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 <bits/stdc++.h>
#define REP(i, n) for (int i=0, i##_limit=(n); i<i##_limit; ++i)
#define FOR(i, a, b) for (int i=(a), i##_limit=(b); i<i##_limit; ++i)
#define pb push_back
#define mp make_pair
using namespace std;
using ll = long long;
using pi = pair<int, int>;

int main()
{
    int N;
    cin >> N;

    int ret = 0;
    for (int i=1; i<=N; ++i)
    {
        if ((int)(log10(i)+1)%2 == 1) ++ret;
    }

    cout << ret << endl;
    return 0;
}