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.util.Scanner;
//
public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
//━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
		int n=sc.nextInt();
		int ans =0;
		for (int i=1;i<=n;i++) {
			int digCount=digCount(i);
			if (digCount%2!=0) {
				ans++;
			}
		}
		System.out.println(ans);
//━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
	}
	//----------------------------------------
	//置き場
	static int digCount(int n) {
		int res = 0;
		while(n > 0) {
			res ++;
			n /= 10;
			}
		return res;
	}
	//----------------------------------------
}