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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream> 
#include<string>

int main(void) {
	int a;
	std::cin >> a;
	int count = 0;
	if (a<10) {
		count = a;
	}
	if (a<100&&a>=10) {
		count = 9;

	}
	if (a > 100&&a<1000) {
		count = 9 + (a - 99);
	}
	if (a >= 1000 && a < 10000) {
		count = 909;
	}
	if (a > 10000 && a < 100000) {
		count = 909 + (a - 9999);
	}
	if (a == 100000) {
		count = 90909;
	}
	std::cout << count;
	return 0;
}