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 <stdlib.h>

int main()
{
	long long N;
	scanf("%lld", &N);
	
	int i, j, k, l, x[10];
	long long X, total = 0, count[10][10] = {};
	for (i = 1; i <= 9; i++) {
		for (j = 1; j <= 9; j++) {
			if (i == j && i <= N) count[i][j]++;
			for (k = 0; k < 10; k++) x[k] = -1;
			x[0] = i;
			x[1] = j;
			X = x[0] + x[1] * 10; 
			while (X <= N) {
				count[i][j]++;
				for (k = 1; x[k] == 9 && x[k+1] != -1; k++);
				if (x[k] != 9 && x[k+1] != -1) {
					for (x[k--]++; k >= 1; k--) x[k] = 0;
				} else {
					x[k+1] = j;
					for (; k >= 1; k--) x[k] = 0;
				}
				for (k = 9, X = 0; k >= 0; k--) if (x[k] >= 0) X = X * 10 + x[k];
			}
		}
	}
	
	for (i = 1; i <= 9; i++) {
		for (j = 1; j <= 9; j++) {
			total += count[i][j] * count[j][i];
		}
	}
	printf("%lld\n", total);
	fflush(stdout);
	return 0;
}