暴力算概率即可。

用这个式子：P(Ai|B)=P(AiB)/P(B)

不过。。貌似有递推的常数做法？

 

#include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;const int MAXN = 20 + 1;int N, R;double p[MAXN], a[MAXN];inline int bitcount(int x){
    return x == 0 ? 0 : (x & 1) + bitcount(x >> 1);}int main(){    int t = 0;    while(cin>>N>>R, (N + R))    {        for(int i = 0; i < N; i++) scanf("%lf", &p[i]);        memset(a, 0, sizeof(a));        printf("Case %d:\n", ++t);        if(R == 0) {            for(int i = 0; i < N; i++) puts("0.000000");            continue;        }        double rp = 0.0;        for(int state = 0; state < (1 << N); state++)        {            int cnt = bitcount(state);            if(cnt != R) continue;            double tmp = 1.0;            for(int i = 0; i < N; i++)                if(state & (1 << i)) tmp *= p[i];                else tmp *= (1 - p[i]);            rp += tmp;            for(int i = 0; i < N; i++)                 if(state & (1 << i)) a[i] += tmp;        }        for(int i = 0; i < N; i++) printf("%.6lf\n", a[i] / rp);    }    return 0;}