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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

int main(void){

  int n;
  scanf("%d", &n);
	int count;
	if (n < 10) {
		count = n;
	} else {
		count = 9;
		if (n >= 100 && n < 1000) {
			count += n-99;
		}
		if (n >= 1000) {
			count += 999-99;
			if (n >= 10000 && n < 100000) {
				count += n-9999;
			}
			if (n == 100000) {
				count += 99999-9999;
			}
		}
	}

	printf("%d\n", count);

  return 0;
}