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Examining the Rooms

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1661 Accepted Submission(s): 1015

Problem Description

A murder happened in the hotel. As the best detective in the town, you should examine all the N rooms of the hotel immediately. However, all the doors of the rooms are locked, and the keys are just locked in the rooms, what a trap! You know that there is exactly one key in each room, and all the possible distributions are of equal possibility. For example, if N = 3, there are 6 possible distributions, the possibility of each is 1/6. For convenience, we number the rooms from 1 to N, and the key for Room 1 is numbered Key 1, the key for Room 2 is Key 2, etc.
To examine all the rooms, you have to destroy some doors by force. But you don’t want to destroy too many, so you take the following strategy: At first, you have no keys in hand, so you randomly destroy a locked door, get into the room, examine it and fetch the key in it. Then maybe you can open another room with the new key, examine it and get the second key. Repeat this until you can’t open any new rooms. If there are still rooms un-examined, you have to randomly pick another unopened door to destroy by force, then repeat the procedure above, until all the rooms are examined.
Now you are only allowed to destroy at most K doors by force. What’s more, there lives a Very Important Person in Room 1. You are not allowed to destroy the doors of Room 1, that is, the only way to examine Room 1 is opening it with the corresponding key. You want to know what is the possibility of that you can examine all the rooms finally.

Input

The first line of the input contains an integer T (T ≤ 200), indicating the number of test cases. Then T cases follow. Each case contains a line with two numbers N and K. (1 < N ≤ 20, 1 ≤ K < N)

Output

Output one line for each case, indicating the corresponding possibility. Four digits after decimal point are preserved by rounding.

Sample Input

3 3 1 3 2 4 2

Sample Output

0.3333 0.6667 0.6250

Hint

Sample Explanation When N = 3, there are 6 possible distributions of keys: Room 1 Room 2 Room 3 Destroy Times #1 Key 1 Key 2 Key 3 Impossible #2 Key 1 Key 3 Key 2 Impossible #3 Key 2 Key 1 Key 3 Two #4 Key 3 Key 2 Key 1 Two #5 Key 2 Key 3 Key 1 One #6 Key 3 Key 1 Key 2 One In the first two distributions, because Key 1 is locked in Room 1 itself and you can’t destroy Room 1, it is impossible to open Room 1. In the third and forth distributions, you have to destroy Room 2 and 3 both. In the last two distributions, you only need to destroy one of Room 2 or Room

Source

2010 Asia Regional Tianjin Site —— Online Contest

第一類Stirling數 s(p,k)

　　

s(p,k)的一個的組合學解釋是：將p個物體排成k個非空循環排列的方法數。

s(p,k)的遞推公式： s(p,k)=(p-1)*s(p-1,k)+s(p-1,k-1) ,1<=k<=p-1

邊界條件：s(p,0)=0 ,p>=1 s(p,p)=1 ,p>=0

遞推關系的說明：

考慮第p個物品，p可以單獨構成一個非空循環排列，這樣前p-1種物品構成k-1個非空循環排列，方法數為s(p-1,k-1)；

也可以前p-1種物品構成k個非空循環排列，而第p個物品插入第i個物品的左邊，這有(p-1)*s(p-1,k)種方法。

 1 //2017-08-05
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <iostream>
 5 #include <algorithm>
 6 #define ll long long 
 7 
 8 using namespace std;
 9 
10 const int N = 22;
11 ll factorial[N], stir1[N][N];//factorial[n]存n的階乘，stir1為第一類斯特林數
12 
13 int main()
14 {
15     int T, n, k;
16     cin >> T;
17     factorial[1] = 1;
18     for(int i = 2; i < N; i++)
19           factorial[i] = factorial[i-1]*i;
20     memset(stir1, 0, sizeof(stir1));
21     stir1[1][1] = 1;
22     stir1[0][0] = 0;
23     for(int i = 2; i < N; i++){
24           for(int j = 1; j <= i; j++)
25               stir1[i][j] = stir1[i-1][j-1] + (i-1)*stir1[i-1][j];
26     }
27     while(T--){
28         cin>>n>>k;
29         ll tmp = 0;
30         for(int i = 1; i <= k; i++)
31               tmp += stir1[n][i] - stir1[n-1][i-1];
32         printf("%.4lf\n", (double)tmp*1.0/factorial[n]);
33     }
34 
35     return 0;
36 }

HDU3625(SummerTrainingDay05-N 第一類斯特林數)