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 <iostream>
#include <vector>
using namespace std;

int maxnum(int n){
  while(n > 9){
    n /= 10;
  }
  return n;
}

int minnum(int n){
  return n % 10;
}

int main(void){
  int N;
  cin >> N;
  
  vector<vector<int>>num(10,vector<int>(10,0));
  for(int i=0;i<=N;i++){
    num[maxnum(i)][minnum(i)]++;
  }
  
  long long ans = 0;
  for(int i=1;i<10;i++){
    for(int j=1;j<10;j++){
      ans += num[i][j] * num[j][i];
    }
  }
  
  cout << ans << endl;
  
  return 0;
}