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;
#define IOS ios::sync_with_stdio(false); cin.tie(0);
#define FOR(i,s,n) for(int i = s; i < (n); i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(n) (n).begin(), (n).end()
#define RALL(n) (n).rbegin(), (n).rend()
#define ATYN(n) cout << ( (n) ? "Yes":"No") << endl;
#define CFYN(n) cout << ( (n) ? "YES":"NO") << endl;
using ll = long long;
using ull = unsigned long long;

int main(void)
{
    IOS
    int n; cin >>n;

    int ans = 0;
    FOR(i,1,n+1) {
        if (to_string(i).size() % 2 == 1) ans++;
    }

    cout << ans << endl;

    return 0;
}