[Usaco2012 Open]Balanced Cow Subsets

Description
Farmer John's owns N cows (2 <= N <= 20), where cow i produces M(i) units of milk each day (1 <= M(i) <= 100,000,000). FJ wants to streamline the process of milking his cows every day, so he installs a brand new milking machine in his barn. Unfortunately, the machine turns out to be far too sensitive: it only works properly if the cows on the left side of the barn have the exact same total milk output as the cows on the right side of the barn! Let us call a subset of cows "balanced" if it can be partitioned into two groups having equal milk output. Since only a balanced subset of cows can make the milking machine work, FJ wonders how many subsets of his N cows are balanced. Please help him compute this quantity.
给出N（1≤N≤20）个数M(i) (1 <= M(i) <= 100,000,000)，在其中选若干个数，如果这几个数可以分成两个和相等的集合，那么方案数加1。问总方案数。

Input
Line 1: The integer N.
Lines 2..1+N: Line i+1 contains M(i).

Output
Line 1: The number of balanced subsets of cows.

Sample Input
4 1 2 3 4

Sample Output
3

---

直接搜复杂度\(O(3^n)\)，显然不行，考虑折半搜索，分成两部分，这样复杂度变为\(O(2*3^{n/2})\)，然后对两部分进行查找即可，细节见代码

/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
inline int frd(){
	int x=0,f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<3)+(x<<1)+ch-'0';
	return x*f;
}
inline int read(){
	int x=0,f=1; char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<3)+(x<<1)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)	putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
const int N=20,M=6e4;
struct S1{
	int val,sta;
	void insert(int v,int s){val=v,sta=s;}
}A[M+10],B[M+10];
int v[N+10],cntA,cntB,n;
bool vis[(1<<N)+10];
bool cmp1(const S1 &x,const S1 &y){return x.val<y.val;}
bool cmp2(const S1 &x,const S1 &y){return x.val>y.val;}
void dfs(int x,int limit,int sta,int sum){
	if (x>limit){
		limit==n>>1?A[++cntA].insert(sum,sta):B[++cntB].insert(sum,sta);
		return;
	}
	dfs(x+1,limit,sta,sum);
	dfs(x+1,limit,sta|(1<<(x-1)),sum+v[x]);
	dfs(x+1,limit,sta|(1<<(x-1)),sum-v[x]);
}
int main(){
	n=read();
	for (int i=1;i<=n;i++)	v[i]=read();
	dfs(1,n>>1,0,0),dfs((n>>1)+1,n,0,0);
	sort(A+1,A+1+cntA,cmp1);
	sort(B+1,B+1+cntB,cmp2);
	int i=1,j=1,Ans=0;
	while (i<=cntA&&j<=cntB){
		while (j<=cntB&&-B[j].val<A[i].val)	j++;
		int tmp=j;
		while (A[i].val+B[j].val==0){
			if (!vis[A[i].sta|B[j].sta])	vis[A[i].sta|B[j].sta]=1,Ans++;
			j++;
		}
		j=tmp,i++;
	}
	printf("%d\n",Ans-1);
}