Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11475 Accepted Submission(s): 8118

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

解題思路：

想了好半天，終於把母函數理解的差不多了。不同的題目需要有不同的母函數，該題的母函數為

(1+x+x^2+x^3...) * (1+x^2+x^4+x^6....) * (1+x^3+x^6+x^9.....)*(1+x^4+x^8+x^12.....)

1選擇0,1,2,3個等 2選擇0,1,2,3個等 3選擇0,1,2,3個等

每個式子前面的1代表選擇0個，即x^0

每個數都可以取無窮多個。

母函數的算法主要就是模擬手動計算多個式子相乘，記錄的是每個指數的系數，比如 c[ 4 ]＝2 指的是 x^4的系數為2

代碼：

#include 
using namespace std;
int c[121],temp[121];
int n;

int main()
{
    while(cin>>n)
    {
        for(int i=0;i<=n;i++)//模擬多個式子相乘
        {
            c[i]=1;//c[]中一開始是第一個式子各個指數的系數，即(1+x+x^2+x^3...)，都為1
            temp[i]=0;
        }
        for(int i=2;i<=n;i++)//表示和第幾個式子相乘
        {
            for(int j=0;j<=n;j++)//c[]中每一項的指數
                for(int k=0;k+j<=n;k+=i)//第i個式子中的每一項的指數
                    temp[j+k]+=c[j];//臨時指數
            for(int j=0;j<=n;j++)
            {
                c[j]=temp[j];//每乘完一個式子，兩個式子相當於合並成一個式子，並把系數存到c[]中
                temp[j]=0;
            }
        }
        cout<


 
參考博客：http://www.wutianqi.com/?p=596
貼一下模板：
#include 
using namespace std;
// Author: Tanky Woo
// www.wutianqi.com
const int _max = 10001;
// c1是保存各項質量砝碼可以組合的數目
// c2是中間量，保存沒一次的情況
int c1[_max], c2[_max];
int main()
{	//int n,i,j,k;
	int nNum;   //
	int i, j, k;

	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;
}


自己手動模擬了一遍才弄懂這個模板的意思。