Input: N = 1
Output: 9
Explanation:
Only 9 is a palindromic number of 1 digit divisible by 9

Input: N = 3
Output: 4995
Explanation:
Three-digit Palindromic Numbers divisible by 9 are –
171, 252, 333, 414, 585, 666, 747, 828, 999

方法：该问题的主要观察结果是，如果一个数字可以被9整除，那么该数字的数字总和也可以被9整除。另一个观察结果是，如果我们使用来自的数字来计算N位数的回文数。 1到9，那么可以观察到

Occurrence of each digit = (count of N-digit numbers / 9)

所以，

1. 首先找到可以被9整除的N位回文数的计数，如下所示：
   $\text{Count of N-digit Palindromic numbers divisible by 9} = \begin{cases} 9^{\frac{N-1}{2}} & \text{ if N is odd} \\ 9^{\frac{N-2}{2}} & \text{ if N is even} \end{cases}$
2. 然后，如果N为1或2，则总和将分别为9和99，因为它们是唯一的1位和2位数字。
3. 如果N> 2，则第N个位数回文数的总和可被9整除为
   $\text{Sum of Nth digit palindromic numbers divisible by 9 }= (\text{sum of }(N-1)^{th}\text{ digit } * 10) + (5*\text{ count of N digit palindromic numbers divisible by 9})$

   下面是上述方法的实现：

   C++

   // C++ implementation to find the sum
   // of all the N digit palindromic
   // numbers divisible by 9
     
   #include 
     
   using namespace std;
     
   // Function for finding count of
   // N digits palindrome which
   // are divisible by 9
   int countPalindrome(int n)
   {
       int count;
     
       // if N is odd
       if (n % 2 == 1) {
           count = pow(9, (n - 1) / 2);
       }
       // if N is even
       else {
           count = pow(9, (n - 2) / 2);
       }
       return count;
   }
     
   // Function for finding sum of N
   // digits palindrome which are
   // divisible by 9
   int sumPalindrome(int n)
   {
       // count the possible
       // number of palindrome
       int count = countPalindrome(n);
     
       int res = 0;
     
       if (n == 1)
           return 9;
       if (n == 2)
           return 99;
     
       for (int i = 0; i < n; i++) {
           res = res * 10 + count * 5;
       }
     
       return res;
   }
     
   // Driver Code
   int main()
   {
       int n = 3;
       cout << sumPalindrome(n);
       return 0;
   }

   Java

   // Java implementation to find the sum
   // of all the N digit palindromic
   // numbers divisible by 9
   import java.util.*;
     
   class GFG{
         
   // Function for finding count of
   // N digits palindrome which
   // are divisible by 9
   static int countPalindrome(int n)
   {
       int count;
     
       // If N is odd
       if (n % 2 == 1)
       {
           count = (int)Math.pow(9, (n - 1) / 2);
       }
         
       // If N is even
       else
       {
           count = (int)Math.pow(9, (n - 2) / 2);
       }
       return count;
   }
     
   // Function for finding sum of N
   // digits palindrome which are
   // divisible by 9
   static int sumPalindrome(int n)
   {
         
       // Count the possible
       // number of palindrome
       int count = countPalindrome(n);
     
       int res = 0;
     
       if (n == 1)
           return 9;
       if (n == 2)
           return 99;
     
       for(int i = 0; i < n; i++)
       {
          res = res * 10 + count * 5;
       }
     
       return res;
   }
     
   // Driver Code
   public static void main(String[] args)
   {
       int n = 3;
         
       System.out.println(sumPalindrome(n));
   }
   }
     
   // This code is contributed by ANKITKUMAR34

   Python3

   # Python3 implementation to find the 
   # sum of all the N digit palindromic
   # numbers divisible by 9
     
   # Function for finding count of
   # N digits palindrome which
   # are divisible by 9
   def countPalindrome(n):
     
       count = 0
         
       # If N is odd
       if (n % 2 == 1):
           count = pow(9, (n - 1) // 2)
             
       # If N is even
       else:
           count = pow(9, (n - 2) // 2)
             
       return count
     
   # Function for finding sum of N
   # digits palindrome which are
   # divisible by 9
   def sumPalindrome(n):
     
       # Count the possible
       # number of palindrome
       count = countPalindrome(n)
     
       res = 0
     
       if (n == 1):
           return 9
       if (n == 2):
           return 99
     
       for i in range(n):
           res = res * 10 + count * 5
     
       return res
     
   # Driver Code
   n = 3
     
   print(sumPalindrome(n))
     
   # This code is contributed by ANKITKUMAR34

   C#

   // C# implementation to find the sum
   // of all the N digit palindromic
   // numbers divisible by 9
   using System;
     
   class GFG{
          
   // Function for finding count of
   // N digits palindrome which
   // are divisible by 9
   static int countPalindrome(int n)
   {
       int count;
      
       // If N is odd
       if (n % 2 == 1)
       {
           count = (int)Math.Pow(9, (n - 1) / 2);
       }
          
       // If N is even
       else
       {
           count = (int)Math.Pow(9, (n - 2) / 2);
       }
       return count;
   }
      
   // Function for finding sum of N
   // digits palindrome which are
   // divisible by 9
   static int sumPalindrome(int n)
   {
          
       // Count the possible
       // number of palindrome
       int count = countPalindrome(n);
      
       int res = 0;
      
       if (n == 1)
           return 9;
       if (n == 2)
           return 99;
      
       for(int i = 0; i < n; i++)
       {
          res = res * 10 + count * 5;
       }
      
       return res;
   }
      
   // Driver Code
   public static void Main(String[] args)
   {
       int n = 3;
          
       Console.WriteLine(sumPalindrome(n));
   }
   }
     
   // This code is contributed by Amit Katiyar

   输出：

   4995