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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
int main(){
  int n,i,j,count=0;
  int a[11][11];
  for(i=0;i<10;i++)
    for(j=0;j<10;j++)
      a[i][j]=0;
  scanf("%d",&n);
  for(i=1;i<=n;i++){
    j=i;
    while(j/10!=0)
      j/=10;
    a[j][i%10]++;
  }
  for(i=1;i<=9;i++){
    for(j=i;j<=9;j++){
      if(i==j)
        count+=a[i][j]*a[i][j];
      else
        count+=2*a[i][j]*a[j][i];
    }
  }
  printf("%d\n",count);
  return 0;
}