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): 11783    Accepted Submission(s): 8343

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