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>
#include<stdlib.h>
#include<string.h>

int main(){
  int n;
  scanf("%d",&n);
  /*
  int matubi=n%10;
  int sento=n;
  while(sento/10 > 0){
  sento/=10;}*/

  int map[9][9];
  for(int i=0;i<9;i++){
    for(int j=0;j<9;j++){
      map[i][j]=0;}}
  int matubi;
  int sento;

  for(int i=1;i<=n;i++){
    matubi=i%10;
    if(matubi==0){
      continue;}
    sento=i;
    while(sento/10 > 0){
      sento/=10;}
    map[sento-1][matubi-1]++;}

  int count=0;
  for(int i=1;i<=9;i++){
    for(int j=1;j<=9;j++){
      count+=map[i-1][j-1]*map[j-1][i-1];}}

  printf("%d\n",count);

  return 0;
}