Saturday, February 20, 2016

连续自然数之和的数 -Beauty of Programming


比咕网 | 关注搜索和自然语言处理技术
知道:
1+2=3;
4+5=9;
2+3+4=9;
等式的左边都是两个或两个以上连续的自然数相加,那么是不是所有的整数都可以写成这样的形式呢?稍微考虑一下,我们发现,4、8等数并不能写成这样的形式。
问题1:写一个程序,对于一个64位正整数,输出它所有可能的连续自然数(两个及以上)之和的算式。
问题2:大家在测试上面程序的过程中,肯定会注意到有一些数字不能表达为一系列连续的自然之和,例如32好像就找不到,那么,这样的数字有什么规律呢?能否证明你的结论?
问题3:在64位正整数范围内,子序列数目最多的数是哪一个?这个问题要用程序蛮力搜索,恐怕要运行很长的时间,能否用数学知识推导出来?
就刚刚那个笨方法,我们明显觉得,很多数试探到a/2的数就差不多了,白费力计算不少不必要的数字,不妨来用数学公式推导一下,看能不能优化一下计算量
http://blog.csdn.net/yutianzuijin/article/details/10300067
将一个正整数表示成连续自然数之和,即N=s+(s+1)+(s+2)+…+(e-1)+e。利用等差数列求和公式,我们有



也即2N=(s+e)(e-s+1),该等式表示可以将2N分解成两个正整数的乘积。我们设x=s+e, y=e-s+1(其中x>y)。利用x、y我们可以求解获得se


因为se都是整数,因而为了使上面的式子能整除,则xy必须一奇一偶。因为2N=xy2N含有偶因子,所以N必须含有奇因子才能使等式成立。这也就证明了N能表示成连续自然数的充要条件:N必须含有奇数因子,问题二得证。

利用公式(2),我们即可获得正整数N的一个连续自然数序列。为了输出所有的连续自然数序列,我们需要获得所有的xy组合,也即求2N的所有因子组合。我们可以利用算法基本定理将2N分解成有限个质数的乘积:
我们将2N按照质因子进行分解,由于xy只有一个偶数,所以2N的分解式中质因子2的所有组合都只能在其中一个数中,否则xy都是偶数。不妨假设x是偶数,则
其中。由此我们可以知道,2N的所有质因子组共有(j+1) (k+1)…组。由于当x等于2N时,y等于1,此时的连续自然数个数为1,小于2,所以满足条件的连续自然数序列个数为:(j+1) (k+1)…-1,这就回答了问题一,我们可以首先获得2N的所有质因子分解,然后组合质因子并满足x>y即可获得所有的序列。

问题三按照网上的资料,有两种理解方式:1)、可以表示为最多个序列的那个数;2)、序列中项最多的那个数。但是按照题意其实应该是第一种理解方式。由于质因子2只能在xy的某一个数中,对质因子分解不起作用,所以为了让序列个数尽可能多,N的质因子中不能包含2。因此若让子序列个数最多,则即(j+1) (k+1)…最大。通过简单分析我们可以获得具有最多序列个数的正整数j、k等质因子指数的两个性质:
性质一可以通过替换来证明,假设当前的最大序列个数不满足性质一,设第一个不满足性质的两个指数为a和b,满足a<b。由于序列个数为(j+1) ∙ (k+1) ∙…,我们通过将两个质因子的指数交换可以获得相同的序列个数。此外,由于前面的质因子小,交换之后总的乘积会变小,从而可能产生一个更大的指数,使序列个数更多。通过这样依次交换不满足性质的相邻两个指数,最终我们会得到满足性质一并且使序列个数最多的指数序列。这也与经验相吻合,要想使序列个数尽可能多,则正整数必须尽量少含有大的质因子。对于性质二,当正整数只包含质因子3时,质因子之和最大,此时为40。而当正整数含有越多较大的质因子时,则质因子之和就越小,在满足性质一的前提下,当正整数等于3 ∙5 ∙…∙53时,和最小,此时共有15个质因子,含有的序列个数为2^15-1。该正整数含有的序列个数已经非常多,但不一定是最多,例如将53替换为27,则序列个数升为5∙2^13-1。如果我们继续替换,将3^4 ∙5 ∙…∙47变为3^3 ∙5^2 ∙…∙47,则序列个数又升为6∙2^13-1。该数可能已经是具有最多序列的数,但是要想获得准确的最大序列个数,我们还需要利用性质一二去遍历所有可能的情况。如果按照第二种理解方式,则我们可以证明具有最多项的数N的最多项必从1开始。假设最多项不从1开始而是从s(s>1)开始,则我们可以将该最多项的每一项均减去s-1,则最多项等价于从1开始的最多项,也即我们至少可以构造和之前的最多项一样长的新序列。此外,由于每一项均减少了s-1,如果此时减少的和大于最多项的最大项,我们还可以添加新的项到最多项的末尾,产生更长的最多项。


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