Bin & Jing in Wonderland (probability, combinatorial math)

problem 2103 Bin & Jing in Wonderlandaccept:201 submit:1048 time limit:1000 mSec Memory limit:32768 KBproblem Description

Bin has a dream that he and Jing is both in a wonderland full of beautiful gifts. Bin wants to choose some gifts for Jing to get in her good graces.

There is N different gifts in the Wonderland, with an ID from 1 to N, and all kinds of these gifts has infinite duplicates. Each time, Bin shouts loudly, ' I love Jing ', and then the Wonderland random drop a gift in front of bin. The dropping probability for gift I (1≤i≤n) is P (i). of cause, P (1) +p (2) +...+p (N) =1. Bin finds the gifts with the higher ID is better. Bin shouts K times and selects R best gifts finally.

That's, firstly Bin gets k gifts, then sorts all these gifts according to their ID, and picks up the largest R gifts at l Ast. Now, if given the final list of the R largest gifts, can do help Bin find out the the probability of the list?

Input

The first line of the input contains an integer T (t≤2,000), indicating number of test cases.

For each test cast, the first line contains 3 integers N, K and R (1≤n≤20, 1≤k≤52, 1≤r≤min (k,25)) as the description above . In the second line, there is N positive float numbers indicates the probability of each gift. There is at most 3 digits after the decimal point. The third line has an r integers ranging from 1 to N indicates the finally list of the "R Best Gifts ' ID.

OutputFor each case, output a float number with 6 digits after the decimal points, which indicates the probability of the final List.Sample Input4 2 3 3 0.3 0.7 1 1 1 2 3 3 0.3 0.7 1 1 2 2 3 3 0.3 0.7 1 2 2 2 3 3 0.3 0.7 2 2 2Sample Output0.027000 0.189000 0.441000 0.343000Source"Higher Education Society Cup" the third Fujian University College Students Program Design contest: There are n kinds of apples, a person in the following shout K times, each fall down an apple, each kind of apple fell the probability known, now give the former r Big Apple species; Ask the probability of the smallest occurrence, find out the probability of a large ratio of MI, To find out the probability of a smaller than MI, enumerate the number of MI occurrences, the probability of the need to multiply the good; code:

#include <iostream>#include<algorithm>#include<cstdio>#include<cmath>#include<cstring>using namespacestd;Doublep[ -], dp[ -];typedefLong LongLL; LL c[ -][ -];intnum[ the];inta[ the];voiddb () {c[1][0] = c[1][1] =1;  for(inti =2; I < -; i++) {c[i][0] = C[i][i] =1;  for(intj =1; J < I; J + +) {C[i][j]= C[i-1][J] + c[i-1][j-1]; }    }}intMain () {intT, N, K, R; scanf ("%d", &T);    DB ();  while(t--) {scanf ("%d%d%d", &n, &k, &R); dp[0] =0;  for(inti =1; I <= N; i++) {scanf ("%LF", p +i); Dp[i]= Dp[i-1] +P[i]; }        DoubleAns =1; memset (num,0,sizeof(num)); intMi =0x3f3f3f3f;  for(inti =1; I <= R; i++) {scanf ("%d", A +i); Num[a[i]]++; Mi=min (mi, a[i]); }        intnow =K;  for(inti = mi +1; I <= N; i++){            if(!num[i])Continue; Ans*= C[now][num[i]] *Pow (p[i], num[i]); now-=Num[i]; }        DoubleANS1 =0;  for(inti = Num[mi]; I <= k-r + num[mi]; i++) {ans1+ = C[k-r + num[mi]][i] * POW (p[mi], i) * POW (DP[MI-1], K-r + Num[mi]-i); } printf ("%.6lf\n", ans *ans1); }    return 0;}

Bin & Jing in Wonderland (probability, combinatorial math)