hdu1028——Ignatius and the Princess III

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16849    Accepted Submission(s): 11857

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
   
   
   
   

 
典型的用母函数解决的问题，寒假学的忘得差不多了，今天又一看，还行哈！

#include<iostream>
#include<iomanip>
#include<queue>
#include<cmath>
#define size 10000
using namespace std;

void solve(int n)
{
	int c1[size],c2[size];
	for(int i=0;i<size;i++)
	{
		c1[i]=1;
		c2[i]=0;
	}
	
	int i,j,k,p;
	
	for(i=2;i<=n;i++)
	{
		for(j=0;j<=n;j++)
		{
			for(k=0;k+j<=n;k+=i)
			{
				c2[j+k]+=c1[j];
			}
		}
		
		for(p=0;p<=n;p++)
		{
			c1[p]=c2[p];
			c2[p]=0;
		}
	}
	cout<<c1[n]<<endl;
}

int main()
{
	int n;
	while(cin>>n)
	{
		solve(n);	
	}
	return 0;
}

加油，加油！