hdoj 1028 Ignatius and the Princess III

//1174393 2009-03-23 16:06:04 Accepted 1028 15MS 152K 566 B 
/*母函数
一个数拆为几个数之和的拆法几种
4=4;4=3+1;4 = 2 + 2;4 = 2 + 1 + 1;4 = 1 + 1 + 1 + 1;有4种
Sample Input        Sample Output
4                    5
10                    42
20                    627
 */
#include <stdio.h>
 int main()
{
    int i,j,k,n;
    int a[121],b[121];
    while(scanf("%d",&n)!=EOF)
    {//n为待拆解的数
         for(i=0;i<=n;i++)
        {
            b[i]=0;//存放结果
             a[i]=1;//中间过程初始化
         }
        for(i=2;i<=n;i++)
        {
            for(j=0;j<=n;j++)
                for(k=0;k+j<=n;k+=i)
                    b[j+k]+=a[j];
            for(j=0;j<=n;j++)
            {
                a[j]=b[j];
                b[j]=0;
            }
        }
        printf("%d\n",a[n]);
    }return 0;
}

 

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

41020

 

 

Sample Output

542627