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 Python (3.4.3) (judge verdict: Wrong Answer)

def bunkai(n:int):
    ls=list(str(n))
    top,bottom=int(ls[0]),int(ls[-1])
    if n<=99:
        middle=0
    else:
        middle=int("".join(ls[1:-1]))
    return top, middle, bottom

n=int(input())
Nt,Nm,Nb = bunkai(n)

count=0
for i in range(1,n+1):
    t,m,b = bunkai(i)
    if b==0:continue
    if t==b:count+=1
    if n<=99:
        if int(b+t)<=n:count+=1
    else:
        keta = len(str(n))
        for x in range(keta-2):
            count+=10**x
        x+=1
        if Nt>b:
            count+=10**x
        elif Nt==b:
            if Nb>=t:
                count+=Nm+1
            else:
                count+=Nm
print(count)