Simple Arithmetics "runtime error (NZEC)"

[egoncasteel]

I have not worked with judge programs programs before, and I am not sure why this code doesn't work.

#http://www.spoj.pl/problems/ARITH/
import string
import re
def parser(lines):
    re_nums = re.compile("[0-9]*")
    re_sign = re.compile("[+,\-,*]")
    for ii in string.split(lines, '\n'):
        nums = re_nums.findall(ii) #finds the 2 numbers
        sign = re_sign.findall(ii) #finds the sign
        if re_sign.search(ii) and len(nums) == 4: #checks that there is 2 numbers and a sign
        if sign[0] == "+":
            add(int(nums[0]),int( nums[2]))
        elif sign[0] == "-":
            subtract(int(nums[0]),int( nums[2]))
        elif sign[0] == "*":
            multiply(int(nums[0]),int(nums[2]))
def multiply(vala, valb):
    product = vala * valb
    rr = max(len(`vala`), len(`valb`) +1, len(`product`)) #how far right to align the calculation
    print
    print string.rjust(`vala`, rr)
    print string.rjust('*' + `valb`, rr)
    print string.rjust('-' * max(len(`vala`), len(`valb`)+1), rr)
    if len(`valb`) > 1: #if there is more then one digit in the multiplyer do split and sum
        for xx in ''.join(reversed(`valb`)): #revese the string to make digits sum correct
            print string.rjust(str(int(xx) * vala), rr)
            rr = rr - 1
        print "-" * len(`product`)
    print product
def add(vala, valb):
    sum = vala + valb
    rr = max(len(`vala`), len(`valb`) +1, len(`sum`)) #how far right to align the calculation
    print 
    print string.rjust(`vala`, rr)
    print string.rjust('+' + `valb`, rr)
    print string.rjust('-' * rr, rr)
    print string.rjust(`sum`, rr)
def subtract(vala, valb):
    sum = vala - valb
    rr = max(len(`vala`), len(`valb`) +1, len(`sum`)) #how far right to align the calculation
    print 
    print string.rjust(`vala`, rr)
    print string.rjust('-' + `valb`, rr)
    print string.rjust('-' * rr, rr)
    print string.rjust(`sum`, rr)   
if __name__ == '__main__':
    input = ''
    items = int(raw_input())
    ii = 0
    while ii < items:
        input = input + str(raw_input()) + '\n'
        ii = ii + 1
    parser(input)

I/O
5
100+5568
2365-8854
656582*6565
55984+23
52525*65
  100
+5568
-----
 5668
 2365
-8854
-----
-6489
     656582
  *6565
 ------
3282910
   3939492
  3282910
 3939492
----------
4310460830
55984
  +23
-----
56007
  52525
    *65
  -----
 262625
315150
-------
3414125

It runs on my computer and I have tried a lot of different cases and the only way I can make it error is to give it a non int on the first line. The judge keeps giving me the "runtime error (NZEC)". Is there a way to get any more information out of the judge, or does any one see the problem?

[Simes_sof]

This might be a silly question, as I don't know python, but will your program handle numbers with 500 digits?

[egoncasteel]

I don't know. Thats a good question I will try modifying it to use long ints and see what happens.

[Simes_sof]

As I said, I don't know python, but I really doubt long ints will be big enough.

[egoncasteel]

Now I just get the wrong answer
Not sure why the output matches perfectly, oh well.

#http://www.spoj.pl/problems/ARITH/
import string
import re
def parser(lines):
    re_nums = re.compile("[0-9]*")
    re_sign = re.compile("[+,\-,*]")
    for ii in string.split(lines, '\n'):
        nums = re_nums.findall(ii) #finds the 2 numbers
        sign = re_sign.findall(ii) #finds the sign
        if len(sign) == 1 and len(nums) == 4: #checks that there is 2 numbers and a sign
            if sign[0] == "+":
                add(long(nums[0]),long( nums[2]))
            elif sign[0] == "-":
                subtract(long(nums[0]),long( nums[2]))
            elif sign[0] == "*":
                multiply(long(nums[0]),long(nums[2]))
def multiply(vala, valb):
    product = vala * valb
    rr = max(len(str(vala)), len(str(valb)) +1, len(str(product))) #how far right to align the calculation
    print
    print string.rjust(str(vala), rr)
    print string.rjust('*' + str(valb), rr)
    print string.rjust('-' * max(len(str(vala)), len(str(valb))+1), rr)
    if len(str(valb)) > 1: #if there is more then one digit in the multiplyer do split and sum
        for xx in ''.join(reversed(str(valb))): #revese the string to make digits sum correct
            print string.rjust(str(long(xx) * vala), rr)
            rr = rr - 1
        print "-" * len(str(product))
    print product
def add(vala, valb):
    sum = vala + valb
    rr = max(len(str(vala)), len(str(valb)) +1, len(str(sum))) #how far right to align the calculation
    print 
    print string.rjust(str(vala), rr)
    print string.rjust('+' + str(valb), rr)
    print string.rjust('-' * rr, rr)
    print string.rjust(str(sum), rr)
def subtract(vala, valb):
    sum = vala - valb
    rr = max(len(str(vala)), len(str(valb)) +1, len(str(sum))) #how far right to align the calculation
    print 
    print string.rjust(str(vala), rr)
    print string.rjust('-' + str(valb), rr)
    print string.rjust('-' * rr, rr)
    print string.rjust(str(sum), rr)   
if __name__ == '__main__':
    items = raw_input()
    is_num =  re.compile('[0-9]+')
    while not is_num.match(items): items = raw_input()
    input = ''
    ii = 0
    while ii < int(items):
        input = input + str(raw_input()) + '\n'
        ii = ii + 1
    parser(input)
    print

 12345
+67890
------
 80235
 324
-111
----
 213
    325
  *4405
  -----
   1625
     0
 1300
1300
-------
1431625

[egoncasteel]

As I said, I don't know python, but I really doubt long ints will be big enough.

heres what the manual said

There are four distinct numeric types: plain integers, long integers, floating point numbers, and complex numbers. In addition, Booleans are a subtype of plain integers. Plain integers (also just called integers) are implemented using long in C, which gives them at least 32 bits of precision (sys.maxint is always set to the maximum plain integer value for the current platform, the minimum value is -sys.maxint - 1). Long integers have unlimited precision. Floating point numbers are implemented using double in C. All bets on their precision are off unless you happen to know the machine you are working with.

I guess long has no limit. I don't know. I am new to python. I am using the site to get some practice.