Changing from int to int32 in Python

Question

so I am doing an exercise from SPOJ, and it's a simple calculator. Everytime I try to submit an answer I get an NZEC error, and I wonder if it's because it should be defined as int32. Here's my code:

import sys
n = input()
n = int(n)


i = 0
while n > i:
    znak, num1, num2 = input().split()
    num1 = int(num1)
    num2 = int(num2)
    if znak == "+":
        b = num1 + num2
        print(b)
    elif znak == "-":
        b = num1 - num2
        print(b)
    elif znak == "*":
        b = num1 * num2
        print(b)
    elif znak == "/":
        b = num1 / num2
        print(b)
    elif znak == "%":
        b = num1 % num2
        print(b)
    i += 1

sys.exit(0)

I've tried a some "solutions" for this NZEC error, but nothing worked.

There's no need to use sys.exit, the program exits automatically. — ForceBru
– ForceBru, Commented Mar 24, 2017 at 10:19
Just wanted to make sure if it works properly for their compilator. — Jessus
– Jessus, Commented Mar 24, 2017 at 10:20
What makes you think you need to switch to int32 (which doesn't exist)? — ForceBru
– ForceBru, Commented Mar 24, 2017 at 10:23
I don't know what a NZEC error is, but an input of / 1 0 will definitely throw an exception — UnholySheep
– UnholySheep, Commented Mar 24, 2017 at 10:24

ForceBru · Accepted Answer · 2017-03-24 16:34:45Z

If the result should be an integer, you should use operator.floordiv, i.e. a // b, not a / b ¹:

from operator import add, sub, mul, floordiv, mod

op = {'+': add, '-': sub, '*': mul, '/': floordiv, '%': mod}

for i in range(int(input())):
    znak, *nums = input().split()

    print(op[znak](*map(int, nums)))

By the way, the code above does exactly the same as yours, but it's around two times shorter!

How?

why check if znak is equal to something with tons of if/else statements, if you could put all the math operators in a dictionary op, and get the operator you need with op[znak]?
you're on Python 3, so you can use that nice a, *b = iterable syntax which extracts the first item of iterable into a and puts other items into a list b, which looks super nice
each op[znak] is a function that accepts two arguments, so you convert nums to integers with map(int, nums) and then pass them as separate arguments with the asterisk: *map(int, nums)
last, but not least, why use this C-style while loop index incrementing if there's the Pythonic way to do it - with range(start, stop, [step])?
finally, you don't really need the variable n, so you can plug it into range right away

Pssst, dude, feeling in need of some craziness? Take a look at how you can squeeze all this into two lines:

from operator import*

sum(0for _ in map(print,((lambda znak,*nums:{'+':add,'-':sub,'*':mul,'/':floordiv,'%':mod}[znak](*map(int,nums)))(*input().split())for _ in range(int(input())))))

Or only one line:

sum(0for _ in map(print,((lambda znak,*nums:{'+':lambda a,b:a+b,'-':lambda a,b:a-b,'*':lambda a,b:a*b,'/':lambda a,b:a//b,'%':lambda a,b:a%b}[znak](*map(int,nums)))(*input().split())for _ in range(int(input())))))

These work exactly as the first version. Now, this is just for fun, to show how powerful Python is and how messy it can be. Don't try this at home :D

¹ True division vs Floor division: 1 / 10 == 0.1, but 1 // 10 == 0.

anekix · Accepted Answer · 2017-03-24 11:14:10Z

0

A simple solution: The idea is to split input charter into tokens preserving their orders . in our case tokens are operators and numbers so it very simple using regular expressions. [+-/*] match any of those operators or any number of digits using \d+

n = input()
import re
while n>0:
    expression = input()
    operator,num1,num2 = re.findall('[+-/*]|\d+',expression)
    if operator == '+': print(int(num1) + int (num2))
    if operator == '-': print(int(num1) - int (num2))
    if operator == '*': print(int(num1) * int (num2))
    if operator == '%': print(int(num1) % int (num2))
    if operator == '/': print(int(num1) // int (num2))

    n = n - 1

edited Mar 24, 2017 at 11:14

answered Mar 24, 2017 at 10:43

anekix

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9 Comments

ForceBru Over a year ago

Still, an integer divided by an integer like a / b will be a floating point number.

anekix Over a year ago

@ForceBru fixed

ForceBru Over a year ago

no, it's not: if you divide a number by a floating-point, you can be even more sure you'll get a floating-point as a result.

ForceBru Over a year ago

well, the OP said this in the comments: "In direction it says that result should be in int32."

ForceBru Over a year ago

you're using Python 2.x, but the OP is on Python 3.x

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