M apples put on n plates problem

Question One

Problem Description: Put m the same apples on n the same plate, allow some tray empty, ask how many different methods? (Note: 5,1,1 and 1,1,5 are the same kind of sub-method)

Problem Solving Analysis:

Set F (m,n) for M apples, N for the number of plates, then the n is discussed first,

When n>m: There must be n-m a plate is always empty, remove them on the number of Apple methods to be placed no effect. i.e. if (n>m) f (m,n) = f (m,m)

When n <= m: Different methods can be divided into two categories: 0 of the program number, not including 0 of the program number

1, contains 0 of the program number, that is, at least one plate empty, that is, the equivalent of F (m,n) =f (m,n-1);

2, does not contain 0 of the program number, that is, all the plates have apples, equivalent to take an apple from each plate, does not affect the number of different methods, namely F (m,n) =f (m-n,n). The total number of put apples is equal to the sum of the two, that is F (m,n) =f (m,n-1) +f (m-n,n )

Description of the recursive export conditions:

When n=1, all apples must be placed on a plate, so return 1;

When the m==0 (no Apple can be placed), defined as 1 kinds of release method;

Recursive

int fun (int m, int n)//m apples on n plates There are several methods {if (m==0 | | n==1) return 1;    if (n>m) return fun (m,m); else return Fun (m,n-1) +fun (m-n,n);}

Dynamic planning

Put the Apple Int main () {    int apple, plate;    if (apple  < 0 | |  apple > 10 | |  plate < 1 | | &NBSP;PLATE&NBSP;&GT;&NBSP;10)     {        cout  << -1 << endl;        return -1;     }    vector<vector<int> > ivec (11,  Vector<int> (11,0));     for (int i=0; i < 11; i++)      {        ivec[0][i] = 1;         ivec[i][1] = 1;    }    for ( Int i = 1; i <= 10; ++i)     {         for (int j&NBSP;=&NBSP;1;&NBSP;J&NBSP;&LT;=&NBSP;10;&NBSP;++J)         {             if (j <= i)                  ivec[i][j] = ivec[i][ J-1] + ivec[i-j][j];            else &NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;IVEC[I][J]  = ivec[i][j];        }    }     cout << ivec[apple][plate] << endl;    return  0;}

Question Two

Problem Description: The sum of integer n divided into k integers and each number greater than or equal to a less than or equal to B, the number of different methods

int Dynamics (int n, int k, int min)//n is divided into k integers, the smallest is greater than or equal to Min, the largest is not more than b{if (n < min) return 0;    When the remainder is smaller than min, it does not meet the requirements, return 0 if (k = = 1) return 1;    int sum = 0;    for (int t = min; t <= B; t++) {sum + = Dynamics (n-t, k-1, T); } return sum;}

Question Three

M----> Same, n----> Same, cannot be empty

How many ways to put m apples into n plates? Note, for example, that the two scenarios, such as 1, 2, and 2 and 1, are a scenario.

Idea, first put each plate of an apple, so the problem is converted to: M-n an apple into the N plate, the plate allows empty, that is, problem 1

M apples put on n plates problem