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: Accepted)

#include <cstdio>
#include <cstring>
#include <cmath>
#include <utility>
#include <iostream>
#include <functional>
#include <bitset>
#include <algorithm>
#include <vector>
#include <forward_list>
#include <set>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#define rep(i, s, g) for ((i) = (s); (i) < (g); ++(i))
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
const ll MOD = 1e9 + 7;
const ll INF = (1ll << 60);

int digit(int n)
{
    int d = 1;

    while (n / 10 > 0)
    {
        d++;
        n /= 10;
    }

    return d;
}

int main(void)
{
    int n;
    int ans = 0;

    cin >> n;

    for (int i = 1; i <= n; i++)
    {
        if (digit(i) % 2 == 1)
        {
            ans++;
        }
    }

    cout << ans << endl;
}