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;
  	scanf("%d", &N);
  	int a,b;
  	int i,j,number = 0;
  	
  	for(a=1;a<=N;a++){
      	if(a%10 != 0){
          	i = a%10;
	    	for(b=1;b<=N;b++){
            	if(b%10 != 0){
                	j=b;
                  	while(j <10){
                    	j %= 10;
                    }
                  	if(i == j){ number++;}
                }
            }
        }
    }
  	printf("%d",number);
	return 0;
}