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>

using namespace std;

typedef long long ll;

#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define FORd(i,a,b) for(int i=a;i>=b;i--)
#define fi first
#define se second
#define II pair<ll,ll>

const int maxn=1e5+10;
const ll maxN=1e18+10;

int main()
{
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int a[10][10],dd[10][10];
  memset(a,0,sizeof(a));
  memset(dd,0,sizeof(dd));
  int n;
  cin >> n;
  FOR(i,1,n)
   if (i%10!=0) {
   	string s=to_string(i);
   	int so=s[0]-'0';
   	a[so][i%10]++;
   }
  ll ans=0;
  FOR(i,1,9)
  FOR(j,1,9) 
  if (dd[i][j]==0)
  {
  	if (i!=j) {
  		 ans+=a[i][j]*a[j][i]*2;
  		 dd[j][i]=1;
  	}
  	else ans+=a[i][j]*a[i][j];
  }
  cout << ans;
}