HDU Ignatius and the Princess III (母函数)

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

Author

Ignatius.L

 1 /*
 2     母函数模板
 3     http://blog.csdn.net/vsooda/article/details/7975485  详解
 4 */
 5 #include<cstdio>
 6 #include<iostream>
 7 #define MAXN 121
 8
 9 using namespace std;
10
11 int c1[MAXN];
12 int c2[MAXN];
13
14 int n;
15
16 int main() {
17     while(~scanf("%d",&n)) {
18         for(int i=0;i<=n;i++) {
19             c1[i]=1;c2[i]=0;
20         }
21         for(int i=2;i<=n;i++) {
22             for(int j=0;j<=n;j++)
23               for(int k=0;k+j<=n;k+=i)
24                 c2[j+k]+=c1[j];
25             for(int j=0;j<=n;j++) {
26                 c1[j]=c2[j];
27                 c2[j]=0;
28             }
29         }
30         printf("%d\n",c1[n]);
31     }
32     return 0;
33 }

代码