bzoj 4036 集合幂级数

 1 /**************************************************************
 2     Problem: 4036
 3     User: idy002
 4     Language: C++
 5     Result: Accepted
 6     Time:3584 ms
 7     Memory:13092 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <cmath>
12 #define eps 1e-10
13 #define N 20
14  
15 int n, bound;
16 long double f[1<<N];
17  
18 int sg( long double x ) { return (x>-eps)-(x<eps); }
19  
20 void trans() {
21     for( int i=0; i<n; i++ ) {
22         int ss=bound^(1<<i);
23         f[ss|(1<<i)] += f[ss];
24         for( int s=(ss-1)&ss; s!=ss; s=(s-1)&ss ) 
25             f[s|(1<<i)] += f[s];
26     }
27 }
28 void inverse() {
29     for( int i=0; i<n; i++ ) {
30         int ss=bound^(1<<i);
31         f[ss|(1<<i)] -= f[ss];
32         for( int s=(ss-1)&ss; s!=ss; s=(s-1)&ss ) 
33             f[s|(1<<i)] -= f[s];
34     }
35 }
36  
37 int main() {
38     scanf( "%d", &n );
39     bound = (1<<n)-1;
40     for( int s=0; s<=bound; s++ ) 
41         scanf( "%Lf", f+s );
42     trans();
43     for( int s=0; s<=bound; s++ ) {
44         if( sg(f[s]-1)==0 ) 
45             f[s] = 0;
46         else
47             f[s] = 1/(f[s]-1);
48     }
49     inverse();
50     if( sg(f[bound])==0 ) 
51         printf( "INF\n" );
52     else
53         printf( "%.10Lf\n", f[bound] );
54 }