bzoj 4036 集合幂级数

集合幂级数其实就是一种集合到数的映射,并且我们针对集合的一些操作(or  xor and specil or )为这种映射定义运算.其中一些东西可以通过某些手段将其复杂度降低.

orz vfk

/**************************************************************
    Problem: 4036
    User: idy002
    Language: C++
    Result: Accepted
    Time:3584 ms
    Memory:13092 kb
****************************************************************/
 
#include <cstdio>
#include <cmath>
#define eps 1e-10
#define N 20
 
int n, bound;
long double f[1<<N];
 
int sg( long double x ) { return (x>-eps)-(x<eps); }
 
void trans() {
    for( int i=0; i<n; i++ ) {
        int ss=bound^(1<<i);
        f[ss|(1<<i)] += f[ss];
        for( int s=(ss-1)&ss; s!=ss; s=(s-1)&ss ) 
            f[s|(1<<i)] += f[s];
    }
}
void inverse() {
    for( int i=0; i<n; i++ ) {
        int ss=bound^(1<<i);
        f[ss|(1<<i)] -= f[ss];
        for( int s=(ss-1)&ss; s!=ss; s=(s-1)&ss ) 
            f[s|(1<<i)] -= f[s];
    }
}
 
int main() {
    scanf( "%d", &n );
    bound = (1<<n)-1;
    for( int s=0; s<=bound; s++ ) 
        scanf( "%Lf", f+s );
    trans();
    for( int s=0; s<=bound; s++ ) {
        if( sg(f[s]-1)==0 ) 
            f[s] = 0;
        else
            f[s] = 1/(f[s]-1);
    }
    inverse();
    if( sg(f[bound])==0 ) 
        printf( "INF
" );
    else
        printf( "%.10Lf
", f[bound] );
}

---

感谢Picks：http://picks.logdown.com/posts/179290-fast-walsh-hadamard-transform

现在可以做or, and, xor, 以及它们的否的卷积了。

#include <cstdio>

const int N = 10;

int n, U;
int a[1<<N], b[1<<N], c[1<<N];

void trans( int a[], int flag ) {
    for( int b=0; b<n; b++ ) {
        int u = U ^ (1<<b);
        for( int s=u,t=1<<(n-1); t; s=(s-1)&u,t-- ) {
            int l=a[s], r=a[s|(1<<b)];
            /*
            NOT AND
            if( flag==1 ) {
                a[s] = l+r;
                a[s|(1<<b)] = r;
            } else {
                a[s] = r;
                a[s|(1<<b)] = l-r;
            }
             */
            /*
            NOT XOR
            if( flag==1 ) {
                a[s] = l+r;
                a[s|(1<<b)] = l-r;
            } else {
                a[s] = (l-r)/2;
                a[s|(1<<b)] = (l+r)/2;
            }
             */
            /*
            NOT OR
            if( flag==1 ) {
                a[s] = l;
                a[s|(1<<b)] = l+r;
            } else {
                a[s] = r-l;
                a[s|(1<<b)] = l;
            }
            */
            /*
            OR
            if( flag==1 ) {
                a[s] = l;
                a[s|(1<<b)] = l+r;
            } else {
                a[s] = l;
                a[s|(1<<b)] = r-l;
            }
            */
            /*
            AND
            if( flag==1 ) {
                a[s] = l+r;
                a[s|(1<<b)] = r;
            } else {
                a[s] = l-r;
                a[s|(1<<b)] = r;
            }
            */
            /*
            XOR
            if( flag==1 ) {
                a[s] = l+r;
                a[s|(1<<b)] = l-r;
            } else {
                a[s] = (l+r)/2;
                a[s|(1<<b)] = (l-r)/2;
            }
            */
        }
    }
}
int main() {
    scanf( "%d", &n );
    U = (1<<n)-1;
    for( int i=0; i<=U; i++ )
        scanf( "%d", a+i );
    for( int i=0; i<=U; i++ )
        scanf( "%d", b+i );
    trans(a,1);
    trans(b,1);
    for( int s=0; s<=U; s++ )
        c[s] = a[s]*b[s];
    trans(c,-1);
    for( int s=0; s<=U; s++ )
        printf( "%d ", c[s] );
    printf( "
" );
}