【发布时间】：2013-09-09 19:54:07
【问题描述】：
我正在解决这个问题：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=286&page=show_problem&problem=3268
我被卡住了，找不到任何提示。
问题：
You will be given an integer n ( n<=10^9 ) now you have to tell how many
distinct sets of integers are there such that each number from 1 to n can
be generated uniquely from a set. Also sum of set should be n. eg for n=5 , one such set is:
{1,2,2} as
1 can be generated only by { 1 }
2 by { 2 }
3 by {1,2} ( note the two 2's are indistinguishable)
4 by {2,2}
5 by {1,2,2}
for generating a number each number of a set can be used only once. ie for above set
we can't do {1,1} to generate 2 as only one 1 is there.
Also the set {1,2,2} is equivalent to {2,1,2} ie sets are unordered.
我的做法：
The conclusion I came to was. Let F(S,k) denote number desired sets of sum S whose
largest element is k.Then to construct a valid set we can take two paths from this
state.Either to F(S+k,k) or to F(2*S+1,S+1).I keep a count of how many times I come
to state where S=n(the desired sum) and do not go further if S becomes > n.This is
clearly bruteforce which I just wrote to see if my logic was correct(which is correct)
.But this will give time limit exceed . How do I improve my approach??I have a
feeling it is done by dp/memoization.
【问题讨论】：
-
当您陈述问题时，答案总是“无限多”。问题是您忽略了源原始问题的条件之一：所有权重必须加起来为
n。如果您忽略该条件，您始终可以向您的集合添加大于n的权重 - 例如，对于n= 5，集合{1,2,2,6}、{1,2,2,7}、{1,2,2,8}等都可以作为好。一个始终有效的集合是采用元素1的n副本，因此我们总是有一个可以扩展的基本情况。 -
（请注意，这个问题实际上并不处理集合，而是处理传统上称为多重集或袋子的对象 - 集合传统上不允许单个元素的多个副本。）
标签： algorithm data-structures dynamic-programming counting