The problem of placing M apples on N Plates

Question 1:

M -----> same, N ---> same, can be empty

How many methods can I put M apples on N plates and leave them empty. At the same time, note that the two solutions, for example, 1, 2, 2, and 1, are one.

 

Ideas:

In fact, this is similar to dividing an integer m into the sum of N integers,

If f [m] [N] is set to the number of schemes that divide m into N portions, and the schemes are not repeated, the values of the first portion of each scheme will not be greater than those of the subsequent schemes.

F [m] [N] = f [m] [n-1] + F [M-N] [N];
= 1 // M = 0 | n = 1
= 0 // m <0

 

F [m] [n-1] is equivalent to 0 on the first plate. You only need to divide the number into n-1 portions.
Because 0 is not greater than any number, it is equivalent to adding a plate of 0 in front of the solution in F [m] [n-1,
And does not violate the definition of F. Therefore, F [m] [n-1] must be part of the f [m] [N] solution,That is, the number of solutions containing 0.
F [M-N] [N] is equivalent to adding 1 to each plate. Because adding a number 1 on each plate does not affect the feasibility of the solution in F [m] [n-1], nor does it affect the definition of F.
Therefore, F [M-N] [N] must be part of the f [m] [N] solution,That is, the number of solutions without 0.

 

 

Question 2:

Problem description: Divide integer N into the sum of K integers and each number is greater than or equal to.
Which of the following methods can be used to calculate a value less than or equal to B?

 

Int dynamics (int n, int K, int min) // divide N into k integers with the minimum value greater than or equal to Min, the maximum value is B <br/>{</P> <p> If (n <min) return 0; // when the remaining value is smaller than min, otherwise, 0 is returned. <br/> If (k = 1) return 1; <br/> int sum = 0; <br/> for (int t = min; T <= B; t ++) <br/> {<br/> sum + = dynamics (n-t, K-1, t ); <br/>}< br/> return sum; </P> <p>}

 

 

Question 3:

M -----> same, N ---> same, cannot be blank

How many methods can I put M apples on N plates. At the same time, note that the two solutions, for example, 1, 2, 2, and 1, are one.

 

Ideas:

Put each one apple first, and the problem becomes: M-N apples are put into N plates, and the plate is allowed to be empty, that is, question 1

 

 

Question 4:

The first Stirling number is positive and negative, and its absolute value isNumber of methods for dividing a set of n elements into k rings.

Recursive Formula,
S (n, 0) = 0, S (1, 1) = 1.
S (n, k) = S (n-1, k-1) + (n-1) S (n-1, K ).

 

The method for dividing a set of n elements into k rings is S (n, k ).

1. The S (n-1, k-1) of the ring can be K-1 by the first n-1 elements; that isThe last element is a single ring, and the first n-1 constitute a K-1 ring;

2. the nth element must not be a single ring. K rings of N-1 elements can place the nth number in any ring to form a new ring! N

The set of elements is divided into k rings. If the set of n elements is divided into k rings, since N is not in a single ring, you can place N in the ring where N is located.

If n-1 elements and K rings are removed, the necessity and adequacy are proved!

Therefore:S (n, k) = S (n-1, k-1) + (n-1) S (n-1, K ).Pass!

 

Question 5:

The second type of Stirling is the number of methods that divide a set containing n elements into k non-empty subsets.
// N-> there is a difference, K-> non-empty, no difference
Recursive Formula,
S (n, n) = S (n, 1) = 1,
S (n, k) = S (n-1, k-1) + KS (n-1, K ).

The above recursive formula can be proved by combination:
On the one hand, if the N element is taken out separately into a set, then the number of methods is S (n-1, k-1 );
On the other hand, if there are more than one element in the set where the nth element is located, You can first divide the remainder n-1 elements and then select another set to put the nth element in, the number of methods is K * s (n-1, k );
Evidence of the principles of Addition

 

Question 6:

Number of Bell and Stirling

B (n) is composed of n elements.Division of SetsThe number of methods.

Set Division: non-empty,

B (0) = 1, B (1) = 1,

B (n) = sum (1, N) S (n, k ).Sum (1, N) indicates the sum of K from 1 to n,

 

Question 7:

When K is different, it is generally multiplied by K's full arrangement on the basis of no difference.