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>

int main(){
    int n;
    scanf("%d",&n);
    int n10=0;
    int tmp=n;
    int nleft,nright;
    nright=n%10;
    while(tmp!=0){
        nleft=tmp;
        tmp=tmp/10;
        n10++;
    }
    unsigned long num=0;
    for(int i=1;i<n+1;i++){
        int left=i;
        int right=left%10;
        while(left/10!=0){
            left=left/10;
        }
        if(left!=0&&right!=0){
            for(int j=1;j<n10;j++){
                if(j==1){
                    if(left==right)num++;
                }else if(j==2){
                    num++;
                }else{
                    tmp=1;
                    for(int k=0;k<j-2;k++){
                        tmp=tmp*10;
                    }
                    num=num+tmp;
                }
            }
            if(nleft>right){
                tmp=1;
                if(n10<3){
                    num++;
                }else{
                    for(int j=0;j<n10-2;j++){
                        tmp=tmp*10;
                    }
                    num=num+tmp;
                }
            }else if(nleft==right){
                if(n10>2){
                    tmp=1;
                    for(int j=0;j<n10-2;j++){
                        tmp=tmp*10;
                    }
                    tmp=n/10-nleft*tmp;
                    if(nright<left){
                        num=num+tmp;
                    }else{
                        num=num+tmp+1;
                    }
                }else if(n10==1){
                    num++;
                }else{
                    if(nright>=left)num++;
                }
            }
        }
    }
    printf("%ld",num);
    return 0;
}