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

 

Author

Ignatius.L

#include<stdio.h>

#include<string.h>

#define max 100+30

int main()

{

	int c1[max],c2[max];

	int n;

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

	{

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

		{

			c1[i]=1;

			c2[i]=0;

		}

		for(int i=2;i<=n;i++)//从第二个多项式开始乘

		{

			for(int j=0;j<=n;j++)//在第一个多项式中的每一项与后边的相乘

			for(int k=0;k+j<=n;k+=i)//在第i个多项式中的每一项与前边的相乘

			c2[k+j]+=c1[j];

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

			{

				c1[j]=c2[j];//更新现在第一个多项式中的每一项的系数

				c2[j]=0;

			}

		}

		printf("%d\n",c1[n]);

	}

	return 0;

}