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  • hdoj--1028--Ignatius and the Princess III(母函数)

    Ignatius and the Princess III

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 16242    Accepted Submission(s): 11445


    Problem Description
    "Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says.

    "The second problem is, given an positive integer N, we define an equation like this:
      N=a[1]+a[2]+a[3]+...+a[m];
      a[i]>0,1<=m<=N;
    My question is how many different equations you can find for a given N.
    For example, assume N is 4, we can find:
      4 = 4;
      4 = 3 + 1;
      4 = 2 + 2;
      4 = 2 + 1 + 1;
      4 = 1 + 1 + 1 + 1;
    so the result is 5 when N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" is the same in this problem. Now, you do it!"
     

    Input
    The input contains several test cases. Each test case contains a positive integer N(1<=N<=120) which is mentioned above. The input is terminated by the end of file.
     

    Output
    For each test case, you have to output a line contains an integer P which indicate the different equations you have found.
     

    Sample Input
    4 10 20
     

    Sample Output
    5 42 627
     

    Author
    Ignatius.L
     
    #include<stdio.h>
    #include<string.h>
    #define max 100+30
    int main()
    {
    	int c1[max],c2[max];
    	int n;
    	while(scanf("%d",&n)!=EOF)
    	{
    		for(int i=0;i<=n;i++)
    		{
    			c1[i]=1;
    			c2[i]=0;
    		}
    		for(int i=2;i<=n;i++)//从第二个多项式开始乘 
    		{
    			for(int j=0;j<=n;j++)//在第一个多项式中的每一项与后边的相乘 
    			for(int k=0;k+j<=n;k+=i)//在第i个多项式中的每一项与前边的相乘 
    			c2[k+j]+=c1[j];
    			for(int j=0;j<=n;j++)
    			{
    				c1[j]=c2[j];//更新现在第一个多项式中的每一项的系数 
    				c2[j]=0;
    			}
    		}
    		printf("%d
    ",c1[n]);
    	}
    	return 0;
    }


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  • 原文地址:https://www.cnblogs.com/playboy307/p/5273765.html
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