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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();

        long ans = 0;
        for(int a = 1 ; a <= n ; a++){
            int tmp = 1;
            while(tmp*10 < a){
                tmp *= 10;
            }
            ans += count(a% 10, a/tmp, n);
        }
        System.out.println(ans);
    }

    static long count(int first, int last, int max){
        if(first == 0 || last == 0){
            return(0);
        }
        int keta = 0;
        int tmp = max;
        while(tmp > 0){
            tmp /= 10;
            keta++;
        }

        long ans = 0;
        if(first == last && max >= first){
            ans = 1;
        }
        if(keta == 1){
            return(ans);
        }

        long tens = 1;
        for(int i = 2 ; i < keta ; i++){
            ans += tens;
            tens *= 10;
        }


        long F = first*tens*10+last;
        long L = first*tens*10+last;
        long tens2 = 10;
        for(int i = 2; i < keta ; i++){
            L += 9*tens2;
            tens2 *= 10;
        }

        if(L <= max){
            ans += tens;
        }else{
            ans += (max - F + 10)/10 < 0 ? 0 : (max - F + 10)/10;
        }

        return(ans);
    }
}