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>
#define    ll   long long int
#define    sz   200010
#define    md   1000000007
#define    py   printf("Yes\n")
#define    pn   printf("No\n")

using namespace std;


ll tani(ll n)
{
    ll i,c=0;
    while(n>0)
    {
        c++;
        n/=10;
    }

    return c;
}


int main()
{


   ll n,c=0,i,j,k;

   cin>>n;

   for(i=1;i<=n;i++)
   {
       k=tani(i);

       if(k%2==1)c++;
   }

   cout<<c<<endl;




    return 0;
}