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[110][110];//[元素数量][分组数量]=方法数 
int n;
void init(){
    int i,j;
    for(i=1;i<=100;i++) f[1][i]=0,f[i][1]=1;
    for(i=2;i<=100;i++)
      for(j=2;j<=i;j++){
          f[i][j]=f[i-1][j-1]+f[i-1][j]*j;
      }
    return;
}
int main(){
    init();
    while(scanf("%d",&n) && n){
        double ans=0;
        for(int i=1;i<=n;i++)ans+=f[n][i];
        printf("%d %.0f\n",n,ans);
    }
    return 0;
}