HDOJ 题目1028Ignatius and the Princess III（母函数模板）

Ignatius and the Princess III

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

Total Submission(s): 12539    Accepted Submission(s): 8851


[align=left]Problem Description[/align]
"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!"

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
For each test case, you have to output a line contains an integer P which indicate the different equations you have found.

 

[align=left]Sample Input[/align]

4
10
20

 

[align=left]Sample Output[/align]

5
42
627

 

[align=left]Author[/align]
Ignatius.L
 
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1028

思路。。http://www.cppblog.com/MiYu/archive/2010/08/05/122290.html
http://wenku.baidu.com/view/d0e5cc4433687e21af45a9a8.html

母函数好神奇啊

#include<stdio.h>
#include<string.h>
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
__int64 c1[125],c2[125];
int i,j,k;
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[k+j]+=c1[j];
}
}
for(j=0;j<=n;j++)
{
c1[j]=c2[j];
c2[j]=0;
}
}
printf("%I64d\n",c1
);
}
}