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>
#include <string>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
typedef long long ll;
#define MOD 1000000007

using namespace std;

int main() {
    int n;
    cin >> n;
    
    vector<vector<int>> v = vector<vector<int>>(9, vector<int>(10, 0));
    for(int i = 1; i <= n; i++) {
        int num = i;
        int matsubi = num % 10;
        
        while(num >= 10) {
            num /= 10;
        }
        int front = num - 1;
        
        v[front][matsubi]++;
    }
    
    ll ans = 0;
    for(int i = 0; i < 9; i++) {
        for(int j = 1; j < 10; j++) {
            // cout << "(i, j) = " << i + 1 << ", " << j << " : " << v[i][j] << " " <<  v[j - 1][i] << endl;
            ans += v[i][j] * v[j - 1][i + 1];
        }
    }
    
    cout << ans << endl;
    
    return 0;
}