hduoj1028 母函数

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19522    Accepted Submission(s): 13654

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

母函数模板问题：url:http://baike.baidu.com/link?url=NfgcbS_kFQBwq1vutjq_IYjEFeuoo4dRnVNlonZgKbLhUWwHpG88x3t-RZ_9zhtXW-EAQHl2ExMFXoUbfZ4zSZzEeRmIgk0ZbnlaPFA3OCsbHaBDZ81hWK59_xw4kuYu

这里给出模板：

#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
const int _max = 10001;

// c1是保存各项质量砝码可以组合的数目

// c2是中间量，保存每一次的情况

int c1[_max], c2[_max];
int main()
{
int n,i,j,k;

int nNum;

while(cin >> nNum)

{

for(i=0; i<=nNum; ++i) // ---- ①

{

c1[i] = 1;

c2[i] = 0;

}

for(i=2; i<=nNum; ++i) // ----- ②

{

for(j=0; j<=nNum; ++j) // ----- ③

for(k=0; k+j<=nNum; k+=i) // ---- ④

{

c2[j+k] += c1[j];

}

for(j=0; j<=nNum; ++j) // ---- ⑤

{

c1[j] = c2[j];

c2[j] = 0;

}

}

cout << c1[nNum] << endl;

}

return 0;

}