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>

int main(){
	int N;
	
	std::cin >> N;
	
	if(N < 10){
		std::cout << N << std::endl;
		return 0;
	}
	if(N < 100){
		std::cout << "9" << std::endl;
	}
	if(N < 1000){
		std::cout << N -90 << std::endl;
		return 0;
	}
	
	if(N < 10000){
		std::cout << "909" <<std::endl;return 0;
	}
	if(N < 100000){
		std::cout << N - 9000 -90 << std::endl;return 0;
	}
	std::cout << "90909"  <<std::endl;
	return 0;
}