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: Wrong Answer)

#include<stdio.h>

int main(){

  int n;
  scanf("%d",&n);
  
  if(1<=n<10){
  	printf("%d",n);
  } else if(10<=n<100){
    printf("9");
  }else if(100<=n<1000){
    printf("%d-90",n);
  }else if(1000<=n<10000){
    printf("909");
  }else if(10000<=n<100000){
    printf("%d-9000",n);
  }else if(n=100000){
    printf("90909");
  }
    return 0;
}