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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>
#include <math.h>


int fstDgt(int n) {
    int nDgt = (int)log10(n);
    return (int)(n/pow(10, nDgt));
}

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


    int cache[9][9];
    // initialize:
    for (int i = 0; i < 9; ++i) {
        for (int j = 0; j < 9; ++j) {
            cache[i][j] = 0;
        }
    }

    int h, t;
    for (int i = 1; i <= n; ++i) {
        h = fstDgt(i);
        t = i % 10;
        if (t == 0) continue;
        cache[h-1][t-1] ++;
    }

    int res = 0;
    for (int i = 0; i < 9; ++i) {
        for (int j = 0; j < 9; ++j) {
            res += cache[i][j] * cache[j][i];
        }
    }
    printf("%d\n", res);
}