HDOJ 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): 9532    Accepted Submission(s): 6722

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 <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int maxn=200;

int c1[maxn],c2[maxn];

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;j+k<=n;k+=i)
                c2[k+j]+=c1[j];
        }

        for(int j=0;j<=n;j++)
        {
            c1[j]=c2[j];  c2[j]=0;
        }
    }
    printf("%d
",c1[n]);
}

    return 0;
}