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
  

 

Author

Ignatius.L

 

#include<stdio.h>
int main()
{
	int i,j,k,n;
    int a[200],b[200];
    while(~scanf("%d",&n))
	{
	 for(i=0;i<=n;i++)
	 {
		a[i]=1;b[i]=0;
	 }
	 for(i=2;i<=n;i++)
	{
	 for(j=0;j<=n;j++)
	 for(k=0;k+j<=n;k+=i)
	 b[k+j]+=a[j];
	 for(int i=0;i<=n;i++)
	 {
	 	a[i]=b[i];b[i]=0;
	 }
    }
    printf("%d\n",a[n]);
    }
	return 0;
} 

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 分析：此题较为简单，母函数模板直接可以

代码 ：