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: Accepted)

#include <stdio.h>

int main(void) {
int N;
int n;
scanf("%d", &N);

if(N>=100000){
	printf("90909");
}
if(N>=10000 && N<100000){
	n=(N-9999)+909;
	printf("%d\n", n);
}
if(N<10000 && N>=1000){
	printf("909");
}
if(N<1000 && N>=100){
	n=(N-99)+9;
	printf("%d\n", n);
}
if(N<100 && N>=10){
	printf("9");
}
if(N<10){
	printf("%d\n",N);
}

	return 0;
}