hdu 1028 Ignatius and the Princess III 好水的整数拆分，零数可重复

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3433    Accepted Submission(s): 2393

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
#include
using namespace std;
int coi[125],cnt[125];
int main()
{
    for(int i=1;i<125;i++) coi[i]=i;
    cnt[0]=1;
    for(int i=1;i<125;i++)
    {
        for(int j=coi[i];j<125;j++)
        {
            cnt[j]+=cnt[j-coi[i]];
        }
    }
    int n;
    while(scanf("%d",&n)==1)
    {
        printf("%d/n",cnt[n]);
    }
    return 0;
}