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;
const int INF = 1001001001;

int main() {
int N;
cin >> N;
if (N < 10) cout << N << endl;
else if (10 <= N && N<100) cout << 9 << endl;
else if (100 <= N && N< 1000) cout << 9+N-99 << endl;
else if (1000 <= N && N<10000) cout << 909 << endl;
else if (10000 <= N && N< 100000) cout << 909 + N -9999 << endl;
else cout << 90909 << endl;


return 0;
}