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<stdlib.h>
#include<stdbool.h>
#define mprint(x) printf(FORMATCONVERSION(x),(x))
#define FORMATCONVERSION(x) _Generic((x),\
                                    int:"%d",unsigned int:"%u",\
                                    double:"%lf",long long int:"%lld",\
                                    unsigned long long int:"%llu",\
                                    char :"%c",char*:"%s")

int mmax_int(int a,int b){
    if(a > b){
        return a;
    }else{
        return b;
    }
}

int mmin_int(int a,int b){
     if(a > b){
        return b;
    }else{
        return a;
    }
}

int flip(int input){
    if(input < 10){
        return input;
    }else if(input < 100){
        return ((input % 10) * 10) + input / 10;
    }else if(input < 1000){
        return ((input % 10) * 100) + (((input / 10) % 10) * 10)+ input / 100;
    }else if(input < 10000){
        return ((input % 10) * 1000) + (((input / 10) % 10) * 100) + (((input / 100) % 10) * 10) + input / 1000;
    }else if(input < 100000){
        return (input % 10) * 10000 + (((input / 10) % 10) * 1000) + (((input / 100) % 10) * 100) + (((input / 1000) % 10) * 10) + input / 10000;
    }else{
        return (input % 10) * 100000 + (((input / 10) % 10) * 10000) + (((input / 100) % 10) * 1000) + (((input / 1000) % 10) * 100) + (((input / 10000) % 10) * 10) + input / 100000;
    }
}

int topnum(int input){
        if(input < 10){
        return input;
    }else if(input < 100){
        return input / 10;
    }else if(input < 1000){
        return input / 100;
    }else if(input < 10000){
        return input / 1000;
    }else if(input < 100000){
        return input / 10000;
    }else{
        return input / 100000;
    }
}

int main(void){
    int n;
    scanf("%d",&n);
    
    unsigned long long ans = 0;
    for(int i = 1;i <= n;i++){
        for(int j = i;j <= n;j++){
            if(topnum(i) == topnum(flip(j)) && topnum(flip(i)) == topnum(j) && (i % 10) != 0 && (j % 10) != 0){
                
                if(i != j){
                    ans+=2;
                }else{
                    ans++;
                }
            }
        }
    }

    mprint(ans);
    return 0;
}