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>
#define repd(i,a,b) for (int i=(a);i<(b);i++)
#define rep(i,n) repd(i,0,n)
#define int long long
typedef long long ll;
using namespace std;
const int MOD = 1000000007;
const int INF = 1010000000;
const double EPS = 1e-10;
const pair<int,int> fd[] = {make_pair(1,0),make_pair(-1,0),make_pair(0,1),make_pair(0,-1)}; 

vector<int> init(int size){
  vector<int> par(size,-1);
  return par;
}

int root(int x, vector<int>& par){
  if(par[x] < 0){
    return x;
  }else{
    return par[x] = root(par[x],par);
  }
}

bool same(int x, int y, vector<int>& par){
  return root(x, par) == root(y, par);
}

void unite(int x, int y, vector<int>& par){
  x = root(x, par), y = root(y, par);
  if(same(x,y, par))return;
  par[x] += par[y];
  par[y] = x;
}

int size(int x, vector<int>& par){
  return -par[root(x, par)];
}

int pow(int x, int n){
  if(n == 0)return 1;
  if(n % 2 == 0)return pow(x * x % MOD, n/2);
  else return x * pow(x, n-1) % MOD;
}

signed main(){
  int n;cin>>n;
  int ans = 0;int a = 100;
  repd(i,1,n+1){
    if(to_string(i).length() % 2 == 1){
      ans++;
    }
  }
  cout << ans << endl;
}