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::str::FromStr;
use std::fmt::Debug;
use std::cmp;
#[allow(dead_code)]
fn read_as_vec<T>() -> Vec<T>
where T: FromStr, <T as FromStr>::Err : Debug{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().split_whitespace().map(|c| T::from_str(c).unwrap()).collect()

}

#[allow(dead_code)]
fn read_as_string() -> String{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

fn get_head_tail(x: usize) -> (usize, usize){
    let mut v = vec![];
    let mut x = x;

    while x > 0{
        v.push(x%10);
        x /= 10;
    }

    return (*v.first().unwrap(), *v.last().unwrap());
}

fn print_v(v: &Vec<Vec<usize>>){
    for row in v{
        println!("{:?}", row);
    }
}

fn main(){
    let n: usize = read_as_vec::<usize>()[0];

    let mut v: Vec<Vec<usize>> = vec![vec![0;10];10];

    for i in 1..n+1{
        let (head, tail) = get_head_tail(i as usize);
        v[tail][head] += 1;
    }

    //print_v(&v);
    let mut ans: usize = 0;

    for head in 0..10{
        for tail in 0..10{
            //println!("(h, t) = ({}, {}): {}", head, tail, v[head][tail] * v[tail][head]);
            ans += v[head][tail] * v[tail][head];
        }
    }

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