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+,);
       vector<int>c2(n+,);
       for(i=;i<=n;i++)
       {
          for(j=;j<=n;j++)
          {
              for(k=;k+j<=n;k+=i)
              {
                c2[j+k]+=c1[j];
              }
          }
          for(j=;j<=n;j++)
          {
              c1[j]=c2[j];
              c2[j]=;
          }
       }
       cout<<c1[n]<<endl;
     }
     return ;
 }