"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!"

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.

For each test case, you have to output a line contains an integer P which indicate the different equations you have found.

4
10
20

5
42
627

Ignatius.L

考察知识点；母函数模板

//母函数模板--整数拆分--无穷多个硬币的情况 
//事实证明：memset函数 不能为数组进行除了0，-1 外的赋值1 
#include<stdio.h>
#include<string.h>
int c1[125],c2[125];
int main()
{
	int n;
	int i,j,k;
	while(~scanf("%d",&n))
	{
		for(i=0;i<=n;++i)
		{
			c1[i]=1;
			c2[i]=0;
		}
		for(i=2;i<=n;++i)//i个表达式 
		{
			for(j=0;j<=n;++j)//前面i个表达式相乘的结果中的第j项 
			{
				for(k=0;k+j<=n;k+=i)//指数 
				{
					c2[k+j]+=c1[j];
				}
			}
			for(j=0;j<=n;++j)
			c1[j]=c2[j],c2[j]=0;
		}
		printf("%d\n",c1[n]);
	}
	return 0;
}