题目链接

https://www.lydsy.com/JudgeOnline/problem.php?id=2679

题解

meet in the middle，枚举左边放在第一个集合，第二个集合还是不放，记录左边能得到的差值和左边的选取状态，右边同理，最后two-pointer扫一下，记录选取状态有没有被考虑过，时间复杂度O(2×3n/2)O(2\times 3^{n/2})O(2×3n/2)。

代码

#include <cstdio>
#include <algorithm>int read()
{int x=0,f=1;char ch=getchar();while((ch<'0')||(ch>'9')){if(ch=='-'){f=-f;}ch=getchar();}while((ch>='0')&&(ch<='9')){x=x*10+ch-'0';ch=getchar();}return x*f;
}const int maxn=20;
const int maxk=59049;
const int maxs=1<<maxn;struct data
{int v,con;bool operator <(const data &other) const{if(v==other.v){return con<other.con;}return v<other.v;}
};data tmp[maxk+10],f[maxk+10],g[maxk+10];
int tot,totf,totg,op[maxn+2],n,v[maxn+2],vis[maxs+10],ans;int search(int x,int mx)
{if(x>mx){int last=0,con=0;for(int i=1; i<=n; ++i){last+=v[i]*op[i];if(op[i]){con|=1<<(i-1);}}tmp[++tot].v=last;tmp[tot].con=con;return 0;}for(int i=-1; i<=1; ++i){op[x]=i;search(x+1,mx);}return 0;
}int main()
{n=read();for(int i=1; i<=n; ++i){v[i]=read();}if(n==1){puts("1");return 0;}search(1,n/2);std::sort(tmp+1,tmp+tot+1);for(int i=1; i<=tot; ++i){f[i]=tmp[i];}totf=tot;tot=0;for(int i=1; i<=n/2; ++i){op[i]=0;}search(n/2+1,n);std::sort(tmp+1,tmp+tot+1);for(int i=1; i<=tot; ++i){g[i]=tmp[i];}totg=tot;int nowf=1,nowg=totg;while((nowf<=totf)&&(nowg>0)){while((nowg>0)&&(f[nowf].v+g[nowg].v>0)){--nowg;}if(f[nowf].v+g[nowg].v==0){int pastf=nowf,pastg=nowg;while((nowf<totf)&&(f[nowf].v==f[nowf+1].v)){++nowf;}while((nowg>1)&&(g[nowg].v==g[nowg-1].v)){--nowg;}for(int i=pastf; i<=nowf; ++i){for(int j=nowg; j<=pastg; ++j){if(!vis[f[i].con|g[j].con]){vis[f[i].con|g[j].con]=1;++ans;}}}}++nowf;}printf("%d\n",ans-1);return 0;
}