HDU 1028 Ignatius and the Princess III(母函数)

题目：HDU-1028 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1028

题目：

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16806    Accepted Submission(s): 11830

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
   
   
   
   

 

这道题目很简单，直接构建母函数就行了，不懂得看上一篇博客讲的原理吧。http://blog.csdn.net/qq_33171970/article/details/50663426

代码~~

#include<iostream>
#include<cstring>
#include<algorithm>
#include<math.h>
#include<cstdio>
using namespace std;
const int maxn= 125;
int a[maxn],b[maxn];
int n;
int main(){
	while(cin>>n){
		for(int i=0;i<=n;i++){            //依旧不限制个数合并起来写
			a[i]=1;
			b[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)
					b[j+k]+=a[j];
			for(int j=0;j<=n;j++){
				a[j]=b[j];
				b[j]=0;
			}
		}
		cout<<a[n]<<endl;
	}
	return 0;
}

啦啦啦，好好学习天天向上啦~