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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.*;
import java.util.*;

class Main {
    static Scanner scanner = new Scanner();

    public static void main(String[]$) {
        int n = scanner.nextInt();
        int ans = 0;
        for (int i = 1; i <= n; i *= 100) {
            if (i * 10 > n) {
                ans += n - i + 1;
                break;
            }
            ans += i * 10 - i;
        }
        System.out.println(ans);
    }

    static class Scanner {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in), 32768);
        StringTokenizer tokenizer;

        String next() {
            try {
                while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                    tokenizer = new StringTokenizer(reader.readLine());
                }
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
            return tokenizer.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}