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,i;
	int cnt=0;
	scanf("%d",&N);
	if(N%10==N)
	{
		for(i=0;i<N;i++)
		cnt++;
		printf("%d",cnt);
	}
	if(N%10!=N && N%100==N)
	printf("9");
	if(N%100!=N && N%1000==N)
	{
		for(i=0;i<(N/100);i++)
		cnt++;
		printf("%d",cnt+9);
	}
	if(N%1000!=N && N%10000==N)
	printf("909");
	if(N%10000!=N && N%100000==N)
	printf("90909");
	return 0;
}