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 floop(i, N) for (ll i = 0; i < N; i++)
#define floop1(i, N) for (ll i = 1; i <= N; i++)
#define out(ans) cout << ans << endl
#define in(n) cin >> n
ll n;
ll ans;

void solve();

int main()
{
    solve();
}

void solve()
{
    in(n);
    if (n < 10)
    {
        ans = n;
    }
    else if (n < 100)
    {
        ans = 9;
    }
    else if (n < 1e3)
    {
        ans = n - 100 + 10;
    }
    else if (n < 1e4)
    {
        ans = 999 - 100 + 10;
    }
    else if (n < 1e5)
    {
        ans = n - 1e4 + 999 - 100 + 11;
    }
    else if (n == 1e5)
    {
        ans = 99999 - 1e4 + 999 - 100 + 11;
    }
    out(ans);
}