[Tutor] crashed with Runtime Error: NZEC (non-zero exit code)

Andrei Petre andrei.petre at gmail.com
Sun Feb 24 13:36:53 CET 2008


I wrote a code for a problem and submitted to an online judge, a program
that runs my program against a set of predefined input data.

I wrote it for exercise my python writing. In don't want to discuss the
algorithm here!

I dare to ask for help because the answer was : crashed with Runtime Error:
NZEC (non-zero exit code). I heard that is a general error for most
interpreters and elsewhere that NZEC error code has to do with exceptions.

So, i'm seeking for a possible problem with the coding, not with the

Here, the code. (l also attached it in a file)
Using characters of the string, the program should construct the maximum
number which divides by fifteen without remainder.




import sys

def write(number): # assamble the number
    # assure the 5 divisibility
    if number[0] != 0:
        number[0] -= 1
        last_d = 0
        number[5] -= 1
        last_d = 5

    zero = True
    for i in reversed(range(1,10)):
         for j in range(digits[i]):
             zero = False

    # leading zeroes should be omitted
    if zero == False:
        for i in range(digits[0]):

    print last_d

def remove_item(d,item):
    if type(item) != list:
        d[item] -= 1
        d[item[0]] -= 1
        d[item[1]] -= 1

def case5(d,item):
    # return True if i try to remove the only five that assure the
divisibility with 5
    return item == 5 and d[0] == 0 and d[item] == 1

def exist(d,item):
    found = False
    if type(item) != list:
        if d[item] != 0 and case5(d,item) == False:
            found = True
        p1 = d[item[0]] != 0 and case5(d,d[item[0]]) == False
        d[item[0]] -= 1
        p2 = d[item[1]] != 0 and case5(d,d[item[1]]) == False
        d[item[0]] += 1
        if p1 == True and p2 == True:
            found = True
    return found

def remove(d, mult):
    found = False
    for item in mult:
        if exist(d,item) == True:
            found = True
    if found == False:
        d = []
    return d

def remove_all(digits,rest):
    # if the number is the form 3k+1, i try to remove one 3k+1 digit or two
3k+2 digits
    # if the number is the form 3k+2, i try to remove one 3k+2 digit or two
3k+2 digits
    one = [1,4,7,[2,2],[2,5],[2,8],[5,5],[5,8],[8,8]]
    two = [2,5,8,[1,1],[1,4],[1,7],[4,4],[4,7],[7,7]]
    if rest == 1:
        d = remove(digits, one)
        d = remove(digits, two)
    return d

def resolve(digits):
    suma = 0
    for i in range(10):
        suma += i*digits[i]
    rest = suma % 3
    if rest != 0: # if it's not divisible with 3 i try to remove some digits
        digits = remove_all(digits,rest)
    return digits

t = int(raw_input())
for tt in range(t):
    line = raw_input()
    digits = [0]*10
    # counts the frequence of digits in the string
    for i in range(len(line)):
        digits[int(line[i])] += 1
    # if it's not divisible with 5
    if digits[0] == 0 and digits[5] == 0:
        number = resolve(digits)
        if(number == []):

Any alternatives for the personal rambling(if it's a wrong way to do it, of

L = [1,2,[3,4],[5,5]]
for item in L:
   if type(item) == list:
         print L[item[0]], L[item[1]]
         print L[item]


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