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 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main()
{
long int n;
scanf("%ld",&n);

if(n<10)
{
printf("%ld\n",n);
}
else if(n>=10 && n<100)
{
printf("9\n");
}
else if(n>=100 && n<1000)
{
printf("%ld\n",10+n-100);
}
else if(n>=1000 && n<10000)
{
printf("909\n");
}
else if(n>=10000 && n<100000)
{
printf("%ld\n",1000+n-10000);
}
else if(n==100000)
{
printf("90909\n");
}
return 0;
}