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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <map>
#include <unordered_map>
#include <set>
#include <queue>
using namespace std;
#define ALL(x) (x).begin(),(x).end()
#define SZ(x) long long((x).size())
#define OUT(x) cout << (x) << endl
typedef long long ll;
typedef vector<ll> V;
typedef vector<vector<ll>> VV;
typedef pair<ll, ll> P;
typedef map<string, ll> M;
typedef unordered_map<ll, ll> HM;
typedef set<ll> S;
typedef multiset<ll> MS;
typedef queue<ll> Q;

int main(){
    int n; cin>>n;
    int ans = 0, k = 1, keta = 1;
    for(int i=9;; i=i*10+9){
        if(keta&1 && i <= n){
            ans += 9*k;
        }
        else if(keta&1 && i > n){
            ans += n-i/10;
        }
        k *= 10;
        keta++;
        if(i>n) break;
    }
    OUT(ans);
}