hdu1028 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): 880    Accepted Submission(s): 652

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

41020

 

Sample Output

542627

 

Author

Ignatius.L

#include<iostream>

using namespace std;

const int m=120;

int c1[m+1],c2[m+1];

int main()

{    

   int n,i,j,k;    

   while(cin>>n)   

    {        

      for(i=0;i<=n;i++)

         {

             c1[i]=1;

             c2[i]=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=0;j<=n;j++)

             {c1[j]=c2[j];c2[j]=0;}

         }

         cout<<c1[n]<<endl;

     }

    return 0;

 }