HDU 1028 Ignatius and the Princess III

Ignatius and the Princess III

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

解：简单的母函数问题

 

#include<iostream>

using namespace std;

#define M 125

int a[M],b[M];

int main()

{

    int n;

    while(scanf("%d",&n)!=EOF)

    {

        int i,j,k;

        for(i=0;i<=n;i++)

         {

             a[i]=1;

             b[i]=0;

         }

         for(i=2;i<=n;i++)

         {

          for(j=0;j<=n;j++)

          {

              for(k=0;k+j<=n;k+=i)

              {

                  b[k+j]+=a[j];

              }

          }

          for(j=0;j<=n;j++)

          {

              a[j]=b[j];

              b[j]=0;

          }

         }

         cout<<a[n]<<endl;

    }

    

    return 0;

}

 

^^^转载请注明出处~~~