[Ybtoj Chapter 1 Example 4] Passing game [Recursion]

Title description

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Problem solving idea
$f(i,j)$ means that the pass passes through j person in the$The$ number of plans in the hands of$i$ classmates. Obviously, the ball of classmate i will be$i-1$ or$i+1\; is$ passed, so the recurrence formula can be obtained:$f(i,j)=f（i-1，j-1）+f(i+1,j-1)$。

PS: Since this question is a ring, 1 and n can be connected, so these two places must be judged specially.

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Code

#include<iostream>
#include<cstdio>
using namespace std;
int n,m,f[30][20]; 
int main(){
    
    
	scanf("%d%d",&n,&m);
	f[1][0]=1;
	for(int j=1;j<=m;j++)
	{
    
    
		for(int i=1;i<=n;i++)
		{
    
    
			if(i==n)
				f[i][j]=f[1][j-1]+f[n-1][j-1];
			else if(i==1)
				f[i][j]=f[n][j-1]+f[2][j-1];
			else f[i][j]=f[i-1][j-1]+f[i+1][j-1];
		}
	}
	printf("%d",f[1][m]);
}