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 N,count=0;
	scanf("%d",&N);
	if(N<10) count=N;
	if(N>=10 && N<100) count=9;
	if(N>=100 && N<1000) count=N-100+1+9;
	if(N>=1000 && N<10000) count=909;
	if(N>=10000 && N<100000) count=N-10000+9+900;
	if(N>=100000) count=90909;
	printf("%d\n",count);
}