MUL in python

[revealer]

n = input()
arr = []
for a in range(n):
	st = raw_input()
	e = st.find(' ')
	n1 = int(st[0:e])
	n2 = int(st[e:])
	mul = n1 * n2
	arr.append(mul)
for b in range(n):
	print arr[b]

I/O is correct, but it has exceeded the given time...
any hints?

[dhoni]

This problem is probably not solvable in python, because of the ultra-tight limits. Try faster languages like C/C++.

[izekia]

import sys
data = [int(x) for x in sys.stdin.read().split()]
for i in xrange(1, len(data) - 40, 2):
    sys.stdout.write("%d\n" % (data[i] * data[i+1]))

try this, after replace 40 with 38 and try again
you can see there that the problem appear at the 20th production before the end. So I haven't any idea how to make this with a standart multiplication
PS: with the psyco I've got the same result.

[izekia]

sorry
I was wrong
this code ran for 1.36

import sys
data = [int(x) for x in sys.stdin.read().split()]
for i in xrange(1, len(data), 2):
    a= data[i] * data[i+1]

this is the conversion problem

[peristarkawan]

The bottleneck is definitely the conversion from long back to str. As can be seen below, the time to format a 20,000 digit number is more than 15 times the combined time to parse and multiply two 10,000 digit numbers!

>>> from timeit import *
>>> a = Timer("x = int(y)", "y = str(10 ** 10000)")
>>> a.timeit(1000)
4.8757819778770681
>>> b = Timer("x = y * y", "y = 10 ** 10000")
>>> b.timeit(1000)
2.4265480668632335
>>> c = Timer("x = str(y)", "y = 10 ** 20000")
>>> c.timeit(1000)
191.91779877051408

Since the formatting code is already in C, it's not surprising that psyco wouldn't make much difference. I agree with dhoni -- this one just isn't currently possible in Python.

[ShiningRay]

is this problem possible in python?