HDUOJ----Ignatius and the Princess III

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9893    Accepted Submission(s): 6996

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

 

母函数.....对于任意一个数你   （1+x+x^2+x^3+x^4+x^5+x^6+x^7...+x^n）*(1+x^2+x^4+x^6+x^8+x^10+.....)*(1+x^3+.....);

#include<iostream>
#include<vector>
using namespace std;
int main()
{
    int n,i,j,k;
    while(cin>>n)
    {
      vector<int>c1(n+1,1);
      vector<int>c2(n+1,0);
      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=1;j<=n;j++)
         {
             c1[j]=c2[j];
             c2[j]=0;
         }
      }
      cout<<c1[n]<<endl;
    }
    return 0;
}