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: Wrong Answer)

import java.util.*;
public class Main {

	public static boolean isOK(int i, int j) {
		char[] str_i = String.valueOf(i).toCharArray();
		char[] str_j = String.valueOf(j).toCharArray();
		
		return str_i[0] == str_j[str_j.length - 1] && str_i[str_i.length -1] == str_j[0];
	}
	
	public static Integer hash(int n) {
		char[] str = String.valueOf(n).toCharArray();
		if(str[0] == '0' || str[str.length -1] == '0') {
			return -1;
		}
		String a = String.valueOf(new char[] {str[0], str[str.length -1]});
		
		
		return Integer.parseInt(a);
	}
	
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt();
		
		Map<Integer, List<Integer>> map = new HashMap<>();
		
		
		for(int i=1; i<=9;i++) {
			for(int j=1; j<=9;j++) {
				map.put(i*10 + j, new ArrayList<Integer>());	
			}
		}
		
		
		for(int i=1; i<=n;i++) {
			int h = hash(i);
			if(h <= 0) {
				continue;
			}
			
			System.out.println(h + "   " + i);
			List<Integer> l = map.get(h);
			l.add(i);
			map.put(h, l);
		}
		
		for(List<Integer> m: map.values()) {
			System.out.println(m);
		}


		int ans = 0;

		for(int i=1; i<=9; i++) {
			for(int j=1; j<=9; j++) {
				System.out.println(i*10 + j);
				int a = map.get(i*10 + j).size();
				int b = map.get(j*10 + i).size();
				ans += a * b;
				
			}
		}
		
		System.out.println(ans);
		
	}
	


}