Question

问题-

https://www.spoj.com/problems/PALIN/

-输入测试用例数

-输入测试用例

-输出与输入的测试用例最接近的回文编号，不包括用例本身

我的尝试-

t=int(input())
while t>0:
    k=int(input())
    rev=0
    while k!=rev:
        k=k+1
        no=k
        rev=0
        while no>0:
            rev=(rev*10)+(no%10)
            no=no//10
        if k==rev:
            print(k)
    t=t-1

问题-

我正在“超出时间限制”，但是我无法思考或找到更快的程序。我发现的答案确实漫长而复杂。

如何修复我的代码？

Answer 1

通过将数字视为字符串，您不仅可以猜测每个可能的数字并检查它是否是回文，还可以节省大量时间：

def next_pal(x):
    s = str(x)
    if len(s) % 2:  # odd
        # take the first half (including the middle digit)
        first_half = s[:len(s)//2+1]
        # construct a number that's that half,
        # plus itself reversed (without the middle digit)
        candidate = int(first_half + first_half[-2::-1])
        # that number could be smaller (e.g. if we started with 245)
        if candidate > x:
            return candidate
        # let's try again: we increment the first half and do the same thing:
        new_first_half = str(int(first_half) + 1)
        # but be careful: we could be adding a digit here (e.g. 999 -> 100001)
        if len(new_first_half) == len(first_half):
            return int(new_first_half + new_first_half[-2::-1])
        # instead, in those cases, we just return the smallest palindrome of that length
        return 10**len(s) + 1
    else:  # even
        # similar dance for even
        first_half = s[:len(s)//2]
        candidate = int(first_half + first_half[::-1])
        if candidate > x:
            return candidate
        new_first_half = str(int(first_half) + 1)
        if len(new_first_half) == len(first_half):
            return int(new_first_half + new_first_half[::-1])
        return 10**len(s) + 1

请注意您有两个循环（我没有计算外部循环，因为没有办法解决），而我没有一个循环。

SPOJ-下一个回文

1 个答案: