杭电1028 Ignatius and the Princess III（母函数）

最新推荐文章于 2020-07-16 21:26:00 发布

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最新推荐文章于 2020-07-16 21:26:00 发布

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分类专栏：母函数模板文章标签：母函数

本文链接：https://blog.youkuaiyun.com/u013634213/article/details/38445421

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本文探讨了一个关于给定正整数N的不同等式数量的问题，通过一个具体的例子说明了如何寻找所有可能的组合，并提供了一段C语言实现的代码来解决这个问题。

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Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12747 Accepted Submission(s): 9010

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//母函数模板
#include<stdio.h>
#include<string.h>
const int MAX=1000;
int c1[MAX],c2[MAX];
int main()
{
	int N;
	while(scanf("%d",&N)!=EOF)
	{
		memset(c1,0,sizeof(c1));
		memset(c2,0,sizeof(c2));
		for(int i=0;i<=N;i++)
			c1[i]=1;
		for(int i=2;i<=N;i++)
		{
			for(int j=0;j<=N;j++)
			{
				for(int k=0;k+j<=N;k+=i)
					c2[j+k]+=c1[j];
			}
			for(int j=0;j<=N;j++)
			{
				c1[j]=c2[j];
				c2[j]=0;
			}
		}
		printf("%d\n",c1[N]);
	}
return 0;
}