hdu 1028 Ignatius and the Princess III

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9729    Accepted Submission(s): 6876

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

 母函数模板题

#include<stdio.h>
int c1[1000],c2[1000];
int main()
{
     int n,i,j,k,c;
     while(scanf("%d",&n)!=EOF)
     {
          for(i=0;i<=n;i++)
          {
               c1[i]=0;
               c2[i]=1;
          }
          for(i=1;i<=n;i++)
          {
               for(j=0;j<=n;j++)
               {
                    for(k=0;k+j<=n;k+=i)
                    {
                         c2[j+k]+=c1[j];
                    }
               }
               for(c=0;c<=n;c++)
               {
                    c1[c]=c2[c];
                    c2[c]=0;
               }
          }
          printf("%d\n",c1[n]);

     }
     return 0;
}