获得总和为n的方法数，且所有小于n的正整数均参与运算。

Question

获得总和为n的方法数，且所有小于n的正整数均参与运算。

7

对于给定的数字n，例如2，使用小于2的数字有多少种方法可以得到总和为2。

1+1 = 2  
so, for 2 - just 1 way.

n = 3   
1+1+1=3  
1+2=3  
so,for 3 - it is 2 ways  
n = 4   
1+1+1+1=4  
1+1+2=4  
1+3=4  
2+2=4  

so, for 4 - it is 4 ways

这个问题是否存在一种通用的模式/解决方案？

- Roger

1个回答

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可以查看英文原文，
原文链接

- Saeed Amiri · Accepted Answer

这个问题被称为划分问题，详见维基百科引用链接中的说明：

One way of getting a handle on the partition function involves an intermediate function p(k, n), which represents the number of partitions of n using only natural numbers at least as large as k. For any given value of k, partitions counted by p(k, n) fit into exactly one of the following categories:
smallest addend is k
smallest addend is strictly greater than k.
The number of partitions meeting the first condition is p(k, n − k). To see this, imagine a list of all the partitions of the number n − k into numbers of size at least k, then imagine appending "+ k" to each partition in the list. Now what is it a list of? As a side note, one can use this to define a sort of recursion relation for the partition function in term of the intermediate function, namely
1+ sum{k=1 to floor (1/2)n} p(k,n-k) = p(n),
The number of partitions meeting the second condition is p(k + 1, n) since a partition into parts of at least k that contains no parts of exactly k must have all parts at least k + 1.

Since the two conditions are mutually exclusive, the number of partitions meeting either condition is p(k + 1, n) + p(k, n − k). The recursively defined function is thus:
p(k, n) = 0 if k > n

p(k, n) = 1 if k = n

p(k, n) = p(k+1, n) + p(k, n − k) otherwise.

实际上，您可以通过记忆化计算所有值，以防止额外的递归调用。

编辑：正如unutbu在他的评论中提到的那样，在计算结束时，您应该减去1以输出结果。也就是说，在最后一步之前，应按照维基百科的建议计算所有P值，但在输出结果之前，您应将其减去1。