POJ 1028 Ignatius and the Princess III（母函数）

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26797    Accepted Submission(s): 18456

 

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

问题描述
“嗯，第一个问题似乎太简单了。我以后会告诉你你有多愚蠢。”冯5166说。
“第二个问题是，给定一个正整数n，我们定义如下方程：
n=a[1]+a[2]+a[3]+…+a[m]；
a[i]＞0,1<＝m <＝n；
我的问题是，对于给定的n，你能找到多少个不同的方程。
例如，假设n为4，我们可以发现：
4＝4；
4＝3＋1；
4＝2＋2；
4＝2＋1＋1；
4=1+1+1+1；
当n为4时，结果是5。注意，“4=3+1”和“4=1+3”在这个问题上是相同的。现在，你来吧！”

 

 

思路：母函数的板子题目。

生成函数就是

code：

#include
#include
#include
#include

using namespace std;
const int N=200;
int sup[N],temp[N];
int x;
int main()
{
	//freopen("51.in","r",stdin);
//	freopen("51.out","w",stdout);
	while(~scanf("%d",&x))
	{
		for(int i=0; i<=x; i++)
		{
			sup[i]=1;
			temp[i]=0;
		}
		for(int i=2; i<=x; i++)
		{
			for(int j=0; j<=x; j++)
			{
				for(int k=0; k+j<=x; k+=i)
				{
					temp[k+j]+=sup[j];
				}
			}
			for(int i=0; i<=x; i++)
			{
				sup[i]=temp[i];
				temp[i]=0;
			}
		}
		printf("%d\n",sup[x]);
	}
	return 0;
}