HDU1028 Ignatius and the Princess III（母函数）_1028 ignatius and the princess iii 母函数-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_41594825/article/details/82937686

本文介绍了一种使用母函数解决HDU1028问题的方法，该问题要求找出给定正整数N的所有不同方程表示方式。通过详细解析代码，展示了如何运用母函数进行组合计算。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26149 Accepted Submission(s): 18059

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

方法：

此题应用母函数来解决，母函数就是常常用来解决组合方面的题目。母函数相关内容讲解：讲解1 讲解2

代码：

#include <iostream>
using namespace std;

int main()
{
	int result[121];
	int temp[121];
	int target;
	
	while (cin >> target)
	{
		for (int i = 0; i <= target; i++)
		{
			result[i] = 1;
			temp[i] = 0;
		}
		
		for (int k = 2; k <= target; k++)
		{
			for (int i = 0; i <= target; i++)
				for (int j = 0; j + i <= target; j += k)
					temp[i+j] += result[i];
			
			for (int i = 0; i <= target; i++)
			{
				result[i] = temp[i];
				temp[i] = 0;
			}
		}
		
		cout << result[target] << endl;
	}
	
	return 0;
}