HDU1028（母函数入门题＋模板）

母函数早就想解决这个问题了，今天终于看明白模板了（自己理解这写的，虽然不对看标准改对的，但...）

 

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

 1 #include<stdio.h>
 2 int save[150],temp[150];
 3 int main()
 4 {
 5     int n;
 6     while(~scanf("%d",&n))
 7     {
 8         for(int i=0;i<=n;i++)
 9         {
10             save[i]=1;
11             temp[i]=0;
12         }
13         for(int i=2;i<=n;i++){
14             for(int j=0;j<=n;j++)
15                 for(int k=0;k+j<=n;k+=i)
16                 temp[k+j]+=save[j];
17             for(int k=0;k<=n;k++)
18             {
19                 save[k]=temp[k];
20                 temp[k]=0;
21             }
22         }
23         printf("%d\n",save[n]);
24     }
25 }

有个人的母函数讲解的很好哦（就是有点问题）我会转一下的