HDU 1028 Ignatius and the Princess III

"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

 

#include <iostream>
#include <cstdio>
using namespace std;
int a[123],b[123];
int main()
{//freopen("in.txt","r",stdin);
 int i,j,k;
 for(i=0;i<=120;i++)
   a[i]=b[i]=1;
    for(k=2;k<=120;k++)
    {
      for(j=k;j<=120;j+=k)  //母函数基本模板
         for(i=0;i+j<=120;i++)
               a[i+j]+=b[i];
      for(i=0;i<=120;i++)
              b[i]=a[i];
    }
while(scanf("%d",&i)!=EOF)
   printf("%d\n",a[i]);
 return 0;
}