2015
09-17

# Bell

What? MMM is learning Combinatorics!?
Looks like she is playing with the bell sequence now:
bell[n] = number of ways to partition of the set {1, 2, 3, …, n}

e.g. n = 3:
{1, 2, 3}
{1} {2 3}
{1, 2} {3}
{1, 3} {2}
{1} {2} {3}
so, bell[3] = 5

MMM wonders how to compute the bell sequence efficiently.

T — number of cases
for each case:
n (1 <= n < 2^31)

T — number of cases
for each case:
n (1 <= n < 2^31)

6
1
2
3
4
5
6

1
2
5
15
52
203

# 斯特灵数[编辑]

Stirling提出的。

## 第一类[编辑]

s(4,2)=11

1. {A,B},{C,D}
2. {A,C},{B,D}
3. {A,D},{B,C}
4. {A},{B,C,D}
5. {A},{B,D,C}
6. {B},{A,C,D}
7. {B},{A,D,C}
8. {C},{A,B,D}
9. {C},{A,D,B}
10. {D},{A,B,C}
11. {D},{A,C,B}

• 给定，有递归关系

## 第二类[编辑]

1. {A,B},{C,D}
2. {A,C},{B,D}
3. {A,D},{B,C}
4. {A},{B,C,D}
5. {B},{A,C,D}
6. {C},{A,B,D}
7. {D},{A,B,C}

• 给定，有递归关系
• 递推关系的说明：考虑第n个物品，n可以单独构成一个非空集合，此时前n-1个物品构成k-1个非空的不可辨别的集合， 方法数为S(n-1,k-1)；也可以前n-1种物品构成k个非空的不可辨别的 集合，第n个物品放入任意一个中，这样有k*S(n-1,k)种方法。

## 两者关系[编辑]

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