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 <iomanip>
#include <string>
#include <sstream>
#include <bitset>
#include <algorithm>
#include <cmath>
#include <numeric>
#include <utility>
#include <set>
#include <map>
#include <tuple> 
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <list>
using namespace  std;
using ll = long long;
using P = pair<ll, ll>;
#define REP(i, n) for(long long (i)=0, i_bound = (long long)(n); (i)<i_bound; ++i)
#define FOR(i, a, b) for(long long (i)=(a); (i)<(b); ++i)
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
template<class T1, class T2> ostream& operator << (ostream &s, pair<T1,T2> P)
{ return s << '<' << P.first << ", " << P.second << '>'; }
template<class T> ostream& operator << (ostream &s, vector<T> P)
{ for (int i = 0; i < P.size(); ++i) { if (i > 0) { s << " "; } s << P[i]; } return s; }
template<class T> ostream& operator << (ostream &s, vector<vector<T> > P)
{ for (int i = 0; i < P.size(); ++i) { s << endl << P[i]; } return s << endl; }
const long long MOD = static_cast<long long>(1e9) + 7LL;
const long long INF = 1234567891234567890LL;
ll pow(ll a, ll n){
    ll res = 1LL;
    while(n){
        if(n%2==1){
            res *= a;
        }
        a *= a;
        n>>=1; 
    }
    return res;
}
signed main(){
    ll N; cin>>N;
    ll n = N;
    ll dig = 0;
    while(n){
        n/=10;
        dig++;
    }
    ll res = 0;
    for(int i=1; i<=dig; i++){
        if(i%2==0){
            continue;
        }
        else{
            if(i==dig) res += N - pow(10, i-1) + 1;
            else res += pow(10, i) - pow(10, i-1);
        }
    }
    cout << res << endl;
    return 0;
}