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

printf("%d",a);
return 0;
}