algorithm - 找到没有。获得所有小于 n 的正整数的和 n 的方法

Question

对于给定的数字 n，说 2 我们可以用小于 2 的数字得到和 2 有多少种方法。

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  

这个问题可以有一个通用的模式/解决方案吗？

Answer 1

此问题称为 Partition Problem ，请参阅来自 wiki 的引用链接中的详细信息:

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.

事实上，您可以通过 memoization 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，在计算结束时，您应该减去 1 以输出结果。 IE。到最后一步的所有 P 值都应该按照 wiki 的建议进行计算，但是在输出结果之前的最后，你应该将它减去 1 .