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: Wrong Answer)

#include <stdio.h>
int main(void){
	int a;
	scanf("%d",&a);
	if(a==100000) printf("%d",9+900+90000);
	else if(a>9999) printf("%d",9+900+a/10000+a%100000);
	else if(a>999) printf("%d",9+900);
	else if(a>99) printf("%d",9+a/100+a%100);
	else if(a>9) printf("%d",9);
	else printf("%d",a);
	return 0;
}