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;

int main(){

int n;
cin>>n;

if(n/10<1) {
    cout<<n<<endl;
    return 0;
}
if(n/10>=1 && n/100<1) {
    cout<<9<<endl;
    return 0;
}
if(n/100>=1 && n/1000<1){
    cout<<9+n-99<<endl;
    return 0;
}
if(n/1000>=1 && n/10000<1){
    cout<<909<<endl;
    return 0;
}
if(n/10000>=1 && n/100000<1){
    cout<<909+n-9999<<endl;
    return 0;
}else{
    cout<<90909<<endl;
}

return 0;
}