hduoj1028！【母函数】

/*
Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12813    Accepted Submission(s): 9066

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
*/
#include<stdio.h>
#include<string.h>
int main()
{
	int i, j, k, n, a[121], b[121];
	for(i = 0; i < 121; i++)
	a[i] = 1;
	for(i = 2;i  < 121; i++)
	{
		memset(b, 0, sizeof(b));
		for(j = 0;j < 121; j++)
		for(k = 0; k < 121; k += i)
		b[k+j]  += a[j];
		for(j =0; j < 121; j++)
		a[j] = b[j];
	}
	while( scanf("%d", &n) != EOF && n)
	printf("%d\n", a[n]);
	return 0;
}

最基础的母函数。