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()
{
  int N;
  scanf("%d",&N);
  if(N>0&&N<10)
  {
    printf("%d",N);
  }
  else if(N>=10&&N<100)
  {
    printf("9");
  }
  else if(N>=100&&N<1000)
  {
    printf("%d",(N-99)+9);
  }
  else if(N>=1000&&N<10000)
  {
    printf("909");
  }
  else if(N>=10000&&N<100000)
  {
    printf("%d",(N-9999)+909);
  }
  else
  {
    printf("90909");
  }
}