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)

#include <stdio.h>
#include <stdlib.h>
 
int main(void) {
 
	int a, b, c, sum;
	a = b = c = sum = 0;
	scanf("%d", &a);
	
	if (a == 10000) {
		sum = 901;
	} else if (a < 10000 && a >= 1000) {
		sum = 900;
	} else if (a < 10000 && a < 1000) {
		sum = a - 101;
	} else if (a < 100 && a >= 10) {
		sum = 9;
	} else if (a < 100 && a < 10) {
		sum = a;
	} else {
		sum = 0;
	}
 
	printf("%d", sum);
	return EXIT_SUCCESS;
}