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>

int Mdigit(const int x) {
	if (99999 < x) {
		return x / 100000;
	}
	else if (9999 < x && x < 100000) {
		return x / 10000;
	}
	else if (999 < x && x < 10000) {
		return x / 1000;
	}
	else if (99 < x && x < 1000) {
		return x / 100;
	}
	else if (9 < x && x < 100) {
		return x / 10;
	}
	else {
		return x;
	}
}

int mdigit(const int x) {
	return x % 10;
}

int main() {
	int N, i, j, AM, Am;
	long ans = 0;
	long c[10][10];
	scanf("%d", &N);
	for (i = 0; i <= 9; i++) {
		for (j = 0; j <= 9; j++) {
			c[i][j] = 0;
		}
	}

	for (i = 1; i <= N; i++) {
		c[Mdigit(i)][mdigit(i)]++;
	}

	//各Aに対して...
	for (i = 1; i <= N; i++) {
		AM = Mdigit(i);
		Am = mdigit(i);
		ans += c[Am][AM];
	}
	printf("%ld", ans);
	return 0;
}