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: Accepted)

#include "bits/stdc++.h"
using namespace std;
int main() {
  	int N,ans;
  	cin >> N;
  	
  	if  (N<10){
    cout << N <<endl;
  	    }
	 else  if(10 <=N && N<100){   
  	cout << 9 <<endl;
  	 }
	else  if (100<=N && N<1000){
  	 cout << N-99+9 <<endl;
  	 }
  	 else if (1000<=N && N <10000){
  	 cout << 909 <<endl;
  	 }
  	 else if (10000<=N && N<10e4){
  	 cout <<N-9999+909 << endl;
  	 }
  	 else if (10e4 <= N && N<10e5){
  	     cout << 90909 << endl;
  	 }
  	 else if ( N ==10e5){
  	 cout <<90910 <<endl;
  	 }
  	}