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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

int main(void)
{
	int N;
	scanf("%d",&N);
	int c,k,e,r;
	int ans = 0;
	c = 10;
	int c_w = 100;
	k = 1000;
	int k_w = 10000;
	e = 100000;
	int e_w = 1000000;
	r = 10000000;
	if(N < c)
	{
		ans = N;
	}
	else if(N >= c && N < c_w)
	{
		ans = 9;
	}
	else if(N >= c_w && N < k)
	{
		ans += 9;
		ans = ans + N - 99;
	}
	else if(N >= k && N < k_w)
	{
		ans += 9;
		ans += 900;
	}
	else if(N >= k_w && N < e)
	{
		ans += 9;
		ans += 900;
		ans = ans + N - 9999;
	}
	else if(N >= e)
	{
		ans += 9;
		ans += 900;
		ans += 90000;
	}
	
	printf("%d",ans);
	return 0;
}