hdu1028

                Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9963    Accepted Submission(s): 7052

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

#include<iostream>
using namespace std;
int main()
{
    int n,i,j,k,c1[10000],c2[10000];
    while(scanf("%d",&n)!=EOF && n)
    {
        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[j+k] += c1[j];
                }
                
            }
            for(j=0;j<=n;j++)
            {
                c1[j] = c2[j];
                c2[j] = 0;
            }
        }
        printf("%d
",c1[n]);
    }
    return 0;
}