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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<string>
#include<utility>
#include<iostream>
#include<queue>
#include<math.h>
#include<algorithm>
#include<vector>
#include<map>
#include<stack>
#include<bitset>
#include<cmath>

using namespace std;
typedef long long ll;
#define REP(i,n) for(int i=0;i<(int)(n);i++)

int main(){
    int n;
    scanf("%d",&n);
    ll rui[200005]={};
    REP(i,n+2){
        if(i==0)rui[0]=0;
        else{
            rui[i]=rui[i-1];
            int one = i%10;
            int last = i;
            int keta = 1;
            while(last>=10){
                last=last/10;
                keta++;
            }
            //printf("%d %d %d\n",last,one,keta);
            if(one<last && keta>=2 && one!=0){
                REP(j,keta-1)rui[i]+=(ll)std::pow(10,j)*2;
            }
            else if(one>last && keta>=3 &&one!=0){
                REP(j,keta-2)rui[i]+=(ll)std::pow(10,j)*2;
            }
            else if(one==last && keta==1)rui[i]+=1;
            else if(one==last && keta==2)rui[i]+=3;
            else if(one==last && keta>2 && one!=0){
                int x = ((i - (ll)std::pow(10, keta-1)*last)/10+1)*2+1;
                rui[i]+=x;
                //printf("%d$$$\n",x);
                REP(j,keta-2)rui[i]+=(int)std::pow(10,j)*2;
            }
            //printf("###[%d %d %d]%d %lld %lld\n",one,last,keta,i, rui[i]-rui[i-1],rui[i]);
        }
    }
    //REP(i,n+2)printf("###%lld %lld\n",i, rui[i]);
    printf("%lld\n",rui[n]);
    return 0;
}