Ignatius and the Princess III

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

　　　　　　　　　　　　　　　　　　     Total Submission(s): 9408    Accepted Submission(s): 6642

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

 

题目链接：

 

题目分析：题目要求的是整数的拆分方案个数，用母函数处理。

 

母函数详解链接：

 

1 #include 
       
         2  3 int N; 4 const int MAX_NUM = 125; 5 int   solNum[MAX_NUM]; 6 int   temp[MAX_NUM]; 7  8 void InitGenerFun() 9 {10     for(int i = 0; i < MAX_NUM; i++)11     {12         solNum[i] = 1;13         temp[i]   = 0;14     }15     for(int i = 2; i < MAX_NUM; i++)16     {17         for(int j = 0; j < MAX_NUM; j++)18         {19             for(int k = 0; k+j < MAX_NUM; k+=i)20             {21                 temp[k+j] += solNum[j];22             }23         }24         for(int k = 0; k < MAX_NUM; k++)25         {26             solNum[k] = temp[k];27             temp[k]   = 0;28         }29     }30 }31 32 int main()33 {34     InitGenerFun();35 36     while(scanf("%d", &N) != EOF)37     {38         printf("%d\n", solNum[N]);39     }40 41     return 0;42 }