原文地址：天津网络赛 Examining the Rooms
作者：又岸
Examining the Rooms
Time Limit: 2000/1000 MS (Java/Others)Total Submission(s): 260
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.
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
The 35th ACM/ICPC Asia Regional Tianjin Site —— Online Contest
这题要用到第一类 striling 数，属于数论上的东西。这是 一个解题报告
如果不考虑那个VIP，N个房间可以被最多K把钥匙打开的情况，实际上就是1..N的置换组成最多K个环的情况，这个就是第一类strling数之和S1[n][1] S1[n][2] ... S1[n][k]。在这些情况里面，如果钥匙1恰好锁在房间1里也是不行的，所以还要减去N-1个房间被最多K-1把钥匙打开的情况数。
//第一类 斯特林数
#include<stdio.h>
#include<string.h>
#include<string.h>
long long fac[25]={1,1};
long long stri[25][25];
long long stri[25][25];
void str()
{
memset(stri,0,sizeof(stri));
stri[1][1]=1;
for(int i=2;i<=20;i )
for(int j=1;j<=i;j )
{
stri[i][j]=stri[i-1][j-1] (i-1)*stri[i-1][j];
}
}
{
}
int main()
{
for(int i=2;i<=20;i )
fac[i]=i*fac[i-1];
int test;
scanf("%d",&test);
str();//打表
while(test--)
{
int n,k;
scanf("%d%d",&n,&k);
long long sum=0;
for(int i=1;i<=k;i )
{
sum =stri[n][i]-stri[n-1][i-1];//这里要看清题
}
printf("%.4lfn",1.0*sum/fac[n]);
{
}