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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;
using static System.Console;
using static System.Math;

class Solver {
    Scanner sc = new Scanner();

    public void Solve() {
        var N = sc.nextInt();
        var ans = 0;
        for (int i = 1; i <= N; i++) {
            var len = i.ToString().Length;
            if (len % 2 == 1) ans++;
        }
        WriteLine(ans);
    }
    
}

class Scanner {
    string[] s;
    int i;

    char[] cs = new char[] { ' ' };

    public Scanner() {
        s = new string[0];
        i = 0;
    }

    public string next() {
        if (i < s.Length) return s[i++];
        string st = Console.ReadLine();
        while (st == "") st = Console.ReadLine();
        s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
        if (s.Length == 0) return next();
        i = 0;
        return s[i++];
    }

    public int nextInt() {
        return int.Parse(next());
    }
    public int[] ArrayInt(int N, int add = 0) {
        int[] Array = new int[N];
        for (int i = 0; i < N; i++) {
            Array[i] = nextInt() + add;
        }
        return Array;
    }

    public long nextLong() {
        return long.Parse(next());
    }

    public long[] ArrayLong(int N, long add = 0) {
        long[] Array = new long[N];
        for (int i = 0; i < N; i++) {
            Array[i] = nextLong() + add;
        }
        return Array;
    }

    public double nextDouble() {
        return double.Parse(next());
    }

    public double[] ArrayDouble(int N, double add = 0) {
        double[] Array = new double[N];
        for (int i = 0; i < N; i++) {
            Array[i] = nextDouble() + add;
        }
        return Array;
    }
}

class Program {
    static void Main(string[] args) {
        Solver s = new Solver();
        s.Solve();
    }
}