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;
#define ll long long
int main()
{
    ll n,t,c=0,ans=0;
    cin>>n;
    t=n;
    while(t>0)
    {
     t/=10;
     c++;
    }
    if(c%2)
    {
        ans+=(n-pow(10,c-1));
        ans++;
        c-=2;
        while(c>0)
        {
            ans+=(9*pow(10,c-1));
            c-=2;
        }
    }
    else
    {
        c--;
        while(c>0)
        {
            ans+=(9*pow(10,c-1));
            c-=2;
        }
    }
    cout<<ans;
    return 0;
}