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>
#include<stdlib.h>


int main(void){
	int N;
	int ans, i, j;

	scanf("%d", &N);

	ans = 0;
	for(i=1;i<N;i*10){
		if(i>N){
			break;
		}else if(i<=N && N<=i*10){
			ans = ans + (N - i + 1);
		}else{
			ans = ans +(i*10-1 - i + 1);
		}
	}

	printf("%d", ans);



	return 0;
}