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 <bits/stdc++.h>
using namespace std;
int keta(int n){
  int ans=0;
  while(n>0){
    ans++;
    n/=10;
  }
  return ans;
}
int main() {
  int n;
  cin>>n;
  int ans=0;
  for(int i=1;i<=n;i++){
    int a=i/pow(10,keta(i)-1);
    int b=i%10;
    int d=n/pow(10,keta(n)-1);
    int c=keta(n)-2;
    if(a!=0 && b!=0){
    for(int i=1;i<=c;i++){
      if(i!=c) ans+=pow(10,i);
      else {
        if(d>b){
          ans+=pow(10,i);
        }
        else if(d==b){
          int aho=pow(10,keta(n));
          int hina=n%aho;
          ans+=hina/10;
        }
      }
    }
    }
  }     
          
          
}