杭电 1028 Ignatius and the Princess III

最新推荐文章于 2020-07-16 21:26:00 发布

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本文介绍了一种使用母函数求解特定组合计数问题的方法，通过三层for循环实现，解决了给定正整数N时，寻找不同方程数量的问题。例如，当N为4时，存在5种不同的方程。

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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

典型母函数的应用，套用模板的三层for循环，本来不懂母函数（看了网上dalao们的博客之后，才明白qwq）不会的话，可以搜一些母函数讲解的博客叭

#include<stdio.h>
int main()
{
	int n,c1[205],c2[205],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+=i)
				c2[j+k]+=c1[j];
			}
			for(j=0;j<=n;j++)
			{
				c1[j]=c2[j];
				c2[j]=0;
			}
		}
		printf("%d\n",c1[n]);
	}
}