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(i = 0; i < n; i++)
using namespace std;

int main() {
    int n;
    cin >> n;
    int keta = 0;
    int temp = n;
    while (temp != 0) {
        temp /= 10;
        keta++;
    }
    // cout << keta << endl;
    if (keta == 1) {
        cout << n << endl;
    } else if (keta == 2) {
        cout << 9 << endl;
    } else if (keta == 3) {
        cout << 9 + (n-99) << endl;
    } else if (keta == 4) {
        cout << 909 << endl;
    } else if (keta == 5) {
        cout << 909 + (n-99999) << endl;
    } else {
        cout << 90909 << endl;
    }

    return 0;
}