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: Time Limit Exceeded)

#include<stdio.h>
int main()
{
  int N,i,j,m;
  long long int k=0;
  int number1,number2;
  int digit1,digit2;
  int oral1,oral2;
  scanf("%d",&N);
  for(i=1;i<N+1;i++){
    digit1=0;
    number1=i;
    oral1=i;
    while(number1!=0){
    number1 = number1 / 10;
    digit1++;
  }
    digit1=digit1-1;
    if(digit1!=0){
    for(m=0;m<digit1;m++){
      oral1=oral1/10;
    }
    }
    
    for(j=1;j<N+1;j++){
     digit2=0;
    number2=j;
      oral2=j;
    while(number2!=0){
    number2 = number2 / 10;
    digit2++;
  }
    digit2=digit2-1;
    if(digit2!=0){
    for(m=0;m<digit2;m++){
      oral2=oral2/10;
    }
    }
      if((i%10==oral2)&&(j%10==oral1)){
        k++;
      }
    }
  }
  printf("%lld",k);
  return 0;
}