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<cmath>
using namespace std;

int GetDigit(int num){
    int digit=0;
    while(num!=0){
        num /= 10;
        digit++;
    }
    return digit;
}

int main(){
    int N,sum=0;
    cin >> N;
    if(GetDigit(N)%2==0){
        for (int i=1; GetDigit(N)-1>=i; i=i+2){
            sum+= 9*pow(10,i-1);
        }
        cout << sum;
    }
    else {
        for (int i=1; GetDigit(N)-2>=i; i=i+2){
            sum+= 9*pow(10,i-1);
        }
        sum+=N-pow(10,GetDigit(N)-1)+1;
        cout << sum;
    }
    return 0;
}