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020         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-124         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-145     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 }