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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::*;
use std::str::*;

fn read<T: FromStr>(s: &mut StdinLock) -> Option<T> {
    let s = s
        .by_ref()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect::<String>();
    s.parse::<T>().ok()
}

fn main() {
    let s = stdin();
    let mut s = s.lock();
    let n: i32 = read(&mut s).unwrap();
    let mut counter: i32 = 0;

    let mut matrix: [[i32; 9]; 9] = [[0; 9]; 9];
    if n < 10 {
        println!("{}", n);
        return;
    }

    for i in 1..n + 1 {
        let ii = i.to_string();
        let ilen = ii.len() - 1;

        let first: usize = ii.chars().nth(0).unwrap() as usize - 48;
        let last: usize = ii.chars().nth(ilen).unwrap() as usize - 48;

        if last == 0 {
            continue;
        }
        matrix[first - 1][last - 1] += 1;
    }

    //println!("{:?}", matrix);
    for i in 0..9 {
        for j in 0..9 {
            if i == j {
                counter += matrix[i][j] * matrix[j][i];
            } else if matrix[i][j] > 0 && matrix[j][i] > 0 {
                counter += matrix[i][j] * matrix[j][i];
            }
        }
    }
    println!("{}", counter);
}