把自然数N分解成若干个互不相同的正整数，使乘积最大；
由于这种分解的数目是有限的，所以最大积存在；
假设最大积的分解为
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}