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: Runtime Error)

#include <stdio.h>
#include <stdlib.h>

int gethead(int n) {
	if(0 < n && n < 10)
		return n;
	return gethead(n/10);
}
int gettail(int n) {
	return n%10;
}

int main(void) {
	int n, i, j;
	scanf("%d", &n);
	int **count = malloc(sizeof(int *)*9);
	for(i = 0; i < 9; i++) {
		*(count + i) = malloc(sizeof(int)*9);
		for(j = 0; j < 9; j++)
			*(*(count + i) + j) = 0;
	}
	for(i = 1; i <= n; i++)
		(*(*(count + gethead(i) - 1) + gettail(i) - 1))++;
	int ans = 0;
	for(i = 0; i < 9; i++)
		for(j = 0; j < 9; j++)
			ans += (*(*(count + i) + j))*(*(*(count + j) + i));
	free(count);
	printf("%d\n", ans);
	return 0;
}