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>
# define ll long long
# define mo 1000000009
using namespace std;
int f1(int x)
{
	int ans;
	while(x!=0)
	{
		ans=x%10;
		x=x/10;
	}
	return ans;
}
int f2(int x)
{
	return x%10;
}
int main()
{
	ll n;
	cin>>n;
	map<pair<int,int>,int> m;
	//cout<<f1(n)<<" "<<f2(n);
	for(ll i=1;i<=n;i++)
	{
		m[{f1(i),f2(i)}]++;
	}
    ll ans=0;
    for(int i=1;i<=9;i++)
    {
    	ans+=1LL*m[{i,i}]*m[{i,i}];
    	for(int j=i+1;j<=9;j++)
    	{
    		ans+=2LL*m[{i,j}]*m[{j,i}];
		}
	}

cout<<ans;
	return 0;
}