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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <inttypes.h>

int ketasu(int hoge) {
	if (hoge < 10) return 1;
	else if (hoge < 100) return 2;
	else if (hoge < 1000) return 3;
	else if (hoge < 10000) return 4;
	else if (hoge < 100000) return 5;
	else if (hoge < 1000000) return 6;
	else if (hoge < 10000000) return 7;
	else if (hoge < 100000000) return 8;
	else if (hoge < 1000000000) return 9;
	else return 10;
}

int delta(int hoge) {
	if (hoge < 10) return 1;
	else if (hoge < 100) return 10;
	else if (hoge < 1000) return 100;
	else if (hoge < 10000) return 1000;
	else if (hoge < 100000) return 10000;
	else if (hoge < 1000000) return 100000;
	else if (hoge < 10000000) return 1000000;
	else if (hoge < 100000000) return 10000000;
	else if (hoge < 1000000000) return 100000000;
	else return 1000000000;
}

int saizyoui(int hoge) {
	return hoge / delta(hoge);
}

int keisan(int enu, int hazime, int owari) {
	if (hazime == 0) {
		return 0;
	} else if(enu < 10) {
		return hazime == owari && hazime <= enu;
	} else {
		int result = (hazime == owari);
		int k = ketasu(enu), s = saizyoui(enu), d = delta(enu);
		int i, sumipe;
		for (i = 2, sumipe = 1; hazime >= s ? (i < k) : (i <= k); i++) {
			result += sumipe;
			sumipe *= 10;
		}
		if (hazime == s) {
			int mamegu = enu - s * d;
			result += (mamegu - owari + 10) / 10;
		}
		return result;
	}
}

int main(void) {
	int N;
	uint64_t kotae = 0;
	int i;
	if (scanf("%d", &N) != 1) return 1;
	for (i = 1; i <= N; i++) {
		kotae += keisan(N, i % 10, saizyoui(i));
	}
	printf("%" PRIu64 "\n", kotae);
	return 0;
}