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);
	if (n < 10)
		printf("%d", n);
	else if (10 <= n && n < 100)
		printf("9");
	else if (100 <= n && n < 1000)
		printf("%d", n - 90);
	else if (1000 <= n && n < 10000)
		printf("909");
	else if (10000 <= n && n < 100000)//83232
		printf("%d", n - 9090);
	else
		printf("90909");
	return 0;
}