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: Runtime Error)

#include <stdio.h>
#include <stdlib.h>

int main(){
  int n;
  int a_low,a_top;
  int b;
  int i,j,k;
  int count=0;
  int pair[200000][200000]={{0},{0}};
  int ten[6]={0,10,100,1000,10000,100000};
  
  scanf("%d",&n);
  
  for(i=1;i<=n;i++){
    a_low=i%10;
    a_top=i;
    for(j=0;j<6;j++){
      if(a_top>9){
        a_top=a_top/10;
      }
    }
    
    for(j=0;j<6;j++){
      b=a_low*ten[j]+a_top;
      if(j>1){
        for(k=0;k<ten[j];k++){
          b=b+k*10;
          if(pair[i][b]==0){
            if(i==b){
              pair[i][b]=1;
              count=count+1;
            } else {
              pair[i][b]=pair[b][i]=1;
              count=count+2;
            }
          }
        }
      }
    }
  }
  printf("%d",count);
}