HDU 1028 Ignatius and the Princess III

/*
中文题意：求组成一个数有多少种方案
中文翻译：4的组成方案有5种，是赤露露的母函数
题目大意：求一个数有多少种组成他的方案
解题思路：直接套用母函数的模板
难点具体解释：见母函数模板的博客
关键点：了解母函数
解题人：lingnichong
解题时间：2014/08/09    9:44
解题感受：母函数的应用，如今还不是很清楚这题是怎样AC的，仅仅是用了母函数的模板，他就AC了
*/

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 12766    Accepted Submission(s): 9024

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
   
     

 

#include<stdio.h>
#include<string.h>
#define MAXN 150
int c1[MAXN],c2[MAXN];
int main()
{
	int n,i,j,k;
	while(~scanf("%d",&n))
	{
		memset(c2,0,sizeof(c2));
		for(i=0;i<=n;i++)
			c1[i]=1;
		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(j=0;j<=n;j++)
			{
				c1[j]=c2[j];
				c2[j]=0;
			}
		}
		printf("%d\n",c1[n]);
	}
	return 0;
}