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 <math.h>
#include <time.h>
#include <ctype.h>
#include <stdio.h>
#include <assert.h>
#include <stdlib.h>
#include <string.h>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <string>
#include <vector>
#include <iostream>
#include <algorithm>
#include<iomanip>

#define MX 10000000000
#define mod 1000000007
#define ll long long int
#define pb(x) push_back(x)
#define MP(x,y) make_pair(x,y)

using namespace std;

ll power(ll a, ll b, ll m)
{
	if (b == 0)
		return(1);
	ll sol = power(a, b / 2, m);
	sol = (sol * sol) % m;
	if (b % 2 == 1)
		sol = (sol * a) % m;
	return(sol);
}

ll gcd(ll a, ll b)
{
	if (b == 0)
		return a;
	else
	{
		return gcd(b, a % b);
	}
}
map<pair<int, int>,ll>m;
void solve()
{
   int n;
   cin >> n;
   for( int i = 1; i <= n; i++ )
   {
      string s = to_string( i );
      int x = s[ 0 ] - '0';
      reverse( s.begin(), s.end() );
      int y = s[ 0 ] - '0';
      m[ MP( x, y ) ]++;
      s.clear();
   }
   ll sol = 0;
   for( int i = 1; i <= 9; i++ )
   {
      for( int j = 1; j <= 9; j++ )
      {
         //cout << m[ MP( i, j ) ] << endl;
         sol += m[ MP(i,j)]*m[MP(j,i)];
      }
   }
   cout << sol;
}
int main()
{
	int t = 1;
//	cin >> t;
	while (t--)
	{
		solve();
	}
}