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 ll long long 
int main(void){
	ll n;
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cin>>n;
	int sum=0;
	int digits=log10(n)+1;
	//<<digits<<endl;
	if(1<digits) sum+=9;
	if(3<digits) sum+=900;
	if(5<digits) sum+=90000;
	if(digits==1) sum+=n;
	if(digits==3) sum+=n-99;
	if(digits==5) sum+=n-9999;
	cout<<sum<<endl;
	return 0;
}