pku1032關于數論的一個結論

把自然數N分解成若干個互不相同的正整數，使乘積最大；

由于這種分解的數目是有限的，所以最大積存在；

假設最大積的分解為

n=a1+a2+a3+...+a[t-2]+a[t-1]+a[t]

(a1<a2<a3<...<a[t-2]<a[t-1]<a[t])

Now, from the five rules above, we could make the mutiple maximum.

to an N, find the integer k, fits

A=2+3+4+...+(k-1)+k <= N < A+(k+1)=B

Suppose N = A + p, (0 <= p < k+1)

1) p=0, then answer is Set A

2) 1<=p<=k-1 then answer is Set B - { k+1-p }

3) p=k, then answer is Set A - {2} + {k+2}