题目125：找出是回文的平方数之和

欧拉计划

Palindromic sums

The palindromic number 595 is interesting because it can be written as the sum of consecutive squares: 6² + 7² + 8² + 9² + 10² + 11² + 12².

There are exactly eleven palindromes below one-thousand that can be written as consecutive square sums, and the sum of these palindromes is 4164. Note that 1 = 0² + 1² has not been included as this problem is concerned with the squares of positive integers.

Find the sum of all the numbers less than 10⁸ that are both palindromic and can be written as the sum of consecutive squares.

题目：

595 这个回文数很有趣，它可以被写作连续数字的平方数之和：6² + 7² + 8² + 9² + 10² + 11² + 12²。

1000 以下一共有 11 个这样可以写作连续数字平方和的回文数，这 11 个数字的和是 4164。注意 1 = 0² + 1² 没有被算作在内，因为该题目只考虑正整数的平方和。

求 10⁸ 以下所有  1). 既是回文； 2). 又可以写成连续数字平方和，的数字之和。

迷雾少年

这个比前面的简单

jerryxjr1220

迷雾少年 发表于 2016-8-31 17:36
这个比前面的简单

确实比较简单

#coding:utf-8
def is_cycle(n):
    return True if str(n) == str(n)[::-1] else False

limit = 10**8
window_s = 1
window_e = int(limit ** 0.5)
lst = []
while window_s<window_e:
    s = window_s*window_s
    for i in range(window_s+1,window_e):
        s += i*i
        if s >= limit:
            break
        if is_cycle(s):
            lst.append(s)
    window_s += 1
print (sum(set(lst)))

输出：
2906969179