神奇的母函数（一） hdoj 1208 Ignatius and the Princess III 【母函数】

感谢TankyWoo大大的详细解说，易懂，全面，详细！！！链接http://www.cppblog.com/MiYu/archive/2010/08/05/122290.html

粘一下题目和代码表示水过~~~ （这道是无限取的）

Ignatius and the Princess III

[b]Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 12505 Accepted Submission(s): 8827

[/b]

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

代码：

#include<stdio.h>
 #define max  150
 int c1[max], c2[max];
 int main()
 {
 	int n, i, j, k;
 	while(scanf("%d", &n) == 1){
 		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 <= n; k+=i )
		 		c2[j+k] += c1[j];
			 }
			 for(k = 0; k <= n; k ++){
			 	c1[k] = c2[k];
			 	c2[k] = 0;
			 }
		 }
		 printf("%d\n", c1
 );
	 }
 	return 0;
 }