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<stdbool.h>
#include<ctype.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")

#define mswap(x,y) _Generic((x),int:swap_int,unsigned int:swap_uint,\
                            long long:swap_ll,unsigned long long:swap_ull)(&(x),&(y))

void swap_int(int *a,int *b){int tmp = *a;*a = *b;*b= tmp;return;}
void swap_uint(unsigned int *a,unsigned int *b){unsigned int tmp = *a;*a = *b;*b= tmp;return;}
void swap_ll(long long *a,long long *b){long long tmp = *a;*a = *b;*b= tmp;return;}
void swap_ull(unsigned long long *a,unsigned long long *b){unsigned long long tmp = *a;*a = *b;*b= tmp;return;}

int main(void){
    int n;
    scanf("%d",&n);

    int handtpair[10][10] = {};

    int temp = 0;
    for(int i = 1;i <= n;i++){
        temp = i;
        while(temp >= 10)temp /= 10;
        if(temp != 0 && i % 10 != 0)++handtpair[temp][i % 10];
    }

    long long int ans = 0;
    temp = 0;
    for(int i = 0;i <= n;i++){
        temp = i;
        while(temp >= 10)temp /= 10;
        if(temp != 0 && i % 10 != 0)ans += handtpair[i % 10][temp];
    }
    mprint(ans);
    return 0;
}