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 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include<cmath>
using namespace std;

int main() {
	int x;
	int y;
	int last_digit;
	int length=0;
	long long count1=0;
	long long count2=0;
	long long count=0;
	cin >>x;
	y=x;
	if (x<10)
	{
		cout <<x;
		return 0;
	}
	while (x>0)
	{
		length++;
		x/=10;
	}
	last_digit = y/(pow(10,length-1));
	x = last_digit*(pow(10,length-1));
	for (int i=1; i<10; i++)
	{
		count1=0;
		for (int k=1; k<=length;k++)
		{
			if (k==1)
			{
				count1++;
			}
			else if (k!=length)
			{
				count1+=pow(10,k-2);
			}
			else
			{
				if (i<last_digit)
				{
					count1+=pow(10,k-2);
				}
				else if (i==last_digit)
				{
					if ((x+i)<=y)
						count1+= ((y-x-i)/10)+1;
				}
			}
		}
		count+=count1*count1;
	}
	
	for (int i=1; i<10; i++)
	{
		for (int j=i+1; j<10;j++)
		{
			count1=0;
			count2=0;
			for (int k=2; k<=length;k++)
			{
				if (k!=length)
				{
					count1+=pow(10,k-2);
					count2+=pow(10,k-2);
				}
				else
				{
					if (i<last_digit)
					{
						count1+=pow(10,k-2);
					}
					else if (i==last_digit)
					{
						if (x+j<=y)
							count1+= ((y-x-j)/10)+1;
					}
					if (j<last_digit)
					{
						count2+=pow(10,k-2);
					}
					else if (j==last_digit)
					{
						if (x+i<=y)
							count2+= ((y-x-i)/10)+1;
					}
				}
			}
			count+=count1*count2*2;
		}
	}
	cout << count;
	return 0;
}