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<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define FOR0(i,n) for(i=0;i<n;i++)
#define FOR(i,j,n) for(i=j;i<n;i++)
#define FORD(i,j,k) for(i=j;i>=k;i--)
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define inf 1000000000
#define ninf -1000000000
#define endl '\n' 
#define she_told_me_dont_worry ios_base::sync_with_stdio (false) ; cin.tie(0) ; cout.tie(0) ;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
// Use cout.flush() for interactive problems.
// Use this for increased stack size: g++ -o a.exe -Wl,--stack=256000000 -O2 source.cpp
inline long long  MAX2(long long  a, long long int b){return (a)>(b)?(a):(b);}
inline long long  MAX3(long long  a, long long  b,long long  c){return (a)>(b)?((a)>(c)?(a):(c)):((b)>(c)?(b):(c));}
inline long long  MIN2(long long  a, long long  b){return (a)<(b)?(a):(b);}
inline long long  MIN3(long long  a, long long b,long long c){return (a)<(b)?((a)<(c)?(a):(c)):((b)<(c)?(b):(c));}
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> ii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector< ii > vii;
ii find_ends(int a)
{
    ii ends;
    int cur = a;
    ends.ss = a%10;
    while( a > 0)
    {
        ends.ff = a;
        a /= 10;
    }
    return ends;
}
int main()
{
    she_told_me_dont_worry
    ll n;
    cin>>n;
    ll i;
    map< ii, ll > m;
    FOR(i, 1, n+1)
    {
        m[find_ends(i)]++;
    }
    ll ans = 0;
    // starts 1 to 9
    map< ii, ll>::iterator it = m.begin(), it2;
    while(it != m.end())
    {
        ii tobe = mp(it->ff.ss, it->ff.ff);
        it2 = m.find(tobe);
        if(it2 != m.end())
            ans += it->ss*it2->ss;
        it++;
    }
    cout<<ans<<endl;
    cerr <<endl<< "Time elapsed : " << clock() * 1000.0 / CLOCKS_PER_SEC << " ms" << '\n';
}