分类：DP，递推，记忆化搜索
作者：ACShiryu
时间：2011-7-15
Subset Sums
JRM
For many sets of consecutive integers from 1 through N (1 <= N <= 39), one can partition the set into two sets whose sums are identical.
For example, if N=3, one can partition the set {1, 2, 3} in one way so that the sums of both subsets are identical:
- {3} and {1,2}
This counts as a single partitioning (i.e., reversing the order counts as the same partitioning and thus does not increase the count of partitions).
If N=7, there are four ways to partition the set {1, 2, 3, ... 7} so that each partition has the same sum:
- {1,6,7} and {2,3,4,5}
- {2,5,7} and {1,3,4,6}
- {3,4,7} and {1,2,5,6}
- {1,2,4,7} and {3,5,6}
Given N, your program should print the number of ways a set containing the integers from 1 through N can be partitioned into two sets whose sums are identical. Print 0 if there are no such ways.
Your program must calculate the answer, not look it up from a table.
PROGRAM NAME: subset
INPUT FORMAT
The input file contains a single line with a single integer representing N, as above.
SAMPLE INPUT (file subset.in)
7
OUTPUT FORMAT
The output file contains a single line with a single integer that tells how many same-sum partitions can be made from the set {1, 2, ..., N}. The output file should contain 0 if there are no ways to make a same-sum partition.
SAMPLE OUTPUT (file subset.out)
4
题目大意就是求讲数据1…N分成两组，使得两组中元素的和加起来相等，求这样分组的情况数
可以利用递推的方法求该题的解，注意：f(k,a)=(f(k-1,a+k)+f(k-1,a-k))/2；其中f(k,a)表示讲1…k元素分两组，使第一组比第二组多a；
因为k只可能分到第一组或第二组这两种情况，如果k加到第一组后使得第一组比第二组多a，则要原来的分组中第一组比第二组多a-k
同理如果k加到第二组后使得第一组比第二组多a，则要原来的分组中第一组比第二组多a+k。
因为交换两组元素后也满足条件，而只算一个解，故后面要除2；
很显然题目要求的是f（N，0）；
为节省递推时间，可使用记忆化搜索，考虑到数据不大，又a有正负之分，可加数组适当开大。
数据分析：N最大为39，32位整数可能存不下，故要使用64位扩展，故要将数据声明为long long .因为使用的是记忆化搜索，很大程度上减少了重复搜索的情况，时间复杂度为O（n^3），远优于O（2^n），不会超时.。
参考代码：
1 /*
2 ID:shiryuw1
3 PROG:subset
4 LANG:G++ //64位整数要用long long
5 */
6 #include<iostream>
7 #include<cstdlib>
8 #include<cstdio>
9 #include<cstring>
10 #include<algorithm>
11 #include<cmath>
12 using namespace std;
13 int sum=0;
14 int n;
15 long long fid[40][2000]; //是否寻找过dp(k,a+1000),当为-1时为否
16 long long dp(int k,int a)
17 {
18 if(k==0)
19 { //当为0时不合要求，故返回0
20 return 0;
21 }
22 if(k==1)
23 { //当为1是可知只有可能两边的集合相差只能为1或-1
24 if(abs(a)==1)
25 return 1; 当相差为1或-1时，则有一种情况符合
26 return 0;
27 }
28 if(k==2)
29 { //理由同k=1时
30 if(abs(a)==1||abs(a)==3)
31 return 1;
32 return 0;
33 }
34 if(fid[k][a+1000]==-1) //如果k,a没有访问过，则访问，并将访问结果存在该数组中，很明显该数组的结果不会再是-1否则直接返回fid[k][a+1000]，避免重复搜索。
35 fid[k][a+1000]=dp(k-1,a+k)+dp(k-1,a-k);
36 return fid[k][a+1000];
37 }
38
39 int main()
40 {
41 freopen("subset.in","r",stdin);
42 freopen("subset.out","w",stdout);
43 cin>>n;
44 memset(fid,-1,sizeof(fid)); //初始化fid数组为-1
45 printf("%lld\n",dp(n,0)/2); 寻找把1……N分成两组后使两组相差0的情况数
46 return 0;
47 }