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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

int main() {
  int n;
  int ans=0;
  int numi,numj;
  vector<int> p(5,0);
  int h,t;

  cin >> n;

  for(int i=1;i<=n;i++) {
    numi = i;
    int ic=0;
    while(numi!=0){
      p[ic]=numi%10;
      numi /= 10;
      ic++;
    }

    h=p[ic-1];
    t=p[0];

    //cout << i <<" "<< h1 <<" "<< t1 << endl;
    for(int j=1;j<=n;j++) {
      //      cout << i <<" "<< j << endl;

      numj=j;
      int jc=0;
      while(numj!=0){
        p[jc]=numj%10;
        numj /= 10;
        jc++;
      }
      if(p[jc-1]!=t) {
        if(jc!=1) j+=pow(10,jc-1)-1;
      }
      else if(j%10==h) {ans++;}
    }
  }

  cout <<ans<< endl;

  return 0;
}