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# Ignatius and the Princess III

** Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K(Java/Others)Total Submission(s): 14254 Accepted Submission(s): 10028
**

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<stdio.h>
#include<iostream>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
#include<algorithm>
#include<vector>
#include<string.h>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<sstream>
#include<time.h>
#include<utility>
#include<malloc.h>
#include<stdexcept>
using namespace std;
int n;
int c1[10000];
int c2[10000];
int main ()
{
int n;
while (scanf ("%d",&n)!=EOF )
{
for (int i=0;i<=n;i++)
{
c1[i]=1;
c2[i]=0;
}
for (int i=2;i<=n;i++)
{
for (int j=0;j<=n;j++)
{
for (int k=0;k+j<=n;k+=i)
{
c2[j+k]+=c1[j];
}
}
for (int j=0;j<=n;j++)
{
c1[j]=c2[j];
c2[j]=0;
}
}
printf("%d\n",c1[n]);
}
return 0;
}
```