hdu- 1028 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
  
  


  
  


  
  

   
最基础的最简单的母函数的模板题目，
如果不会的话，最起码先要把模板给记住。。。
代码：
#include<stdio.h>
#include<string.h>
int main(){
	int a[150];
    int b[200];
	int N;
	int i,j,k,t;
	while(~scanf("%d",&N)){
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		for(i=0;i<=N;i++){
			a[i]=1;
		}
		for(i=2;i<=N;i++){
			for(j=0;j<=N;j++)
				for(k=0;k+j<=N;k+=i)
				   b[j+k]+=a[j];
			
			for(t=0;t<=N;t++){
			a[t]=b[t];
            b[t]=0;
		}
	}
	//	int sum=0;
	
		printf("%d\n",a[N]);
	}
	return 0;
}