USACO2.2.2--Subset Sums

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
题解：看完题目果断想到了搜索，因为这个题目本质上和Healthy Holsteins是一样的。都是求集合的子集问题。对于每个元素都有两种选择，即选或者不选，这样的话用搜索是非常优美的。所以果断写了个DFS。
这是代码：

#include<stdio.h>
long maxlen,count,total,n;
void dfs(long step)
{
  if(count<0) return;
  if(step>n)
  {

    if(count==0) total++;
    return;
}
    count-=step;
    dfs(step+1);
    count+=step;
    dfs(step+1);
}
int main(void)
{
    freopen("subset.in","r",stdin);
    freopen("subset.out","w",stdout);
    scanf("%ld",&n);
    total=0;
    if((n*(n+1)%4)!=0)
    {
        printf("0\n");
        return 0;
    }
    else
   {
        count=n*(n+1)/4;
        dfs(1);
   }
    printf("%ld\n",total/2);
    return 0;
}

不过在USACO上提交之后，第四个点N=31直接超时了。突然才想到N=39时，搜索量是多么的恐怖！！！2^39=549755813888。。。。然后继续想，因为对于每个数来说，就两种状态嘛，选或者不选，突然联想到0-1背包，因为它对于第K个物品，要么选，要么不选。只是0-1背包求的是最大价值方案。而此题是从N个元素中选取K个元素，使得背包装满的方案数。所以两者还是有点区别的，但思想是一样的。不过当时直接是套01背包的公式，所以果断WA了。。。需要对01背包稍微修改一下才行。。
状态转移方程为：
f[i][j]=f[i-1][j]+f[i-1][j-i],j>=i;
f[i][j]=f[i-1][j],j<i;

/*
ID:spcjv51
PROG:subset
LANG:C
*/
#include<stdio.h>
int main(void)
{
    freopen("subset.in","r",stdin);
    freopen("subset.out","w",stdout);
    int n,i,j,count;
    long long f[40][1000];
    scanf("%d",&n);
    if((n*(n+1)%4)!=0)
    {
        printf("0\n");
        return 0;
    }
    count=n*(n+1)/4;
    memset(f,0,sizeof(f));
    f[1][0]=1;
    f[1][1]=1;
    for(i=2; i<=n; i++)
        for(j=count; j>=0; j--)
        if(j>=i)
           f[i][j]=f[i-1][j]+f[i-1][j-i];
        else
          f[i][j]=f[i-1][j];
    printf("%lld\n",f[n][count]/2);
    return 0;
}