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>
int main(void) {
	int n;
	scanf("%d", &n);
	int cnt = 0;
	int ans = 0;
	if (n >= 10) {
		ans = 9;
		for (int i = 100; i <= n; i=i*100) {
			cnt = 0;
			for (int j = i; j < i*10; j++) {
				cnt++;
				if (j == n) {
					break;
				}
			}
			ans = ans + cnt;
		}
	}
	else {
		ans = n;
	}
	printf("%d\n", ans);
	return 0;
}