HDU/HDOJ 1028 母函数 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): 4800Accepted Submission(s): 3362

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

这个题有人说是DP，但是我最近在看母函数，所以用母函数过掉了

这个应该是母函数的入门题目了吧。。

我的代码：

#include<stdio.h>

int c1[125];
int c2[125];

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