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;
int main(){int i,n,s,e,k;
long long int sum=0;
cin>>n;
//sum=n;
int cnt[10][10];
memset(cnt,0,sizeof(cnt));
for(i=1;i<=n;i++){
    k=i;
    e=k%10;
    while(k>=10){k=k/10;}
    s=k;
    cnt[s][e]++;
    sum+=2*cnt[e][s];
    if(s==e){sum--;}
 //   cout<<cnt[e][s]<<" ";
}cout<<sum;
    return 0;
}