hdu（1028）（母函数）

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20764    Accepted Submission(s): 14494

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

题意：输一个书n，求n有多少种组成方式

思路：基础母函数模板，不懂的可以看看上一篇博客

#include 
using namespace std;
int main()
{
    long long int c1[1500],c2[1500];  //别开小了 
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<=n;i++)
        {
            c1[i]=1;               //最终的值（x的i次方的系数） 
            c2[i]=0;              //临时的情况 
        }
        for(int i=2;i<=n;i++)    // 每次的增量 
        {
            for(int j=0;j<=n;j++)  
            {
                for(int k=0;k+j<=n;k+=i)
                c2[k+j]=c2[k+j]+c1[j];
            }
            for(int j=0;j<=n;j++)
            {
                c1[j]=c2[j];
                c2[j]=0;
            }
        }
        
        printf("%lld\n",c1[n]);
    } 
}