第一类斯特林数 hdu 3625

Examining the Rooms

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1602    Accepted Submission(s): 975

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

33 13 24 2

 

Sample Output

0.33330.66670.6250
Hint
Sample ExplanationWhen N = 3, there are 6 possible distributions of keys:Room 1Room 2Room 3Destroy Times#1Key 1Key 2Key 3Impossible#2Key 1Key 3Key 2Impossible#3Key 2Key 1Key 3Two#4Key 3Key 2Key 1Two#5Key 2Key 3Key 1One#6Key 3Key 1Key 2OneIn 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

模板题：

就是求n个元素最多组成k个环的概率（n个元素可以组合成n个环）

且第一个元素不能单独成环，那么久用第一类斯特林数就好了

第一类斯特林数表示的是将n个不同元素分成k个不同的环的方案数

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define mod 1000000007#define MAXN 2010#define LL long longLL a[MAXN][MAXN],c[MAXN];void get_num1_stl(){    for(int i=1;i<=20;i++){        a[i][0]=0,a[i][i]=1;        for(int j=1;j<i;j++)            a[i][j]=(((i-1)*a[i-1][j])+a[i-1][j-1]);    }}void init(){    c[1]=1;    for(int i=2;i<=20;i++)       c[i]=c[i-1]*i;}int main(){    init();    get_num1_stl();    int n,k,T;    freopen("in.txt","r",stdin);    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&k);        LL temp=0;        for(int i=1;i<=k;i++)            temp+=a[n][i]-a[n-1][i-1];        double ans=((double)(temp)/c[n]);        printf("%0.4lf\n",ans);    }    return 0;}