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 <vector>
#include <algorithm>
#include <cmath>


int main()
{
    long N;
    std::cin >> N;
    long count=1;
    long tempN=N;
    while(tempN>=10){
        tempN=tempN/10;
        count++;
    }
    int flag=0;
    if(count%2!=0){
        count=count/2;
        flag=1;
    }else{
        count=count/2;
    }
    long num=0;
    long temp=0;
    for(long i=0; i<count;i++){
        num+=9*pow(100,i);
        temp=pow(100,i);
    }
    if(flag==1){
        num+=N-temp*100+1;
    }
    std::cout << num << std::endl;
    return 0;
}