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))
#define all(x) (x).begin(),(x).end()
#define dump(x) cout << #x << " = " << (x) << endl
#define YES(n) cout << ((n) ? "YES" : "NO") << endl
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl
#define yes(n) cout << ((n) ? "yes" : "no") << endl

using namespace std;
using ll = long long;
using P = pair<int, int>;

const int INF = 1001001001;
const vector<int> di = {-1,  0, 1, 0};
const vector<int> dj = { 0, -1, 0, 1};

void chmin(int &a, int b) { if (a > b) a = b; }

ll GCD(ll a, ll b) {
    if (b == 0) return a;
    else return GCD(b, a % b);
}
ll LCM(ll a, ll b) {
    ll g = GCD(a, b);
    return a / g * b;
}


int main() {
    int n; cin >> n;

    int ans = 0;
    rep(i, n) {
        int cnt = 0;
        while (n>0) {
            n /= 10;
            cnt++;
        }
        if (cnt % 2 == 1) ans++;
    }
    cout << ans << endl;
}