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>

using namespace std;
typedef long long ll;

#define rep(i, n) for (int i = 0; i < (n); i++)
#define rep2(i, a, b) for (int i = (a); i < (b); ++i)
#define all(a) (a).begin(), (a).end()
#define all2(a, b) (a).begin(), (a).begin() + (b)
#define debug(vari) cerr << #vari << " = " << (vari) << endl;

int keta(int N)
{
        int cnt;
        if (N < 10)
        {
                cnt = N;
        }
        else if (N < 1000)
        {
                if (N - 100 + 1 > 0)
                {
                        cnt = 9 + (N - 100 + 1);
                }
                else
                {
                        cnt = 9;
                }
        }
        else if (N < 100000)
        {
                if (N - 10000 + 1 > 0)
                {
                        cnt = 9 + 900 + (N - 10000 + 1);
                }
                else
                {
                        cnt = 9 + 900;
                }
        }
        else
        {
                cnt = 9 + 900 + 90000;
        }

        return cnt;
}

int main()
{
        int N;
        cin >> N;
        int ans = keta(N);
        cout << ans << endl;
        return 0;
}