ACM_The third part of the recursive topic series put apples

The third of the recursive topic series, put apples

Time Limit: 2000/1000ms (Java/Others)

Problem Description:

Put M identical apples on N identical plates, and allow some plates to be left empty. How many different ways are there in total? (represented by K) 5, 1, 1 and 1, 5, 1 are the same division.

Input:

The first row is the number t of test data (0 <= t <= 20). Each of the following lines contains two integers M and N, separated by spaces. 1<=M, N<=10.

Output:

For each set of input data M and N, use one line to output the corresponding K.

Sample Input:

2
8 6
7 3

Sample Output:

20
8 Problem- 
solving ideas: 
Let f(m,n) be m apples and the number of ways to place n plates, then discuss n first: 
①When m<n: there must be nm plates that are always empty, remove them and The number of ways to place apples has no effect. That is, if(n>m) f(m,n) = f(m,m);　　
②When m>=n: different placement methods can be divided into two categories: 
1. At least one plate is empty, which is equivalent to f (m,n) = f(m,n-1); 
2. All plates have apples, which is equivalent to removing one apple from each plate without affecting the number of different placement methods, that is, f(m,n ) = f(mn,n); 
and the total number of ways to put apples is equal to the sum of the two, that is, f(m,n) =f(m,n-1)+f(mn,n). 
Recursive exit condition description: 
when n=1, all apples must be placed in a plate, so return 1; 
when no apples can be placed, it is defined as one way of placing; 
two recursive paths, the first n will be Gradually decrease, it will eventually reach the exit n==1; 
the second m will gradually decrease, because when n>m, it will return f(m,m), so it will eventually reach the exit m==0. 
Why export m==0? Because we always let m>=n to solve, so mn>=0, let m=0 end, if it is changed to m=1, 
there may be a situation where mn=0 and the correct solution cannot be obtained. 
AC code:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int fun(int m,int n){
 4     if(m==0 || n==1)return 1;
 5     if(m<n)return fun(m,m);
 6     else return fun(m,n-1)+fun(m-n,n);
 7 }
 8 int main(){
 9     int t,m,n;
10     cin>>t;
11     while(t--){
12         cin>>m>>n;
13         cout<<fun(m,n)<<endl;
14     }
15     return 0;
16 }