HDU 1028 Ignatius and the Princess III 母函数

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

code:

#include<stdio.h>
int main()
{
    int c1[200],c2[200]; 
    int n,i,j,k;
    while(scanf("%d",&n)!=EOF)
    {
        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;
            }
        }
        printf("%d\n",c1[n]);
            
    }

    return 0;
}