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;
	scanf("%d",&N);
	int c,k,e,r;
	int ans = 0;
	c = 10;
	k = 1000;
	e = 100000;
	r = 10000000;
	if(N >= c){
		ans += 9;
		if(N >= k){
			ans += 900;
			if(N >= e){
				ans += 90000;
				if( N >= r){
					ans += 9000000;
				}
				else{
					ans = ans+N-999999;
				}
			}
			else{
				ans = ans + N - 9999;
			}
		}
		else{
			ans = ans + N - 99;
		}
	}
	else{
		ans = N;
	}
	printf("%d",ans);
	return 0;
}