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 N;
	scanf("%d",&N);

	int i,cnt=0;
	
	//10000�`99999
	if(N==100000){
		cnt += 99999 - 10000 +1;
	}
	else if(N<100000 && N>=10000)
	{
		cnt += N%10000 + 1;
	}
	// 100�`999
	if(N>=1000){
		cnt += 999-100+1;
	}
	else if(N<1000&&N>=100)cnt += N%100+1;
	
	// 1~9
	if(N>=10){
		cnt += 9;
	}
	else if(N<10)cnt += N%10;
	
	

	printf("%d\n",cnt);
 return 0;
}