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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>

int get_head (int num) {
    int head;
    while ( num != 0 ) {
        head = num;
        num /= 10;
    }
    return head;
}

int get_tail (int num) {
    return num % 10;
}


int main (void) {
    int n, ans = 0, i, j;
    int head, tail;
    int table[10][10];
    memset(table, 0, sizeof(table));
    scanf("%d", &n);

    for ( i = 1; i <= n; i++ ) {
        head = get_head(i);
        tail = get_tail(i);
        table[head][tail]++;
    }

    for ( i = 1; i <= 9; i++ ) {
        for ( j = 1; j <= 9; j++ )
            ans += table[i][j] * table[j][i];
    }

    printf("%d\n", ans);

    return 0;
}