Bin & Jing in wonderland(概率,组合数学)
Accept: 201 Submit: 1048 Time Limit: 1000 mSec Memory Limit : 32768 KB
Problem Description
Bin has a dream that he and Jing are both in a wonderland full of beautiful gifts. Bin wants to choose some gifts for Jing to get in her good graces.
There are N different gifts in the wonderland, with ID from 1 to N, and all kinds of these gifts have 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 that the gifts with the higher ID are better. Bin shouts k times and selects r best gifts finally.
That is, firstly Bin gets k gifts, then sorts all these gifts according to their ID, and picks up the largest r gifts at last. Now, if given the final list of the r largest gifts, can you help Bin find out 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 are N positive float numbers indicates the probability of each gift. There are at most 3 digits after the decimal point. The third line has r integers ranging from 1 to N indicates the finally list of the r best gifts’ ID.
Output
Sample Input
Sample Output
Source
“高教社杯”第三届福建省大学生程序设计竞赛#include<iostream> #include<algorithm> #include<cstdio> #include<cmath> #include<cstring> using namespace std; double p[25], dp[25]; typedef long long LL; LL C[53][53]; int num[110]; int a[110]; void db(){ C[1][0] = C[1][1] = 1; for(int i = 2; i < 53; i++){ C[i][0] = C[i][i] = 1; for(int j = 1; j < i; j++){ C[i][j] = C[i - 1][j] + C[i - 1][j - 1]; } } } int main(){ int T, N, k, r; scanf("%d", &T); db(); while(T--){ scanf("%d%d%d", &N, &k, &r); dp[0] = 0; for(int i = 1; i <= N; i++){ scanf("%lf", p + i); dp[i] = dp[i - 1] + p[i]; } double ans = 1; memset(num, 0, sizeof(num)); int mi = 0x3f3f3f3f; for(int i = 1; i <= r; i++){ scanf("%d", a + i); num[a[i]]++; mi = min(mi, a[i]); } int now = k; for(int i = mi + 1; i <= N; i++){ if(!num[i])continue; ans *= C[now][num[i]] * pow(p[i], num[i]); now -= num[i]; } double ans1 = 0; for(int i = 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; }