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;
#define rep(i, n) for(int i = 0; i < (int)(n); i++)

int c[10][10];

int main(){
    int n;
    cin >> n;

    rep(i,10)rep(j,10) c[i][j] = 0;

    int first, last;
    for(int i = 1; i <= n; i++){
        int j = i;
        last = j % 10;
        while(j > 10){
            j /= 10;
        }
        first = j;
        c[first][last]++;
    }

    long long ans = 0;
    for(int i = 0; i <= 9; i++){
        for(int j = 0; j <= 9; j++){
            ans += c[i][j] * c[j][i];
        }
    }

    cout << ans << endl;

    return 0;
}