Ignatius and the Princess III HDU 1028

Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10267    Accepted Submission(s): 7281

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

#include <iostream>using namespace std;int main(){    int n,c1[121],c2[121];    while(cin>>n&&n)    {        for(int i=0;i<=n;i++)        {            c1[i]=1;c2[i]=0;        }        for(int i=2;i<=n;i++)        {            for(int j=0;j<=n;j++)            {                for(int k=0;k+j<=n;k+=i)                {                    c2[k+j]+=c1[j];                }            }            for(int j=0;j<=n;j++)            {                c1[j]=c2[j];c2[j]=0;            }        }        cout<<c1[n]<<endl;    }    return 0;}