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<string.h>

int main() {
	
	char check[1000000] = "";
	long N;
	int len = 0;
	long ans = 0;

	scanf("%ld", &N);
	

	for (int i = 1; i <= N; i++) {
		sprintf(check, "%d", i);
		len = strlen(check);
		if (len % 2 != 0) {
			ans += 1;
		}
	}

	printf("%ld", ans);

	return 0;
}