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 a;
    scanf("%d",&a);
    if(0<a && a<10){
        printf("%d",a);
    }
    else if(9<a && a<100 ){
        printf("9");
    }
    else if(99<a && a<1000){
        printf("%d", (a-90));
    }
    else if(999<a && a<10000){
        printf("909");
    }
    else if(99999<a && a<100000){
        printf("%d",(a-9090));
    }
    else if(a==100000){
        printf("90909");
    }
    return 0;
}