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 (void){
    int n;
    scanf("%d",&n);
    int b=0,c=0,d=0;
    int a[100000];
    for (int i=0;i<n;i++){
        a[i]=i+1;
    }
    for (int i=0;i<n;i++){
        if(a[i]<10){
            b+=1;
        }
        if(a[i]<1000&&a[i]>99){
            c+=1;
        }
        if(a[i]<10000&&a[i]>9999){
            d+=1;
        }
    }
    int sum=0;
    sum=b+c+d;
    printf("%d",sum);
    return 0;
}