hdu1028

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5903    Accepted Submission(s): 4166

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
   
   
   
   

 

这个是一个母函数问题。可以看我博客里买你转载的母函数详解。。

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
    long long  n;
    int c1[120+1];
    int c2[120+1];
    while(cin>>n)
    {
        for(int j=0;j<=n;j++)
        {
            c1[j]=1;
            c2[j]=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[j+k]+=c1[j];
                }
            }
            for(int j=0;j<=n;j++)
            {
                c1[j]=c2[j];
                c2[j]=0;
            }
        }
        cout<<c1[n]<<endl;
    }
    return 0;
}