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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
typedef long long ll;

int keta(int n)
{
	int cnt = 1;
	while (n > 9)
	{
		n /= 10;
		cnt++;
	}
	return cnt;
}

int main()
{
	int n;
	scanf("%d", &n);
	// printf ("%d\n", keta(n));
	int i = 1;
	int cnt = 0;
	while (i <= n)
	{
		if (keta(i) % 2)
			cnt++;
		i++;
	}
	printf("%d", cnt);
}