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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main() {
	int N, i;
	int ans = 0;
	int cnt = 0;
	double a;
	scanf("%d", &N);

	if (N < 10) {
		printf("%d", N);
		return 0;
	}
	else {
		ans += 9;
	}

	for (i = 10; i <= N; i++) {
		cnt = 0;
		a = i;
		while(a >= 10){
			a /= 10;
			cnt++;
		}
		cnt++;
		if (cnt % 2 == 1) {
			ans++;
		}
	}

	printf("%d\n", ans);

	return 0;
}