[Ybt] [Basic calculation recurrence lesson example 3] Division of numbers

Division of numbers

Topic link: division of numbers

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Title description

Problem solving ideas

设 $f_{i,j}$He said it would $i\; is$ divided intojjNumber of cases of$j$ copies.
Obviously, when$i=j$ 时 $There$ is only one case for$f$$($$i$$)$$($$j$$)$ , which is$1$ .
If$j>i$ , even$1$ is not enough, so it must be illegal, as$0$ .
Consider the normal situation:

• If we add 1 1 after the original sequence$1$ , then the original total is$i-1$ , the original number of copies is$j-1$。
• If we add 1 1 to each number on the basis of the original sequence$1$ , then the original total is$i-j$ , the original number of copies is$j$。

Then the recurrence formula is:
$$f_{i,j}=f_{i-1,j-1}+f_{i-j,j}$$

code

#include<iostream>
#include<cstdio>
using namespace std;

int n,m;
int f[210][10];

int main()
{
    
    
	cin>>n>>m;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		{
    
    
			if(i==j)
				f[i][j]=1;
			else if(j>i)
				f[i][j]=0;
			else
				f[i][j]=f[i-1][j-1]+f[i-j][j];
		}
	cout<<f[n][m];
}