Question

我试图使用Python 2.7.9作为我的编码语言来解决SPOJ中的TWOSQRS问题，并设计了一个程序来实现这一点。由于代码运行完美而没有在我的系统上使用给定的测试用例输入引发任何异常。如果有人可以提供更多测试用例或在我的代码中找到错误，那将会很有帮助。

问题链接：http://www.spoj.com/problems/TWOSQRS/

    # -*- coding: utf-8 -*-
'''To solve the problem 2 primary condition that should be met are:
    1. No should not be of the form 4k+3 as for sum of square of two nos will always be of form 4k or 4k+1
    2. All the prime factors of form 4k+3 should have even power from the Fermat thorem.
   Steps involved in solving the problem are:
    1. Sieve a list of prime nos. upto 1000001 as in problem. 
    2. Check if all the prime factors has even powers.
    3. Check if the no is not of form 4k+3'''

import numpy

def sieve(n):
    """ An implementation that sieves separately 
    for primes of the form 6i−1 and 6i+1, due to Robert William Hanks"""

    prime = numpy.ones(n//3 + (n%6==2), dtype=numpy.bool)
    for i in range(3, int(n**.5) + 1, 3):
        if prime[i // 3]:
            p = (i + 1) | 1
            prime[       p*p//3     ::2*p] = False
            prime[p*(p-2*(i&1)+4)//3::2*p] = False
    result = (3 * prime.nonzero()[0] + 1) | 1
    result[0] = 3
    return numpy.r_[2,result]

primes=sieve(10**6+1)            #List of all the prime upto 10**6

def main():
    noOfCase=input()
    for i in range(noOfCase):
        N=input()
        is_multiple= True
        i = 0
        while(primes[i]*primes[i] <= N):
       count = 0;
       while (N % primes[i] == 0):
           count+=1;
           N/= primes[i];
           if (primes[i]%4 == 3 and count%2 == 1):
               is_multiple = False;
               break;
           i+=1

    if (N%4 == 3):
       is_multiple = False
    if(is_multiple):
       print "Yes"
    else:
       print "No"


main()

Answer 1

您的问题几乎可以肯定，您使用的是NumPy，但选择的是不支持NumPy的Python群集/版本。

正如PYTHON: What you should know所解释的那样，Pyramid集群没有任何Python版本的NumPy，Cube集群适用于所有CPython版本但没有PyPy版本。

使用Python 2.7.9的TWOSQRS SPOJ给出了运行时错误（NZEC）

1 个答案: