我一生都无法在 Hackerrank 上解决这个挑战。我得到的最接近的是 4/6 次传球。规则:在公历中,必须考虑三个标准来确定闰年:
The year can be evenly divided by 4, is a leap year, unless:
The year can be evenly divided by 100, it is NOT a leap year, unless:
The year is also evenly divisible by 400. Then it is a leap year.
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代码:
The year can be evenly divided by 4, is a leap year, unless:
The year can be evenly divided by 100, it is NOT a leap year, unless:
The year is also …
Run Code Online (Sandbox Code Playgroud) 我正在尝试为每个用户制作个人资料页面.我添加了一个代码,用于检查用户是否已登录并执行重定向(请参阅下面代码的第12行).
from django.shortcuts import render
from django.contrib.auth import login, authenticate
from django.contrib.auth.forms import UserCreationForm
from django.http import HttpResponseRedirect, HttpResponse
from .models import Account, ForSale, WTB
from mysite.forms import MyRegistrationForm
def signup(request):
if request.user.is_authenticated():
return HttpResponseRedirect('/user/')
else:
if request.method == 'POST':
form = MyRegistrationForm(request.POST)
if form.is_valid():
form.save()
return HttpResponseRedirect('/user/')
context = {}
context.update(csrf(request))
context['form'] = MyRegistrationForm()
return render(request, 'signup.html', context)
def index(request):
return render(request, 'index.html')
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但是,在访问/注册/网站时,我收到以下调试消息:
TypeError at /signup/
'bool' object is not callable
Request Method: GET
Request URL: http://url:8000/signup/
Django …
Run Code Online (Sandbox Code Playgroud) 我有一些简单的斐波那契序列函数,我正在练习单元测试并使用Travis-CI/Docker构建:
fib_recursive.py:
from fib.fib import benchmark, fib_rec_memo
@benchmark
def print_fib(n):
for x in range(0, n):
print(fib_rec_memo(x))
print_fib(100)
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这是fib.fib导入源代码:
from time import time
from functools import wraps
def benchmark(func):
@wraps(func)
def wrapper(*args, **kwargs):
t = time()
func(*args, **kwargs)
print(func.__name__, 'took:', time() - t)
return func(*args, **kwargs)
return wrapper
def fib_rec_memo(n, hash = {0:1, 1:1}):
if n not in hash:
hash[n] = fib_rec_memo(n-1) + fib_rec_memo(n-2)
return hash[n]
@benchmark
def fib_standard(num):
a, b = 0, 1
c = []
while a < num: …
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