标签：
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 1776 |
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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
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
1 /**/
2 #include<iostream>
3 #include<cstdio>
4 #include<cmath>
5 #include<cstring>
6 #include<algorithm>
7 using namespace std;
8 double f[110][110];//[元素数量][分组数量]=方法数
9 int n;
10 void init(){
11 int i,j;
12 for(i=1;i<=100;i++) f[1][i]=0,f[i][1]=1;
13 for(i=2;i<=100;i++)
14 for(j=2;j<=i;j++){
15 f[i][j]=f[i-1][j-1]+f[i-1][j]*j;
16 }
17 return;
18 }
19 int main(){
20 init();
21 while(scanf("%d",&n) && n){
22 double ans=0;
23 for(int i=1;i<=n;i++)ans+=f[n][i];
24 printf("%d %.0f\n",n,ans);
25 }
26 return 0;
27 }
POJ1671 Rhyme Schemes
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原文地址：http://www.cnblogs.com/SilverNebula/p/5734348.html
