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)

/*
 * main.c
 *
 *  Created on: 2019/07/21
 *      Author: family
 */

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

int main()
{
	int N = 0, ans = 0;
	scanf("%d", &N);
	if(N < 100) {
		if (N < 10) {
			ans = N;
		} else {
			ans = 9;
		}
	} else if (N < 1000) {
		ans = 9 + ((N+1) - 100);
	} else if (N < 100000) {
		if(N < 10000) {
			ans = 909;
		} else {
			ans = 909 + ((N+1) - 10000);
		}
	} else {

	}
	printf("%d\n",ans);
    return 0;
}