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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define vsort(v) sort(v.begin(), v.end())
#define rvsort(v) sort(v.begin(), v.end(), greater<ll>())
#define prvsort(v) sort(v.begin(), v.end(), greater<pair<ll, ll>>())
#define ll long long
#define ld long double
#define fixset(val) fixed << setprecision(10) << val 
#define rep(i, s, g) for (ll i = (s); i < (g); i++)
#define mod 1000000007
#define vfind(v, target) find(v.begin(), v.end(), target)
#define notFound(v, target) (vfind(v, target) == v.end())

using namespace std;

int main()
{
  ll N, a = 1, ans = 0; cin >> N;
  vector<vector<ll>> mtrx(10, vector<ll>(10, 0));
  rep(k, 1, N + 1) {
    if (k / a == 10) a *= 10;
  	mtrx[k / a][k % 10]++;
  }
  rep(i, 0, 10) {
  	rep(j, 0, 10) {
      	ans += mtrx[i][j] * mtrx[j][i];
    }
  }
  cout << ans << endl;
}