hdu 1028

A - Ignatius and the Princess III

Time Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u

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

41020 

 

Sample Output

542627 

        这道题方法很多，但是这里主要提一下母函数的概念。母函数可以快速得出所给条件的方案数，其他的博客都有介绍，这里综合我的理解来说明一下。所谓母函数即是将条件中用于组合的单个数目进行情况的列举。比如我可以取无数个3，那么得出的子函数就是1+x^3+x^6....，那么这里就例举出了3的所有情况（注意如果不取的1情况），那么母函数就是将这些组合数的子函数乘起来，展开后找到需要的那一项（即指数与目标一致的项）取其系数就是方案数，这里就不展示代码了。

代码引用其他的博客：http://blog.sina.com.cn/s/blog_79b832820100vc2d.html