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)

#pragma warning(disable:4996)
#include<stdio.h>


int main(void) {
	int n;
	scanf("%d", &n);

	int ans = 0;
	int i = 1;
	int w = n / i;
	int count = 0;
	while (w) {
		i *= 10;
		++count;
		w = n / i;
	}

	switch (count) {
	case 6:
		ans = 90909;
		break;

	case 5:
		ans = 909;
		ans += n - 9999;
		break;

	case 4:
		ans = 910;
		break;

	case 3:
		ans = 9;
		ans += n - 99;

	case 2:
		ans = 9;
		break;

	case 1:
		ans += n - 9;
		break;

	default:
		ans = 1;
		break;
	}
//	printf("%d\n", count); test
	printf("%d\n", ans);
	return 0;
}