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)

fn keta(x: usize) -> usize {
    let mut keta = 0;
    let mut y = x;
    while y > 0 {
        keta += 1;
        y /= 10;
    }
    keta
}

fn top(x: usize) -> usize {
    let mut top = 0;
    let mut y = x;
    while y > 0 {
        top = y;
        y /= 10;
    }
    top
}

fn main() {
    let n = read::<usize>();
    let k = keta(n);
    //let t = top(n);
    //let b = n % 10;

    let mut count = 0;
    for i in 1..n + 1 {
        if i % 10 == 0 {
            continue;
        }
        for j in 1..k + 1 {
            //println!("{} {}", i, j);
            if j == 1 {
                if top(i) == i % 10 {
                    count += 1;
                }
            } else if j == 2 {
                if top(i) + i % 10 * 10 <= n {
                    count += 1;
                }
            } else if j == k {
                if i % 10 < top(n) {
                    count += 10usize.pow((j - 2) as u32);
                } else if i % 10 == top(n) {
                    let a = 10usize.pow((k - 1) as u32) * (i % 10);
                    count += (n - a) / 10;
                    if top(i) <= n % 10 {
                        count += 1;
                    }
                }
            } else {
                count += 10usize.pow((j - 2) as u32);
            }
        }
    }
    println!("{}", count);
}

#[allow(dead_code)]
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()
}
#[allow(dead_code)]
fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}
#[allow(dead_code)]
fn read_vec2<T: std::str::FromStr>(n: u32) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}
#[allow(dead_code)]
fn yn(result: bool) {
    if result {
        println!("Yes");
    } else {
        println!("No");
    }
}