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.*;
public class Main { 
    public static void main(String[] args) {
       Scanner sc=new Scanner(System.in);
       int o=0;
       int n=sc.nextInt();
       int[][] v=new int[9][9];
      for(int i=1;i<=n;i++){
      	String str=String.valueOf(i);
		String str1 = str.substring(0, 1);          
		int x = Integer.parseInt(str1);
        int y=i%10;
        //System.out.println(x+""+y);
        if(x!=0&&y!=0)v[x-1][y-1]++;
      }
      for(int i=0;i<9;i++){
      	for(int j=0;j<9;j++){
          //System.out.println(i+1+" " +(int)(j+1)+ " " +v[i][j]);
          o+=v[i][j]*v[j][i];
        }
      }
      System.out.println(o);
    }
}