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  • BZOJ4762 最小集合(动态规划+容斥原理)

      https://www.cnblogs.com/AwD-/p/6600650.html

    #include<iostream> 
    #include<cstdio>
    #include<cmath>
    #include<cstdlib>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    #define ll long long
    #define P 1000000007
    #define N 1024
    char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
    int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
    int read()
    {
    	int x=0,f=1;char c=getchar();
    	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    	return x*f;
    }
    int n,a[N],f[2][N][N];
    void inc(int &x,int y){x+=y;if (x>=P) x-=P;}
    int main()
    {
    #ifndef ONLINE_JUDGE
    	freopen("a.in","r",stdin);
    	freopen("a.out","w",stdout);
    	const char LL[]="%I64d
    ";
    #else
    	const char LL[]="%lld
    ";
    #endif
    	n=read();
    	for (int i=1;i<=n;i++) a[i]=read();
    	f[0][1023][1023]=1;
    	for (int i=0;i<n;i++)
    	{
    		for (int j=0;j<N;j++)
    			for (int k=N-1^j;k>=0;k=k==0?-1:(k-1&(N-1^j)))
    			f[i&1^1][j][k|j]=f[i&1][j][k|j];
    		for (int j=0;j<N;j++)
    			for (int k=N-1^j;k>=0;k=k==0?-1:(k-1&(N-1^j)))
    			inc(f[i&1^1][j&a[i+1]][(k|j)&a[i+1]],f[i&1][j][k|j]),
    			inc(f[i&1^1][j&a[i+1]][((k|j)&a[i+1])|j],P-f[i&1][j][k|j]);
    	}
    	cout<<f[n&1][0][0]<<endl;
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/Gloid/p/10697754.html
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