P3175 [HAOI2015] Bitwise OR

Initially, you have the number $0$. Every second, you randomly choose a number from $[0,2^n-1]$ and apply bitwise OR with your current number (C++, C: `|`, Pascal: `or`). The probability of choosing number $i$ is $p_i$. It is guaranteed that $0\leq p_i \leq 1$, $\sum p_i=1$. Find the expected number of seconds until your current number becomes $2^n-1$.

The first line contains $n$, the number of elements. The second line contains $2^n$ numbers; the $i$-th number is the probability of selecting $i-1$.

Output a single number representing the answer. An absolute or relative error not exceeding $10^{-6}$ is accepted. If there is no solution, output `INF`.

For $100\%$ of the testdata, $n\leq 20$. The following is the SPJ source code. ```cpp //liuchenrui #include #include #include #include #include #define AC {fclose(fstd),fclose(fuser);return 0;} #define WA {fclose(fstd),fclose(fuser);return 1;} #define PE {fclose(fstd),fclose(fuser);return 5;} #define eps 1e-6 int main(int const argc, char*const argv[]){ FILE *fstd,*fuser; fstd=fopen(argv[2],"r"); fuser=fopen(argv[3],"r"); //fstd=fopen("x1.in","r"); //fuser=fopen("x2.in","r"); char s[30],t[30]; if(fscanf(fuser,"%s",s+1)==-1)WA; fscanf(fstd,"%s",t+1); if(s[1]=='I' && t[1]=='I')AC; if(s[1]=='I' || t[1]=='I')WA; double p,q; sscanf(s+1,"%lf",&p); sscanf(t+1,"%lf",&q); if(fabs(p-q)