1 solutions

  • 0
    @ 2024-10-15 14:55:39

    这题可以使用数论,我这里用dp去解决

    #include<bits/stdc++.h>
    typedef long long LL;
    using namespace std;
    const int N = 30;
    int dp[N][N],n;
    /*
    这里dp[i][j]当中 i 表示不大于i,j表示排列排j个,即dp[4][3] 就是最大数是
    4,取三个来排 dp[4][3] = 8,dp[4][2] = 12,dp[4][4] = 2;
    最后加起来就是答案 
    */ 
    int main ()
    {
    	cin >> n;
    	for(int j = 1;j <= n;++ j)
    	{
    		for(int i = 1;i <= n;++ i)
    		{
    			if(j == 1)dp[i][j] = i * 2;//这里将所有dp[i][1]变成2 * i; 
    			else dp[i][j] = dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1];//状态转移方程; 
    		}
    	 } 
    	  
    	LL ans = 0;
    	for(int i = 2;i <= n;++ i)
    	{
    		ans += dp[n][i];
    	} 
    	cout << ans;
    	
    	
    	
    	return 0;
    }
    
    
    • 1

    Information

    ID
    1183
    Time
    1000ms
    Memory
    512MiB
    Difficulty
    8
    Tags
    # Submissions
    31
    Accepted
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    Uploaded By