母函数 HDU 1028

Ignatius and the Princess III

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 100   Accepted Submission(s) : 83

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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

Author

Ignatius.L

#include<iostream>
 #include<stdio.h>
 using namespace std;
 int main()
 {
     int i,k,j,n;
     int a[121],b[121];
    while(scanf("%d",&n)!=EOF)
     {
        for(i=0;i<=n;i++)
        {
            a[i]=1;
            b[i]=0;
        }
        for(i=2;i<=n;i++)
        {
            for(j=0;j<=n;j++)//(1+x+x2+x3..xn)
            for(k=0;k+j<=n;k+=i)//(1+x2+x4+x6..) 两层for循环　是用第一个循环中ｘ的ｊ次　来乘第二个循环中ｘ的ｋ次
                 b[j+k]+=a[j];
            for(j=0;j<=n;j++)
            {
                a[j]=b[j];
                b[j]=0;
            }
        }
        printf("%d\n",a[n]);
    }
    return 0;
}