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 <cstdio>
#include <algorithm>
#include <cstring>
#include <set>
#include <map>
#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#define ull unsigned long long
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
void close(){
       std::ios::sync_with_stdio(false);
       std::cin.tie(0);
}
ll a[10][10];
ll find(ll x){
	while(x){
		if(x/10==0) return x;
		x/=10;
	}
}
int main(){
	ll n;
	cin>>n;
	for(ll i=1;i<=n;i++){
		a[find(i)][i%10]++;
	}
	ll ans=0;
	for(ll i=1;i<=9;i++){
		for(ll j=1;j<=9;j++){
			ans+=(a[i][j]*a[j][i]);
		}
	}
	cout<<ans<<endl;
}