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 (Clang 3.8.0) (judge verdict: Accepted)

#include <cstdio>
#include <cmath>
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <utility>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <iterator>
#define loop(i, a, b) for(int i = (int)(a); i < (int)(b); i++)
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define debug(x)  cout << #x << " = " << (x) << endl;

using namespace std;
using LL = long long;
using ULL = unsigned long long;

int count_digit(long long n)
{
    int count = 0;
    while (n != 0) {
        n = n / 10;
        ++count;
    }
    return count;
}

int main() {
    int n; cin >> n;
    vector<vector<int>> table(9, vector<int>(9, 0));

    for (int i = 1; i <= n; i++) {
        if (i % 10 == 0) continue;
        if (i <= 9) {
            table[i - 1][i - 1] = 1;
            continue;
        }
        int count = count_digit(i);
        int most = i / (int)pow(10, count - 1);
        int least = i % 10;
        table[most - 1][least - 1]++;
    }

    long long ans = 0;
    // rep(i, 9) {
    //     rep(j, 9) {
    //         cout << table[i][j] << " ";
    //     }
    //     cout << endl;
    // }
    rep(i, 9) {
        for (int j = i; j < 9; j++) {
            if (i == j) {
                if (table[i][j] == 1) ans += 1;
                else ans += table[i][j] * table[i][j];
            } else {
                ans += table[i][j] * table[j][i] * 2;
            }
        }
        cout << endl;
    }
    cout << ans << endl;
}