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 <iostream>
using namespace std;

int lk(int x){
    
    while(x>=10){
        x/=10;
    }
    
    return x;
}

int main(void){
    
    int n;
    long long ans=0;
    
    cin >> n;
    
    for(int i=1;i<=n;i++){
        for(int j=i;j<=n;j++){
            int f=i%10;
            int l=lk(j);
            
            if(f!=l) continue;
            
            f=j%10;
            l=lk(i);
            
            if(f!=l) continue;
            if(i==j) ans++;
            else ans+=2;
        
        }
    }
    
    cout << ans << endl;
    
    return 0;
    
}