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 int long long
typedef long long ll;
using namespace std;

ll N, DP[200001]={0}, ans=0;
map<pair<ll, ll>, ll> M;
signed main(){
    cin >> N;
    for(int i=1;i<=N;i++) {
        if(i%10==0) continue;
        int a = log10(i);
        int topd = i/pow(10,a);
        int botd = i%10;
        pair<ll, ll> p = make_pair(topd, botd);
        M[p]++;
    }
    for(int i=1;i<=N;i++) {
        if(i%10==0) continue;
        int a = log10(i);
        int topd = i/pow(10,a);
        int botd = i%10;
        pair<ll, ll> p = make_pair(botd, topd);
        ans += M[p];
    }
    cout << ans << endl;
    return 0;
}