Ignatius and the Princess III

http://acm.hdu.edu.cn/showproblem.php?pid=1028

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5126    Accepted Submission(s): 3602

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

 

典型的母函数题目。

#include
int main()
{
    int n;
    int i,j,k;
    int c1[130],c2[130];   //c1,c2 都是系数
    while(scanf("%d",&n)!=EOF)
    {
        for(i=0;i<=n;i++)
        {
            c1[i]=1;c2[i]=0;  //这首先对c1初始化，由第一个表达式(1+x+x2+..xn)初始化，把质量从0到n的所有砝码都初始化为1. 
        }
        for(i=2;i<=n;i++) //i从2到n遍历，这里i就是指第i个表达式，上面给出的第二种母函数关系式里，每一个括号括起来的就是一个表达式。 

        {
            for(j=0;j<=n;j++) //j 从0到n遍历，这里j就是(前面i個表达式累乘的表达式)里第j个变量

                for(k=0;j+k<=n;k+=i)  //  k表示的是第j个指数，所以k每次增i（因为第i个表达式的增量是i）
                    c2[j+k]+=c1[j]; 
            for(j=0;j<=n;j++)
            {
                c1[j]=c2[j];    //把c2的值赋给c1,而把c2初始化为0，因为c2每次是从一个表达式中开始的 

                c2[j]=0;   
            }
        }
        printf("%d\n",c1[n]);
    }
    return 0;
}

//这里的值是这样变化的,1,2,3,4,……n，那么第i个表达式的增量是i, 可以这么理解

第一个表达式是(1+x+x^2+x^3+……  )  ，幂数： 1*0, 1*2, 1*3, 1*4, ……

第二个表达式是(1+x^2+x^4+x^6+……) ，幂数：2*0 , 2*1,2*2,2*3,……

……

如果是这样变化的，1,4,9,……n^2 ，那么第i个表达式的增量是i^2;

其他的请看下篇。