与运算 递归

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>

#define ll long long
#define il inline
#define db double

using namespace std;

il int gi()
{
	int x=0,y=1;
	char ch=getchar();
	while(ch<‘0‘||ch>‘9‘)
		{
			if(ch==‘-‘)
				y=-1;
			ch=getchar();
		}
	while(ch>=‘0‘&&ch<=‘9‘)
		{
			x=x*10+ch-‘0‘;
			ch=getchar();
		}
	return x*y;
}

int point[100045];

int stackk[100045];

int stack[100045];

int ci[45];

int num[31][100005];

int s[31];

int er[100005][31];

int check(int now,int tot)//now是当前2的几次方，tot是有多少数满足
{
	//printf("now=%d tot=%d\n",now,tot);
	if(now==0)
		return 1;
	for(int next=1;now-next>=0;next++)
		{
			int cnt=0;
			for(int i=1;i<=tot;i++)
				if(er[stack[i]][now-next])
					stackk[++cnt]=stack[i];
			if(cnt>=2)
				{
					for(int i=1;i<=cnt;i++)
						stack[i]=stackk[i];
					return ci[now]+check(now-next,cnt);
				}
		}
	return ci[now];
}

int main()
{
	freopen("and.in","r",stdin);
	freopen("and.out","w",stdout);

	int n=gi();

	ci[0]=1;
	for(int i=1;i<=30;i++)
		ci[i]=ci[i-1]*2;

	int x;
	bool flag=0;
	for(int i=1;i<=n;i++)
		{
			x=gi();
			if(x!=0)
				flag=1;
			point[i]=x;
			for(int j=30;j>=0;j--)
				if(x>=ci[j])
					{
						x-=ci[j];
						er[i][j]=1;
						num[j][++s[j]]=i;
					}
		}
   
	int now=30;
	while(s[now]<2&&now>=0)
		{
			now--;
		}

	for(int i=1;i<=s[now];i++)
		stack[i]=num[now][i];

	if(flag)
		printf("%d\n",check(now,s[now]));
	else
		printf("0\n");

	return 0;
}