Codewars算法題（8）

第26題：

問題：

問題描述：

給定一個列表，和一個整數s，找出列表中兩數之和爲s,的兩個數，並返回，若有若干對這樣的數，那麼選擇相距較短的，在相距較短的中再找出index小的。

評分較高答案：

def sum_pairs(lst, s):
    cache = set()
    for i in lst:
        if s - i in cache:
            return [s - i, i]
        cache.add(i)

該方法用減法推倒~~~~，大神就是大神啊

第27題：

問題：

Divisors of 42 are : 1, 2, 3, 6, 7, 14, 21, 42. These divisors squared are: 1, 4, 9, 36, 49, 196, 441, 1764. The sum of the squared divisors is 2500 which is 50 * 50, a square!

Given two integers m, n (1 <= m <= n) we want to find all integers between m and n whose sum of squared divisors is itself a square. 42 is such a number.

The result will be an array of arrays(in C an array of Pair), each subarray having two elements, first the number whose squared divisors is a square and then the sum of the squared divisors.

#Examples:

list_squared(1, 250) --> [[1, 1], [42, 2500], [246, 84100]]
list_squared(42, 250) --> [[42, 2500], [246, 84100]]

問題描述：

 給定一個區間[m, n] 求這個區間內符合條件的 數對， [i, f[i]] 即 f[i] 爲 i的所有因子的平方和且f[i]爲平方數

評分較高代碼：

CACHE={}
def squared_cache(number):
    if number not in CACHE:
        divisors = [x for x in range(1, number + 1) if number % x == 0]
        CACHE[number] = sum([x * x for x in divisors])
    return CACHE[number]

def list_squared(m, n):
    ret = []
    for number in range(m, n + 1):
        divisors_sum = squared_cache(number)
        if (divisors_sum ** 0.5).is_integer():
            ret.append([number, divisors_sum])
    return ret

其中：squared_cache函數是求整數的因子的平方和

自己代碼：（有誤）

def factors(p):
    divisors = []
    for k in range(1, p + 1):
        if p % k == 0:
            divisors.append(k)
    return divisors
def list_squared(m, n+1):
    output = []
    sum = 0
    for num in (m, n):
        for i in factors(num):
            sum += i*i
            if (sum ** 0.5).is_integer()

                return output.append([num, sum])

現在我的代碼是錯誤的，但是我仍找不出原因。函數factors（）是求一個數的所有因子，並以列表形式返回。希望看到帖子的大神幫幫忙~~~，菜鳥兒在此感謝了！

201707281111持續更新~~~~