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>
#include <string.h>

#define MAX (200000)

int main(){
    int n;
    int a,b;
    int i, j;
    int a_len,b_len;
    int a_top, b_top;
    int a_last, b_last;
    char str_a[MAX], str_b[MAX];
    int count=0;

    scanf("%d", &n);

    for(i=1;i<=n;i++){
        snprintf(str_a, MAX, "%d", i);
        a_len = strlen(str_a);
        a_top = (int)str_a[0];
        a_last = (int)str_a[a_len-1];
        for(j=1;j<=n;j++){
            snprintf(str_b, MAX, "%d", j);
            b_len = strlen(str_b);
            b_top = (int)str_b[0];
            b_last = (int)str_b[b_len-1];
            if(a_top == b_last && a_last == b_top)
                count++;
        }
    }

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

    return 0;
}