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.*;
 
import java.io.*;
 
public class Main{
    
	public static void main(String[] $){
		Scanner s = new Scanner(System.in);
        int N = s.nextInt();
        HashMap<String, Integer> normalSet = new HashMap<String, Integer>();
        for(int i = 0; i <= 9; i++){
            for(int j = 0; j <= 9; j++){
                normalSet.put(String.valueOf(i)+String.valueOf(j), 0);
            }
        }

        for(int i = 1; i <= N; i++){
            String normal = String.valueOf(i);
            String normalmoji = String.valueOf(normal.charAt(0)) + String.valueOf(normal.charAt(normal.length()-1));
            normalSet.put(normalmoji, normalSet.get(normalmoji)+1);
        }

        long result = 0;
        for(int i = 0; i <= 9; i++){
            for(int j = 0; j <= 9; j++){
                result += normalSet.get(String.valueOf(i)+String.valueOf(j))*normalSet.get(String.valueOf(j)+String.valueOf(i));
                
            }
        }
        
        System.out.println(result);
	}
}