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 <iostream>
#include <string>

using namespace std;

int main(){
    int N,count=0;
    cin >>N;
    string a;
    a=to_string(N);
    int A=a.length();
    if(A==2){
        cout <<9;
    }else if(A==4){
        cout <<909;
    }else if(A==6){
        cout <<90909;
    }else if(A==1){
        cout <<N;
    }else if(A==3){
        cout <<9+N-99;
    }else if(A==5){
        cout <<9+900+N-9999;
    }
    return 0;
}