1 solutions

  • 0
    @ 2025-5-9 19:18:31
    #include<bits/stdc++.h>
    using namespace std;
    const int N=25;
    int f[N]={1,1}; //f[i]表示i+2条边的方案数 
    int main()
    {
    	for(int i=2;i<N;i++) //每次多一条边 
    	{
    		for(int j=0;j<i;j++) //左边的三角形方案数 
    		{
    			f[i]+=f[j]*f[i-j-1];  //组合计数 
    		}
    	}
    	int n;
    	cin>>n;
    	cout<<f[n-2]; //n条边的总方案数 
    	return 0;
    }
    
    • 1

    Information

    ID
    194
    Time
    1000ms
    Memory
    256MiB
    Difficulty
    3
    Tags
    (None)
    # Submissions
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