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>

#define all(v) (v).begin(), (v).end()
#define sortv(v) sort(all(v))
#define uniqv(v) (v).erase(unique(all(v)), (v).end())
#define pb push_back
#define FI first
#define SE second
#define lb lower_bound
#define ub upper_bound
#define mp make_pair
#define test 1
#define TEST if(test)

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;

const int MOD = 1000000007; // 998244353
const int INF = 2e9;
const ll INFLL = 1e18;
const int MAX_N = 1;

int N;
ll ans;
ll cnt[10][10];

int main(){
	cin>>N;
	for(int i=1; i<=N; i++){
		int t = i%10, n=i;
		while(n>=10)	n/=10;
		cnt[n][t]++;
	}
	for(int i=0; i<=9; i++){
		for(int j=0; j<=9; j++){
			ans += cnt[i][j]*cnt[j][i];
		}
	}
	cout<<ans;
	return 0;
}