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 <math.h>
#include <stdlib.h>
#include <string.h>

int combination(int n, int r) {
	int a;
	for (int i = 1; i <= r; i++) {
		a *= (n - i + 1) / i;
	}
	return a;
}

int main() {
	int n;
	long int total = 0;
	int lsd, msd;
	int pair[10][10] = { 0 };
	int count[11][11][200] = { 0 };
	scanf("%d", &n);
	//printf("%d\n", n);
	for (int i = 1; i <= n; i++) {
		//printf("i=%d", i);
		lsd = i % 10;
		if (lsd != 0) {
			msd = i;
			for (int j = 0; msd >= 10; j++) {
				msd /= 10;
			}
			pair[msd][lsd]++;
		}
		int a = pair[msd][lsd];
		count[msd][lsd][a] = i;
		//printf("%d:pair[%d][%d]=%d\n",i, msd, lsd, pair[msd][lsd]);
	}

	for (int i = 1; i <= 9; i++) {
		for (int j = 1; j <= 9; j++) {
			total += pair[i][j] * pair[j][i];
			/*
			printf("\n\npair[%d][%d]=%d\n", i, j, pair[i][j]);
			for (int k = 1; k <= pair[i][j]; k++) {
				//if(count[i][j][k]>0)
				//printf("count[%d][%d][%d]=%d   ", i, j, k,count[i][j][k]);
			}
			*/
		}
	}
	printf("%ld", total);
	return 0;
}