[Combined mathematics] Class 1, Class 2 Stirling number, Bell number

I. The second type of Stirling number

Theorem: The second type of Stirling number S (p, k) counts the number of Division of the p element set into k undistinguished boxes without empty boxes.

Proof: it is not important for elements to take some boxes. The only thing that matters is what is installed in each box, no matter which box is loaded.

Recursive formulas include: S (p, p) = 1 (p> = 0) S (p, 0) = 0 (p> = 1) S (p, k) = k * S (P-1 k) + S (PM, K-1) (1 <= k <= PM ). Consider the first p positive integer, 1, 2,... p as the set to be divided

{1, 2,... p} is divided into k non-empty and non-differentiated boxes in two cases:

(1) those which make p separate in a box, there are S (P-1 K-1) division number

(2) the division that makes p not separate in a box exists with k * S (P-1 k) division numbers.

In the second case, p is not in a single box, that is, p and other elements are in a set, that is, before p is put, p-1 has been divided into k non-empty and undistinguished boxes.

The number of Shards is S (MA-1, k). Now the question is which box should p be put in. There are k options, so k * S (P-1 k) exists ).

Template:

Long s [maxn] [maxn]; // Stirling number required for storage const long mod = 1e9 + 7; // modulo void init () // preprocessing {memset (s, 0, sizeof (s); s [1] [1] = 1; for (int I = 2; I <= maxn-1; I ++) for (int j = 1; j <= I; j ++) {s [I] [j] = s [i-1] [j-1] + j * s [i-1] [j]; if (s [I] [j]> = mod) s [I] [j] % = mod ;}}

Note: Use the long type. If the number of elements is greater than 20, the value of the int type is exceeded.

Extension: k! * S (p, k) counts the number of bins that divide the p element set into k differentiated boxes without empty boxes.

 

II. Number of Bell

Theorem: Bell number B (p) refers to the number of bins that divide the p element set into non-empty and non-differentiated boxes (not in a few boxes ).

B (p) = S (p, 0) + S (p, 1) +... + S (p, k)

Therefore, for the number of Bell, the second type of Stiring must be obtained first.

III. First Stirling number

Theorem: The First Stirling number s (p, k) counts the number of methods that arrange p objects in k non-empty loops.

Proof: The circular arrangement in the preceding theorem is called a circle. Recurrence formula:

S (p, p) = 1 (p> = 0) there are p and p circles, and each circle has only one person

S (p, 0) = 0 (p> = 1) if at least one person exists, any arrangement should contain at least one circle.

S (p, k) = (P-1) * s (PM, k) + s (PM, K-1)

The person was marked as 1, 2,... p. There are two situations in which the p individual is arranged into k circles. The first sorting method is only p in a circle of the people themselves, the sorting method has s (1, K-1. In the second sort method, p should at least be in the same

In a circle. You can sort the values of 1, 2 .... p-1 is arranged in k circles and then p is placed in 1, 2 .... the left side of any person in type 1-1 is obtained. Therefore, the second type of sorting method is p-1) * s (P-1, k.

In proof, what we do is to divide {1, 2,..., p} into k non-empty and non-differentiated boxes, and then arrange the elements in each box in a circular arrangement.

Template:

Long s [maxn] [maxn]; // The first Stirling number required for storage. const long mod = 1e9 + 7; // modulo void init () // preprocessing {memset (s, 0, sizeof (s); s [1] [1] = 1; for (int I = 2; I <= maxn-1; I ++) for (int j = 1; j <= I; j ++) {s [I] [j] = s [i-1] [j-1] + (i-1) * s [i-1] [j]; if (s [I] [j]> = mod) s [I] [j] % = mod ;}}

 

      

[Combined mathematics] Class 1, Class 2 Stirling number, Bell number