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: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
int int_cmp(const int*a, const int *b){
  if(*a < *b){
    return -1;
  }
  else if(*a > *b){
    return 1;
  }
  else {
    return 0;
  }
}

int main(void){
  int n;
  int ans=0;
  char s[10],t[10];
  int i,j;
  char maxs,mins;
  char maxt,mint;
  scanf("%d",&n);
  for(i=0;i<n;i++){
    sprintf(s,"%d",i);
    maxs=s[strlen(s)-1];
    mins=s[0];
    for(j=0;j<n;j++){
      sprintf(t,"%d",j);
      maxt=t[strlen(t)-1];
      mint=t[0];
      if(maxt==mins&&maxs==mint){
        ans++;
      }
    }

  }
  if(ans!=1){
    ans--;
  }
  printf("%d",ans);


  return 0;
}