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::FromStr;
use std::collections::HashMap;

fn read<T: FromStr>() -> Option<T> {
    let stdin = stdin();
    let s = stdin
        .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 n: i32 = read().unwrap();
    let mut map: HashMap<(i32, i32), i32> = HashMap::new();

    for k in 1..n + 1 {
        let num_str = k.to_string();
        let i: i32 = num_str[0..1].parse().unwrap();
        let j: i32 = num_str[num_str.len() - 1..].parse().unwrap();
        let count = map.entry((i, j)).or_insert(0);
        *count += 1;
    }

    let mut total = 0;
    for i in 0..10 {
        for j in 0..10 {
            let pro = map.get(&(i, j)).unwrap_or(&0) * map.get(&(j, i)).unwrap_or(&0);
            total += pro;
            // println!("({}, {}) {}", i, j, total);
        }
    }
    println!("{}", total);
}