hdu-1028 Ignatius and the Princess III 【母函数】

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
);
    }
    return 0;
}