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>
#include <stdlib.h>

int main(void){
	int i,cnt=0,ans=0,num=0;
	char n[100000];
	scanf("%s",n);
	
	for (i = 0; n[i] != '\0'; ++i) cnt++;
	
	if (cnt == 6){
		printf("90909\n");
	}else if (cnt == 2){
		printf("9\n");
	}else if (cnt == 4){
		printf("909\n");
	}else{
		num = atoi(n);
		if (cnt == 1)
		{
			for (int i = 0; i < num; ++i)
			{
				ans++;
			}
		}
		else if (cnt == 3){
			ans = 9;
			for (int i = 100; i <= num; ++i)
			{
				ans++;
			}
		}
		else if (cnt == 5){
			ans = 909;
			for (int i = 10000; i <= num; ++i)
			{
				ans++;
			}
		}
		printf("%d\n",ans);
	}
	
	
	return 0;
}