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 <bits/stdc++.h>
#define rep(i,n) for(int i=0; i< (n); i++)
using namespace std;

int main()
{
    int n; cin >> n;
    int count = 0;
    int ans = 0;
    int k = n;
    
        
    while(n){
        count++;
        n /= 10;
    }
    if(count == 1){
        cout << k << endl;
        return 0;
    }else if(count == 2){
        cout << 9 << endl;
        return 0;
    }else if(count == 3){
        cout << (k-100+ 9 +1) << endl;
        return 0; 
    }else if(count == 4){
        cout << 909 << endl;
        return 0;
    }else {
        cout << 90909 << endl;
        return 0;
    }
    
    return 0;
}