Python：Project Euler 145

Question

我今天的问题是，我是否沿着正确的欧拉145道路行驶，并且是否有点效率。我的大部分精力都在下降，只有我的一个Defs给我带来了麻烦：int（str（numb）[：i]）％2 == 0进行均匀检查。我的代码如下。第10行是问题点def reversed(reg): # to flip the number around    fliped = str(reg)[::-1];    return(int(fliped)); # Return it as a int. def allEvenDigits(numb): # This is the issue one    hasEvenNumb = False;    for i in range(0, len(str(numb))):        if int(str(numb)[:i])%2 == 0:  # if int of the string numb's char at i is even            hasEvenNumb = True; ## return that it is true            break; # why go on if we found a even.     return(hasEvenNumb);for i in range(1, 1000): # its 1000 to save a few minutes    revNumb = reversed(i);    total = revNumb+i;    if(allEvenDigits(total)):        print(i, "+" , revNumb, "=",Total);

Answer 1

def sumrevers(x):

    summation = x + int(str(x)[::-1])

    if summation % 2 != 0: return summation

def checknum(x):

    if not (str(x)[-1] == "0") or (str(x)[0] == "0"):

        if type(sumrevers(x)) == int:

            num = str(sumrevers(x))

            checklis = [k for k in str(num)]

            if all(int(i) % 2 != 0 for i in checklis): return True

cnt = 0

for i in xrange(1, 1000000001):

    if checknum(i):

        cnt += 1

print cnt