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 main() {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    let n: usize = s.trim().parse().unwrap();
    s.clear();
    let mut count: Vec<Vec<usize>> = vec![vec![0; 10]; 10];
    for i in 1..10 {
        for j in 1..10 {
            let mut cnt = 0;
            if i == j && i <= n {
                cnt += 1;
            }
            if i * 10 + j <= n {
                cnt += 1;
            }
            let mut pow_ten = 100;
            while i * pow_ten + j <= n {
                cnt += std::cmp::min(pow_ten / 10, (n - (i * pow_ten + j)) / 10 + 1);
                pow_ten *= 10;
            }
            count[i][j] = cnt;
        }
    }
    let mut ans = 0;
    for i in 1..10 {
        for j in 1..10 {
            ans += count[i][j] * count[j][i];
        }
    }
    println!("{}", ans);
}