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

 

/*
    母函数模板
    http://blog.csdn.net/vsooda/article/details/7975485  详解 
*/
#include<cstdio>
#include<iostream>
#define MAXN 121

using namespace std;

int c1[MAXN];
int c2[MAXN];

int n;

int main() {
    while(~scanf("%d",&n)) {
        for(int i=0;i<=n;i++) {
            c1[i]=1;c2[i]=0;
        }
        for(int i=2;i<=n;i++) {
            for(int j=0;j<=n;j++)
              for(int k=0;k+j<=n;k+=i)
                c2[j+k]+=c1[j];
            for(int j=0;j<=n;j++) {
                c1[j]=c2[j];
                c2[j]=0;
            }
        }
        printf("%d
",c1[n]);
    }
    return 0;
}