Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 13553    Accepted Submission(s): 9590

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

#include
       
        int a[130],s[130];int main(){	int n,i,j,k;	while(scanf("%d",&n)!=EOF)	{		for(i=0;i<=n;i++)		{			a[i]=0;			s[i]=1;		}		for(i=2;i<=n;i++)		{			for(j=0;j<=n;j++)			{				for(k=0;k+j<=n;k+=i)				a[k+j]+=s[j];			}			for(j=0;j<=n;j++)			{				s[j]=a[j];				a[j]=0;			}		}		printf("%d\n",s[n]);	}	return 0;}