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,D,E,F;

     scanf("%d",&N);

     if(N<10){
        D=N;
       printf("%d\n",D);
        return 0;
     }

     if((N>=100)&&(N<1000)){
        E=9+(N-99);
       printf("%d\n",E);
        return 0;
     }

     if((N>=10000)&&(N<100000)){
        F=908+(N-9999);
       printf("%d\n",F);
        return 0;
     }

     if((N>=10)&&(N<100)){
       printf("9\n"); return 0;
     }

     if((N>=1000)&&(N<10000)){
       printf("908\n"); return 0;
     }

     if(N==100000) { printf("90909\n"); return 0; }

 }