問(wèn)題

和為n 連續(xù)正數(shù)序列

問(wèn)題分析

此問(wèn)題為等差序列求和的逆計(jì)算
等差學(xué)列求和公式為:S = na + dn(n-1)/2 （n為序列個(gè)數(shù),a為序列最小值,d為等差值，此處d=1）
轉(zhuǎn)化后可得 a = (2S-n(n-1))/2n
即可得條件((2S-n(n-1))/2n) % 1 == 0 && (2S-n(n-1))/2n > 0 && n > 1 && a > 0

算法實(shí)現(xiàn)

def getContinuousSequence(number=3):
    if number <1:
        raise Exception, "參數(shù)錯(cuò)誤"

    data = []
    for n in range(2, number):

        a = (number-(n*(n-1)/2))*1.00/n
        if a % 1 == 0 and a > 0:
            a = int(a)
            data.append(range(a, a+n))
    return data


if __name__ == '__main__':
    print getContinuousSequence(13)