母函数模板 HDU-1028

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9464    Accepted Submission(s): 6671

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

代码模板：

    
    
    
    
#include <iostream>
using namespace std;
const int lmax=300;
int c1[lmax+1],c2[lmax+1];
int main()
{
int n,i,j,k;
while (cin>>n)
{
   for (i=0;i<=n;i++)
   {
       c1[i]=1;
       c2[i]=0;
   }
   for (i=2; i<=n; i++)
   {
     for (j=0; j<=n; j++)
       for (k = 0; k + j <= n; k += i)
         {
          c2[j + k] += c1[j];
         }
       for (j = 0; j <= n; j++)
            {
            c1[j] = c2[j];
            c2[j]=0;
         }
    }
 cout<<c1[n]<<endl;
}
return 0;
}