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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main()
{
	int N;
	cin >> N;

	vector<vector<int>> tbl( 10, vector<int>( 10 ) );
	for( int i = 1; i < N + 1; i++ ) {
		char buff[20000+5];
		sprintf( buff, "%d", i );
		int len = strlen( buff );
		int st = buff[0] - '0';
		int end = buff[len - 1] - '0';
		tbl[st][end]++;
	}
	long long ans = 0;
	for( int i = 1; i < 10; i++ ) {
		for( int j = 1; j < 10; j++ ) {
			ans += tbl[i][j] * tbl[j][i];
		}
	}

	cout << ans << endl;
}