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): 7012    Accepted Submission(s): 4968

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 */

#include<iostream>
#include<stdio.h>
using namespace std;
//const int max=10001;//注意不能用这个
#define max 10000000
int c1[max];
int c2[max];
int main()
{
    int i,j,k,n;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=0;i<=n;i++)
            c1[i]=1;
        for(i=2;i<=n;i++)
        {
            for(j=0;j<=n;j++)
                for(k=0;j+k<=n;k+=i)
                c2[j+k]+=c1[j];
            for(j=0;j<=n;j++)
            {
                c1[j]=c2[j];
                c2[j]=0;
            }
        }
        printf("%d\n",c1[n]);
    }
    return 0;
}