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.

输入样例

4
10
20
 

输出样例

5
42
627

利用幂次运算代表组合
附ac代码

#include<bits/stdc++.h>
using namespace std;
const int N=130;
int c1[N],c2[N];
int main()
{
    int n;
    while(scanf("%d",&n)==1)
    {
        for(int i=0;i<=n;++i)
        {
            c1[i]=1;c2[i]=0;//c1代表当前式子各次系数
        }
        for(int i=2;i<=n;++i)//进行n-1次合并
        {
        for(int j=0;j<=n;++j)//各次幂更新
          for(int k=0;k+j<=n;k+=i)//每次更新的限制
              c2[j+k]+=c1[j];
          for(int j=0;j<=n;++j)
           {
               c1[j]=c2[j];c2[j]=0;//更新完毕保存
           }
         }
        printf("%d\n",c1[n]);
    }
    return 0;
}