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)

pub fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

pub fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

pub fn read_vec2<T: std::str::FromStr>(n: usize) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}

pub fn read_col<T: std::str::FromStr>(n: usize) -> Vec<T> {
    (0..n).map(|_| read()).collect()
}

fn head(n: u32) -> u32 {
    let s: String = n.to_string();
    let c: Vec<char> = s.chars().collect();
    c[0] as u32 - '0' as u32
}

fn tail(n: u32) -> u32 {
    n % 10
}

fn middle(n: u32) -> u32 {
    let s: String = n.to_string();
    let c: Vec<char> = s.chars().collect();
    let mut ret = 0;
    for i in 1..c.len() - 1 {
        ret *= 10;
        ret += c[i] as u32 - '0' as u32;
    }
    ret
}

fn pow10(l: usize) -> u32 {
    let mut ret = 1;
    for _ in 0..l {
        ret *= 10;
    }
    ret
}

fn main() {
    let n: u32 = read();

    let mut ans = 0;

    let nh = head(n);
    let nt = tail(n);
    let nl = n.to_string().len();

    for a in 1..n + 1 {
        let bt = head(a);
        let bh = tail(a);

        if bh == 0 {
            continue;
        }
        if bh == bt {
            ans += 1;
        }
        if bh * 10 + bt <= n {
            ans += 1;
        }

        for bl in 3..nl + 1 {
            if bl == nl {
                if bh < nh {
                    ans += pow10(bl - 2);
                } else if bh == nh {
                    if bt <= nt {
                        ans += middle(n) + 1;
                    } else {
                        ans += middle(n);
                    }
                }
            } else {
                ans += pow10(bl - 2);
            }
        }
    }

    println!("{}", ans);
}