母函数入门 HUD 1028 整数拆分

最新推荐文章于 2023-06-20 14:35:00 发布

最新推荐文章于 2023-06-20 14:35:00 发布 · 80 阅读

本文介绍了一种计算给定正整数N的不同分解方程数量的方法。通过使用动态规划技巧，实现了一个O(n^2)的时间复杂度解决方案。实例演示了当N为4时，可以找到5种不同的等式。

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
 

http://www.wutianqi.com/?p=596

这篇博文讲的很好！

#include <stdio.h>
int n;
int arr[130],tmp[130],i,j,k;
int main(){
	//freopen("in.txt","r",stdin);
	n = 120;
	for(i=0; i<=n; i++)
	{
		arr[i] = 1;
		tmp[i] = 0;
	}
	for(i=2; i<=n; i++)
	{
		for(j=0; j<=n; j++)
		{
			for(k=0; k+j <=n;  k += i)
			{
				tmp[k+j] += arr[j];
			}
		}
		for(j=0; j<=n; j++){
			arr[j] = tmp[j];
			tmp[j] = 0;
		}
		}
	while(scanf("%d",&n) != EOF){
	
		printf("%d\n",arr[n]);
	}

	return 0;
}