HDU 1028 Ignatius and the Princess III （母函数模板）

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本文介绍了一个编程问题“公主III”，该问题是关于给定正整数N的不同等式的数量计算。通过一个C++代码示例展示了如何使用动态规划的方法解决这个问题，提供了完整的源代码实现。

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22724 Accepted Submission(s): 15867

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

Sample Output

分析：这是一个模板题，母函数的基础只是可以详见如下链接

1、点击打开链接

2、点击打开链接

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
const int MAXN=125;
int c1[MAXN],c2[MAXN];

int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		for(int i=0;i<=n;i++)//初始化第一个表达式 
		{
			c1[i]=1;
			c2[i]=0;
		}
		for(int i=2;i<=n;i++)//从第二个表达式开始 
		{
			for(int j=0;j<=n;j++)//每个表达式中第j个变量 
			{
				for(int k=0;k+j<=n;k+=i)//第i个表达式都是有i^k表示的，所以这里表示的指数每次加 i； 
				{
					c2[j+k] += c1[j];	//暴力两个表达式的所有组合，并把能组合的情况保存下； 
				}
			}
			for(int j=0;j<=n;j++)	//将本次的两个表达式运算结果转存。 
			{
				c1[j]=c2[j];
				c2[j]=0;
			}
		}
		printf("%d\n",c1[n]);	//最后将下标 n 位置所存的系数输出。 
	}
	return 0;
}