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 <stdio.h>
#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <cmath>
#include <iomanip>
#include <queue> 
 
using namespace std;
 
int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
 
	int N;
 
	cin >> N;

	int base = 1;
 
	vector< vector<int> > K(9, vector<int>(9));

	for(int i = 1; i <= N; i++) {
		int x = i % 10;
		if(x == 0) continue;
		if(i >= base * 10) base *= 10;
		int y = i / base;
		K[x - 1][y - 1]++;
	}

	int count = 0;

	for(int i = 0; i < 9; i++) {
		for(int j = i; j < 9; j++) {
			if(i != j) {
				count += K[i][j] * K[j][i] * 2;
			} else {
				count += K[i][j] * K[i][j];
			}
		}
	}
/*
	for(int i = 0; i < 9; i++) {
		for(int j = 0; j < 9; j++) {
			cout << K[i][j] << " ";
		}
		cout << endl;
	}
*/
	cout << count << endl;
}