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\n",c1[n]);
	}
	return 0;
}