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<iostream>
#include<cassert>
#include<cctype>
#include<cerrno>
#include<cfloat>
#include<vector>
#include<algorithm>
#include<string>
#include<map>
#include<math.h>
#include<queue>
#include<deque>
#include<stack>
#include<cstdio>
#include<utility>
#include<set>
#include<list>
#include<cmath>
#include<stdio.h>
#include<string.h>
#include<iomanip>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<bitset>
#include<functional>
#include<iterator>
#include<complex>
#include<fstream>
#include<random>
#include<ctype.h>
#include<stdio.h>
using namespace std;
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define REP(i, n) FOR(i, 0, n - 1)
#define NREP(i, n) FOR(i, 1, n)
using ll = long long;
using pii = pair<int, int>;
using PII = pair<ll, ll>;
using piii = pair<pii, pii>;
using Pi = pair<int, pii>;
using Graph = vector<vector<int>>;
const int dx[4] = { 0, -1, 1, 0 };
const int dy[4] = { -1, 0, 0, 1 };
bool check(int x, int y) {
	if (0 <= x && x<55 && 0 <= y && y<55)return true;
	else return false;
}
const ll INF = 1e9+7;
int gcd(int x, int y) {
	if (x < y)swap(x, y);
	if (y == 0)return x;
	return gcd(y, x%y);
}
void mul(ll a, ll b) {
	a = a * b % INF;
}
using Graph = vector<vector<int>>;

//iの逆元mod inf
long long modinv(long long a, long long m) {
	long long b = m, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}
const double PI = 3.14159265358979323846;
const int MAX = 510000;
ll fac[MAX], finv[MAX], inv[MAX];
//テーブルをつくる前処理
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;//mod pにおける1,2,,,nの逆元
	for (int i = 2; i < MAX; ++i) {
		fac[i] = fac[i - 1] * i%INF;
		inv[i] = INF - inv[INF%i] * (INF / i) % INF;
		finv[i] = finv[i - 1] * inv[i] % INF;
	}
}

ll COM(int n, int k) {
	if (n < k)return 0;
	if (n < 0 || k < 0)return 0;
	return fac[n] * (finv[k] * finv[n - k] % INF) % INF;
}
double Euclidean_distance(double x1, double y1, double x2, double y2) {
	return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}

int prime[1001000];//i番目の素数
bool is_prime[1001010];//is_prime[i]がtrueならiは素数
					   //n以下の素数の数を返す
int sieve(int n) {
	int p = 0;
	for (int i = 0; i <= n; ++i) {
		is_prime[i] = true;
	}
	is_prime[0] = false;
	is_prime[1] = false;
	for (int i = 2; i <= n; ++i) {
		if (is_prime[i]) {
			prime[p] = i;
			p++;
			for (int j = 2 * i; j <= n; j += i) {
				is_prime[j] = false;
			}
		}
	}
	return p;
}

//素因数分解
map<int, int>prime_factor(int n) {
	map<int, int>res;
	for (int i = 2; i*i <= n; ++i) {
		while (n%i == 0) {
			++res[i];
			n /= i;
		}
	}
	if (n != 1)res[n] = 1;
	return res;
}


ll powmod(ll a, ll k,ll mod) {
	ll ap = a, ans = 1;
	while (k) {
		if (k & 1) {
			ans *= ap;
			ans %= mod;
		}
		ap = ap * ap;
		ap %= mod;
		k >>= 1;
	}
	return ans;
}
ll invi(ll a,ll mod) {
	return powmod(a, mod - 2,mod);
}
//逆元Aを足したときのmodで割った余りは
//+=invi(A)
//Yで割ることはY^(10^9+7)-2を掛けることは同値
///////////////////////////////////////


int main() {
	int N; cin >> N;
	vector<vector<int>>V[10];
	REP(i, 10)V[i].resize(10);
	Graph G1, G2;
	G1.resize(10);
	G2.resize(10);
	//G1[i]は先頭がiである数G2[i]は末尾がiである数
	for (int i = 1; i <= N; ++i) {
		int cnt = -1;
		int copy = i;
		while (copy != 0) {
			copy /= 10;
			cnt++;
		}
		int key = i / powmod(10, cnt, INF);
		V[key][i % 10].push_back(i);
	}
	int ans = 0;
	REP(i, 10) {
		REP(j, 10) {
			for (int k = 0; k < V[i][j].size(); ++k) {
				ans += V[j][i].size();
			}
		}
	}
	cout << ans << endl;
	return 0;
}