hdoj 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): 14538 Accepted Submission(s): 10236



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

第一道母函数：

#include<stdio.h>
#include<string.h>
#define max 100+30
int main()
{
    int c1[max],c2[max];
    int n,i,j,k;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=0;i<=n;i++)
        {
            c1[i]=1;
            c2[i]=0;
        }
        for(i=2;i<=n;i++)//从第二个到最后一个 多项式 
        {
            for(j=0;j<=n;j++)//更新第一个多项式的 n+1个代数式的系数 
            {
                for(k=0;k+j<=n;k+=i)//限***用   i为当前多项式步长 (比如说:1 x2 x4 x6 步长就为2) 
                {
                    c2[k+j]+=c1[j];
                }
            }
            for(j=0;j<=n;j++)
            {
                c1[j]=c2[j];//更新 
                c2[j]=0;//初始化 
            }
        }
        printf("%d\n",c1
); 
    }
    return 0;
}