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: Wrong Answer)

#include<stdio.h>
#include<math.h>

long int f(int n,int i,int j)
{
	int k=0;
	long int ans=0;
	int num=0;
	
	if(n<10){
		if(i==j && i<=n)
			ans=1;
		else
			ans=0;
	}
	else
	{
		ans=(pow(10,(int)log10(n)-1)-1)/9;

		while((pow(10,(int)log10(n))*i + j + k*10<n) && (k*10<pow(10,(int)log10(n))))
		{
			ans ++;
			k++;
		}
		
		if(i==j)
			ans++;
	}
	return ans;
}

int main(void)
{
	int n;
	scanf("%d",&n);
	long int ans2=0;
	int s,t;
	
	for(s=1;s<=9;s++){
		for(t=1;t<=9;t++){
			ans2 += f(n,s,t)*f(n,t,s);
		}
	}
	
	printf("%ld",ans2);
	return 0;
}