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: Wrong Answer)

#include <cmath>
#include <iostream>
#include <vector>
using namespace std;
using ll = long long;

vector<int> keta(int n) {
  vector<int> res;
  while (n > 0) {
    res.push_back(n % 10);
    n /= 10;
  }
  return res;
}

int main() {
  int n;
  cin >> n;
  ll ans = 0;
  auto vn = keta(n);
  int lndn = vn.size();
  int nn = (n - vn.back() * pow(10, lndn - 1)) / 10;
  for (int i = 1; i <= n; i++) {
    auto v = keta(i);
    int l = v.back();
    int h = v[0];
    if (h == 0)
      continue;
    if (h == l)
      ans++;
    for (int j = 2; j < lndn; j++) {
      ans += pow(10, j - 2);
    }
    if (h > vn.back())
      continue;
    if (h == vn.back()) {
      if (l > vn[0]) {
        ans += nn;
      } else {
        ans += nn + 1;
      }
    } else {
      ans += pow(10, lndn - 2);
    }
  }
  cout << ans << endl;
}