Problem:
Score : 400 points

Problem Statement

Given is a positive integer N.

Find the number of pairs (A, B) of positive integers not greater than N that satisfy the following condition:

When A and B are written in base ten without leading zeros, the last digit of A is equal to the first digit of B, and the first digit of A is equal to the last digit of B.

Constraints

1 \leq N \leq 2 \times 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

25

Sample Output 1

17

The following 17 pairs satisfy the condition: (1,1), (1,11), (2,2), (2,22), (3,3), (4,4), (5,5), (6,6), (7,7), (8,8), (9,9), (11,1), (11,11), (12,21), (21,12), (22,2), and (22,22).

Sample Input 2

1

Sample Output 2

1

Sample Input 3

100

Sample Output 3

108

Sample Input 4

2020

Sample Output 4

40812

Sample Input 5

200000

Sample Output 5

400000008

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;

namespace ConsoleApp2
{
    public class Program
    {
        public static void Main(string[] args)
        {
            int N = int.Parse(Console.ReadLine());

            int count = 0;
            for (int l = 1; l < 9; l++)
            {
                for (int r = l + 1; r <= 9; r++)
                {
                    if (l != r)
                    {
                        int[] ab = GetIntsCounts(N, l, r);
                        count += ab[0]*ab[1];
                    }
                }
            }

            count *= 2;

            for (int i = 1; i <= 9; i++)
            {
                int ab = GetIntsCount(N, i);
                count += (int)Math.Pow(ab, 2);
            }

            Console.WriteLine(count);
            Console.ReadLine();
        }

        static int[] GetIntsCounts(int limit,int L,int R)
        {
            int[] ans = {0,0 };
            for (int i = 1; i <= limit; i++)
            {
                int r = i % 10;
                int digit = (int)Math.Log10(i) + 1;
                int l = i / (int)(Math.Pow(10, digit - 1));
                if (r == R && l == L) ans[0]++;
                if (r == L && l == R) ans[1]++;

            }
            return ans;
        }

        static int GetIntsCount(int limit, int n)
        {
            int ans =0;
            for (int i = 1; i <= limit; i++)
            {
                int r = i % 10;
                int digit = (int)Math.Log10(i) + 1;
                int l = i / (int)(Math.Pow(10, digit - 1));
                if (r == n && l == n) ans++;

            }
            return ans;
        }
    }
}