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;
#define REP(i, n) for(int i=0, i##_len=(n); i<i##_len; ++i)
typedef long long ll;
#define pb push_back
void Yes() {cout << "Yes" << endl;}
void No() {cout << "No" << endl;}
void YES() {cout << "YES" << endl;}
void NO() {cout << "NO" << endl;}
const double PI=acos(-1);

void solve(){
    int n; cin >> n;
    vector<int> a(100001);
    for(int i=1; i<=100001; i++){
        if((1<=i && i<=9) || (100<=i && i<=999) || (10000<=i && i<=99999)){
            a[i] = a[i-1] + 1;
        }
        else{
            a[i] = a[i-1];
        }
    }
    cout << a[n] << endl;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    solve();
}