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 <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define rep2(i, x, n) for (int i = x; i < (n); i++)
#define all(n) begin(n), end(n)
struct cww
{
    cww()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
    }
} star;
const long long INF = numeric_limits<long long>::max();
typedef long long ll;
typedef vector<int> vint;
typedef vector<char> vchar;
typedef vector<vector<int>> vvint;
typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef unsigned long long ull;
const int MAX = 510000;
const int MOD = 1000000007;
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit()
{
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++)
    {
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
long long COM(int n, int k)
{
    if (n < k)
        return 0;
    if (n < 0 || k < 0)
        return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
int main()
{
    ll ans = 0;
    int N;
    cin >> N;
    int ketaN = log10(N) + 1;
    rep2(i, 1, N + 1)
    {
        if (i > 10)
        {
            int ti = 0;
        }
        int a0 = i % 10;
        int ak = to_string(i).front() - '0';

        int keta = to_string(i).size();
        int tmp = 0;

        for (int nowketa = 1; nowketa <= ketaN; nowketa++)
        {
            if (a0 != 0)
            {
                if (nowketa == 1)
                {
                    if ((a0 == ak) and a0 <= N)
                    {
                        ans++;
                    }
                }
                else if (nowketa == 2)
                {
                    if (a0 * 10 + ak <= N)
                    {
                        ans++;
                    }
                }
                else
                {

                    tmp = a0 * pow(10, nowketa - 1) + ak;
                    if (N - tmp >= 0)
                    {
                        ans += min((N - tmp) / 10, (int)pow(10, nowketa - 2) - 1) + 1;
                    }
                }
            }
        }
    }

    cout << ans;
    return 0;
}