HDOJ 1028 Ignatius and the Princess III （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): 9532    Accepted Submission(s): 6722

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 <iostream>
 #include <cstdio>
 #include <cstring>

 using namespace std;

 const int maxn=;

 int c1[maxn],c2[maxn];

 int main()
 {
     int n;
 while(scanf("%d",&n)!=EOF)
 {
     for(int i=;i<=n;i++)
     {
         c1[i]=; c2[i]=;
     }

     for(int i=;i<=n;i++)
     {
         for(int j=;j<=n;j++)
         {
             for(int k=;j+k<=n;k+=i)
                 c2[k+j]+=c1[j];
         }

         for(int j=;j<=n;j++)
         {
             c1[j]=c2[j];  c2[j]=;
         }
     }
     printf("%d\n",c1[n]);
 }

     return ;
 }