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>
#include <math.h>
int main(void) {

	int N;
	scanf("%d", &N);
	if (N < 10) {
		printf("%d", N);
	}
	else if (N >= 10 && N < 100) {
		printf("%d", 9);
	}
	else  if (N >= 100 && N < 1000) {
		printf("%d",N-100 +1+ 9);
	}
	else if (N >= 1000 && N < 10000) {
		printf("%d",909);
	}
	else if (N >= 10000 && N < 100000) {
		printf("%d", N -10000 +1+ 9 + 900);
	}
	else if (N == 100000) {
		printf("%d",9+900+90000 );
	}


	return 0;

}