我 580085 ,为什么会出错?还有另一种方法来检查数字是否是回文?
def reverse(str, aux=''):
count = len(str)
while count > 0:
aux += str[count-1]
count -= 1
return aux
def palindrome(num):
j = str(num)
if j == reverse(j):
return 1
else:
return 0
i = 999
found = 0
while i > 99 and not found:
j = 999
while j > 99 and not found:
if palindrome(i * j):
found = 1
print(i * j)
else:
j -= 1
if not found:
i -= 1
我的帖子主要是代码,但我没有别的话要说。
答案 0 :(得分:1)
花了一些时间研究这个问题。我尝试采用反向方法(从回文数字开始,找到它们的分隔符(如果有的话))。在 Win10 上使用 Py354 。
code.py :
import time
from math import sqrt
def reverse(str, aux=''):
count = len(str)
while count > 0:
aux += str[count-1]
count -= 1
return aux
def palindrome(num):
j = str(num)
if j == reverse(j):
return 1
else:
return 0
def question_makhfi_0():
ret = 0
i = 999
found = 0
while i > 99 and not found:
j = 999
while j > 99 and not found:
if palindrome(i * j):
found = 1
ret = i * j
else:
j -= 1
if not found:
i -= 1
return ret, i, j
def answer_makhfi_0():
i = 999
_max = 0
while i > 99:
j = 999
while j > 99:
num = i * j
if palindrome(num):
if num > _max:
_max = num
factor0, factor1 = i, j
j -= 1
i -= 1
return _max, factor0, factor1
"""
def answer_makhfi__0_improved_0():
i = j = 999
prod = i * j
step = 0
while prod > 100000:
if step % 2:
i -= 1
else:
j -= 1
prod = i * j
prod_str = str(prod)
if prod_str == prod_str[::-1]:
return prod
step += 1
"""
def answer_cfati_0():
pal = 999999
while pal >= 900009:
if pal % 10 == 9:
pal_str = str(pal)
if pal_str == pal_str[::-1]:
pal_sqrt = sqrt(pal)
for factor0 in range(101, int(pal_sqrt) + 1):
if pal % factor0 == 0:
factor1 = int(pal / factor0)
if 100 <= factor1 <= 999:
return pal, factor0, factor1
pal -= 10
#return pal
def time_func(f, *args):
t0 = time.time()
res = f(*args)
t1 = time.time()
return t1 - t0, res
if __name__ == "__main__":
for func in [question_makhfi_0, answer_makhfi_0, answer_cfati_0]:
print("\nStarting: {}".format(func.__name__))
#print(func.__name__)
res = time_func(func)
print(" Time: {:.6f}\n Result: {}".format(*res))
备注强>:
answer_makhfi_0
:
max
替换为_max
以避免隐藏内置名称return
语句(效率极低,无论如何)answer_cfati_0
:
[900009, 999999]
范围内的回文数,可以表示为2(3位)数字乘积。假设(测试支持这一点)是公平的。但是,如果针对防弹代码,这个表单不会这样做,我将不得不稍微调整一下(这将是性能损失)<强>输出强>:
"e\Work\Dev\StackOverflow\q47999634>"e:\Work\Dev\VEnvs\py35x64_test\Scripts\python.exe" code.py Starting: question_makhfi_0 Time: 0.008551 Result: (580085, 995, 583) Starting: answer_makhfi_0 Time: 1.457818 Result: (906609, 993, 913) Starting: answer_cfati_0 Time: 0.012599 Result: (906609, 913, 993)
根据输出,原始和新代码之间存在差异(对于一次运行,时间会有所不同):
的 @ EDIT0 强> :
答案 1 :(得分:0)
新代码:
def reverse(str, aux=''):
count = len(str)
while count > 0:
aux += str[count-1]
count -= 1
return aux
def palindrome(num):
j = str(num)
if j == reverse(j):
return 1
else:
return 0
i = 999
max = 0
while i > 99:
j = 999
while j > 99:
num = i * j
if palindrome(num):
if num > max:
max = num
print(max)
j -= 1
i -= 1