【杭电oj】1028 - Ignatius and the Princess III（母函数打表）

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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): 18586 Accepted Submission(s): 13020

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

Author

Ignatius.L

代码如下：

#include <cstdio>
#include <cstring>
#define MAX 120
int main()
{
	int c[MAX+10];
	int t[MAX+10];
	memset (c,0,sizeof (c));
	memset (t,0,sizeof (t));
	for (int i = 0 ; i <= MAX ; i++)
		c[i] = 1;
	for (int i = 2 ; i <= MAX ; i++)
	{
		for (int j = 0 ; j <= MAX ; j++)
			for (int k = 0 ; k+j <= MAX ; k += i)
				t[j+k] += c[j];
		for (int j = 0 ; j <= MAX ; j++)
		{
			c[j] = t[j];
			t[j] = 0;
		}
	}
	int n;
	while (~scanf ("%d",&n))
		printf ("%d\n",c[n]);
	return 0;
}