POJ1671 Rhyme Schemes

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 1776		Accepted: 984		Special Judge

Description

The rhyme scheme for a poem (or stanza of a longer poem) tells which lines of the poem rhyme with which other lines. For example, a limerick such as If computers that you build are quantum
Then spies of all factions will want 'em

Our codes will all fail

And they'll read our email

`Til we've crypto that's quantum and daunt 'em

Jennifer and Peter Shor (http://www.research.att.com/~shor/notapoet.html)

Has a rhyme scheme of aabba, indicating that the first, second and fifth lines rhyme and the third and fourth lines rhyme.

For a poem or stanza of four lines, there are 15 possible rhyme schemes:

aaaa, aaab, aaba, aabb, aabc, abaa, abab, abac, abba, abbb, abbc, abca, a bcb, abcc, and abcd.

Write a program to compute the number of rhyme schemes for a poem or stanza of N lines where N is an input value.

Input

Input
will consist of a sequence of integers N, one per line, ending with a 0
(zero) to indicate the end of the data. N is the number of lines in a
poem.

Output

For
each input integer N, your program should output the value of N,
followed by a space, followed by the number of rhyme schemes for a poem
with N lines as a decimal integer with at least 12 correct significant
digits (use double precision floating point for your computations).

Sample Input

1
2
3
4
20
30
10
0

Sample Output

1 1
2 2
3 5
4 15
20 51724158235372
30 846749014511809120000000
10 115975

Source

Greater New York 2003

 

按照题目所说，double的精度就可以过

第二类Stirling数：将n个不同的元素分成m个集合的问题。

度娘百科：http://baike.baidu.com/link?url=Gf9ql9PnQNNjCZVUgI6SH_o1DgFwpL5yOFalDr_baNqKmrr0unKZvaDNU5RzSGmMQIbKW3Efivp0GPOlz3tcga

别人简洁的题解：http://blog.csdn.net/nvfumayx/article/details/12356847

 /**/
 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 double f[][];//[元素数量][分组数量]=方法数
 int n;
 void init(){
     int i,j;
     for(i=;i<=;i++) f[][i]=,f[i][]=;
     for(i=;i<=;i++)
       for(j=;j<=i;j++){
           f[i][j]=f[i-][j-]+f[i-][j]*j;
       }
     return;
 }
 int main(){
     init();
     while(scanf("%d",&n) && n){
         double ans=;
         for(int i=;i<=n;i++)ans+=f[n][i];
         printf("%d %.0f\n",n,ans);
     }
     return ;
 }