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(A) (A).begin(),(A).end()
#define ll long long
#define rep(i,n) for(int i=0;i<(n);i++)

const ll mod = 1e9+7;
const ll INF = -1*((1LL<<63)+1);
const int inf = -1*((1<<31)+1);

using namespace std;

ll cnt[10][10]; // cnt[a][b] := 先頭がaで末尾がbの数の個数
int main(void){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout<<fixed<<setprecision(20);
    int N;
    cin >> N;
    for(int i=1;i<=N;i++){
        int head = to_string(i)[0] - '0';
        int end  = i%10;
        cnt[head][end]++;
    }
    ll ans = 0;
    for(int i=1;i<=9;i++)for(int j=1;j<=9;j++){
        ans += cnt[i][j] * cnt[j][i];
    }
    cout << ans << endl;
}