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 rep0(i, n) for (long long i = 0; i < (long long)(n); i++)
#define rep1(i,n) for(long long i=1;i<=(long long)(n);i++)
#define all(v) v.begin(),v.end()
#define Sort(v) sort(all(v))
#define Reverse(v) reverse(all(v))
#define Rsort(v) sort(all(v),greater)
#define pub(v) push_back(v)
#define pob(v) pop_back(v)
typedef long long ll;
typedef vector<ll> vel;
typedef vector<vector<ll>> vel2d;
typedef string str;
typedef pair<ll,ll> pal;
typedef vector<str> ves;


int main() {
    ll N;
    cin>>N;
    
    if(N==100000){cout<<9+999-99+99999-9999<<endl;}
    if(10000<=N&&N<100000){cout<<9+999-99+N-9999<<endl;}
    if(1000<=N&&N<10000){cout<<9+999-99<<endl;}
    if(100<=N&&N<1000){cout<<9+N-99<<endl;}
    if(10<=N&&N<100){cout<<9<<endl;}
    if(1<=N&&N<10){cout<<N<<endl;}
}