ACM--steps--2.2.8--第一类stirling数
2015-02-03 22:21
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Examining the Rooms |
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
Total Submission(s): 652 Accepted Submission(s): 370 |
[align=left]Problem Description[/align] 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. |
[align=left]Input[/align] 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) |
[align=left]Output[/align] Output one line for each case, indicating the corresponding possibility. Four digits after decimal point are preserved by rounding. |
[align=left]Sample Input[/align]3 3 1 3 2 4 2 |
[align=left]Sample Output[/align]0.3333 0.6667 0.6250 <div style="" font-size:="" 14px;="" border-top:="" #b7cbff="" 1px="" dashed;="" font-family:="" times"="">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 |
[align=left]Source[/align] The 35th ACM/ICPC Asia Regional Tianjin Site —— Online Contest |
[align=left]Recommend[/align] lcy 参考博文:/article/2565982.html 建立的数组一定要用long long。 第一类stirling给出恰好含有m个圈的n个数的排列数目。又一次感到身为数学系的学生,数学都学到猪身上去了 。 #include<iostream> #include<iomanip> using namespace std; long long dyx[21];//用以表示房间的一共可能的数目。 long long wyx[21][21];//stirling数。 int main() { //第一类stirling数的递推公式为;s(p,k)=(p-1)*s(p-1,k)+s(p-1,k-1); //第一类stirling数给出恰好包含k个圈的p个数的排列数目。 int i,j; i=j=0; //先计算出房间里钥匙的一共的可能性。我们可以得出是房间数目的阶乘; dyx[0]=1; for(i=1;i<21;i++) { dyx[i]=dyx[i-1]*i; } //第一类stirling数的边界条件 wyx[0][0]=0; for(i=1;i<21;i++) { wyx[i][0]=0; wyx[i][i]=1; } //进行stirling数的计算。 for(i=1;i<=20;i++) { for(j=1;j<i;j++) { wyx[i][j]=(i-1)*wyx[i-1][j]+wyx[i-1][j-1]; } } int n,k,T; cin>>T; while(T--) { cin>>n>>k; if(n==1||k==0) { cout<<"0.0000"<<endl; continue; } //k表示最大破门数,即为最大圈数。 long long sum=0; for(j=1;j<=k;j++) { sum+=wyx [j]-wyx[n-1][j-1];//减去1自成1个圈的情况。 } cout<<fixed<<setprecision(4)<<(double)sum/dyx <<endl;; } return 0; } |
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