ACM hdu 1028 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): 11692    Accepted Submission(s): 8275

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

没有分析，这就是明显用母函数的最简单的题，用模板就可以过

#include <iostream>
using namespace std;
const int _max = 10001;
int c1[_max], c2[_max];
int main()
{

    int nNum;   
    int i, j, k;
    while(cin >> nNum)
    {
        for(i=0; i<=nNum; ++i)
        {
            c1[i] = 1;
            c2[i] = 0;
        }
        for(i=2; i<=nNum; ++i)
        {
            for(j=0; j<=nNum; ++j) 
                for(k=0; k+j<=nNum; k+=i)  
                {
                    c2[j+k] += c1[j];
                }
            for(j=0; j<=nNum; ++j) 
                c1[j] = c2[j];
                c2[j] = 0;
            }
        }
        cout << c1[nNum] << endl;
    }
    return 0;
}