HDU 1028 Ignatius and the Princess III——————数字拆分，母函数用法

Ignatius and the Princess III

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

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问题是给出一个N,可以找出多少个不同的等式

例如，当N = 4时，我们可以找到：

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 133;
int C1[MAXN];//
int C2[MAXN];//临时存放
int n;
int main()
{
        // freopen("../in.txt","r",stdin);
        // freopen("../out.txt","w",stdout);
        while(~scanf("%d",&n))
        {
                memset(C2,0,sizeof(C2));
                for(int i=0;i<=n;++i)   C1[i] = 1;
                for(int i=2;i<=n;++i)//从第2个多项式开始相乘
                {
                        for(int j=0;j<=n;++j)// 已经得出结果的前i项
                                for(int k=0;k+j<=n;k+=i)// k 表示下一项的指数，指数每次增加i
                                        C2[k+j] += C1[j];
                        memcpy(C1,C2,sizeof(C2));
                        memset(C2,0,sizeof(C2));
                }
                cout<<C1[n]<<endl;
        }
        return 0;
}