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[<<N];

 int sg( long double x ) { return (x>-eps)-(x<eps); }

 void trans() {
     for( int i=; i<n; i++ ) {
         int ss=bound^(<<i);
         f[ss|(<<i)] += f[ss];
         for( int s=(ss-)&ss; s!=ss; s=(s-)&ss )
             f[s|(<<i)] += f[s];
     }
 }
 void inverse() {
     for( int i=; i<n; i++ ) {
         int ss=bound^(<<i);
         f[ss|(<<i)] -= f[ss];
         for( int s=(ss-)&ss; s!=ss; s=(s-)&ss )
             f[s|(<<i)] -= f[s];
     }
 }

 int main() {
     scanf( "%d", &n );
     bound = (<<n)-;
     for( int s=; s<=bound; s++ )
         scanf( "%Lf", f+s );
     trans();
     for( int s=; s<=bound; s++ ) {
         if( sg(f[s]-)== )
             f[s] = ;
         else
             f[s] = /(f[s]-);
     }
     inverse();
     if( sg(f[bound])== )
         printf( "INF\n" );
     else
         printf( "%.10Lf\n", 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 = ;

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

 void trans( int a[], int flag ) {
     for( int b=; b<n; b++ ) {
         int u = U ^ (<<b);
         for( int s=u,t=<<(n-); t; s=(s-)&u,t-- ) {
             int l=a[s], r=a[s|(<<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 = (<<n)-;
     for( int i=; i<=U; i++ )
         scanf( "%d", a+i );
     for( int i=; i<=U; i++ )
         scanf( "%d", b+i );
     trans(a,);
     trans(b,);
     for( int s=; s<=U; s++ )
         c[s] = a[s]*b[s];
     trans(c,-);
     for( int s=; s<=U; s++ )
         printf( "%d ", c[s] );
     printf( "\n" );
 }