Convert string to decimal integer

Enter a to # Ending string , This problem requires that all non hexadecimal characters be filtered out （ Case insensitive ）, Form a new string of hexadecimal digits , Then convert it to decimal number and output . If there is a character before the first hexadecimal character “-”, It means that the number is negative .

Input format ：

The input gives one in a line to # The ending non empty string .

Output format ：

Output the converted decimal number on one line . Ensure that the output is in the long integer range .

sample input ：

+-P-xf4+-1!#

sample output ：

-3905

Code ： 

#include <stdio.h>
int main() {
	char a[50];
	int b[50];
	int i=0,sum=0,now=0,past=1,k,number=0,m=0;
	while((a[i]=getchar())!='#') {
		if(a[i]>='0'&&a[i]<='9') {
			b[m++]=a[i]-'0';
			number++;
		} else if(a[i]>='A'&&a[i]<='F') {
			b[m++]=a[i]-'7';
			number++;
		} else if(a[i]>='a'&&a[i]<='f') {
			b[m++]=a[i]-'W';
			number++;
		}
		i++;
	}
	now=i;
	k=i;
	sum=0;
	for(i=0; i<number; i++) {
		if(i>0) {
			sum=sum*16+b[i];
		} else if(i==0) {
			sum=sum+b[i];
		}
	}
	for(i=0; i<now; i++) {
		if((a[i]>='0'&&a[i]<='9')||(a[i]>='A'&&a[i]<='F')||(a[i]>='a'&&a[i]<='f')) {
			now=i;
			break;
		}
	}
	for(i=0; i<k; i++) {
		if(a[i]=='-') {
			past=i;
			break;
		}
	}
	if(past>now) {
		printf("%d",sum);
	} else {
		printf("%d",-sum);
	}
}