hdu 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): 12758    Accepted Submission(s): 9016

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

 

/*题解：
母函数问题， 这里可视为母函数模板. 
*/

#include<cstdio>#include<cstring>int main(){    int i,j,k,n,c1[1010],c2[1010];    while(scanf("%d",&n)!=EOF)    {        memset(c2,0,sizeof(c2));        for(i=0; i<=n; i++)        c1[i]=1;                      //在未乘时，每个未知数前的系数为1         for(i=2; i<=n; i++)           //括号数         {                      for(j=0; j<=n; j++)        //j和k均为相乘时双方的指数              {                   for(k=0; j+k<=n; k+=i)                {                    c2[j+k]+=c1[j];   //c1,c2存储的是系数                 }            }            for(j=0; j<=n; j++)            {                c1[j]=c2[j];         //c2只是临时储存，在这里释放，将c2值赋给c1;                 c2[j]=0;            }         }        printf("%d\n",c1[n]);    }    return 0;}