HDU Ignatius and the Princess III (母函数)

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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

41020

Sample Output

542627

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