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
#define PB push_back
#define MP make_pair
#define MOD (LL)(1e9+7)

int main(){
	int N;
	cin >> N;
	int ans = 0;
	for(int i=1; i<=N; i++){
		string S = to_string(i);
		if(S.size()%2 == 1) ans++;
	}
	cout << ans << endl;
}