Problem:
Score : 200 points

Problem Statement

Given is an integer N. Find the number of positive integers less than or equal to N that have an odd number of digits (in base ten without leading zeros).

Constraints

1 \leq N \leq 10^5

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of positive integers less than or equal to N that have an odd number of digits.

Sample Input 1

11

Sample Output 1

9

Among the positive integers less than or equal to 11, nine integers have an odd number of digits: 1, 2, \ldots, 9.

Sample Input 2

136

Sample Output 2

46

In addition to 1, 2, \ldots, 9, another 37 integers also have an odd number of digits: 100, 101, \ldots, 136.

Sample Input 3

100000

Sample Output 3

90909

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include "bits/stdc++.h"

#define REP(i,n) for(ll i=0;i<ll(n);++i)
#define RREP(i,n) for(ll i=ll(n)-1;i>=0;--i)
#define FOR(i,m,n) for(ll i=m;i<ll(n);++i)
#define RFOR(i,m,n) for(ll i=ll(n)-1;i>=ll(m);--i)
#define ALL(v) (v).begin(),(v).end()
#define UNIQUE(v) v.erase(unique(ALL(v)),v.end());
#define INF 1000000001ll
#define MOD 1000000007ll
#define EPS 1e-9

constexpr int dx[8] = { 1,1,0,-1,-1,-1,0,1 };
constexpr int dy[8] = { 0,1,1,1,0,-1,-1,-1 };


using namespace std;

using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
using vvi = vector<vi>;
using vvl = vector<vl>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;

template <class T> bool chmin(T& a, const T& b) { if (a > b) { a = b; return true; } return false; }
template <class T> bool chmax(T& a, const T& b) { if (a < b) { a = b; return true; } return false; }


int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n; cin >> n;
	int ans = 0;
	REP(i, n) {
		int a = i + 1;
		int b = log10(a) + EPS;
		if (b % 2 == 1)ans++;
	}
	cout << ans << endl;
}