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;
typedef long long ll;
int n;
int main() {
    ios::sync_with_stdio(false); cin.tie(0);
    cin >> n;
    int x = 0;
    int tmp = n, t, p = 1;
    while(tmp) x++, t = tmp % 10, tmp /= 10, p *= 10;
    p /= 10;
    int ans = 0;
    for(int i = 1; i < x; i++) {
        if(i == 1) ans += 9;
        if(i == 3) ans += 900;  //100~999
        if(i == 5) ans += 90000;
    }
    if(x & 1) {
        ans += n - p + 1;
    }
    cout << ans;
    return 0;
}