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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
int asc(const void *a, const void *b)
{
    return *(long int *)a - *(long int *)b;
}
int dis(const void *a, const void *b)
{
    return *(long int *)b - *(long int *)a;
}
int main()
{
    int n;
    long long int ans = 0;
    long long int a[10][10] = {};
    scanf("%d", &n);
    for (int l = 1; l <= 9; l++)
    {
        for (int r = 1; r <= 9; r++)
        {
            if (l == r && l <= n)
            {
                a[l][r]++;
            }
            int o;
            for (o = 10; (l + 1) * o <= n; o *= 10)
            {
                a[l][r] += o / 10;
            }
            if (n + 10 > l * o + r)
            {
                a[l][r] += (n + 10 - (l * o + r)) / 10;
            }
        }
    }
    for (int l = 1; l <= 9; l++)
    {
        for (int r = 1; r <= 9; r++)
        {
            ans += a[l][r] * a[r][l];
        }
    }
    for (int i = 1; i <= 9; i++)
    {
        for (int j = 1; j <= 9; j++)
        {
            printf("%lld ", a[i][j]);
        }
        printf("\n");
    }

    printf("%lld", ans);
    return 0;
}