HDOJ1028Ignatius and the Princess III（整数划分的母函数做法）

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9077    Accepted Submission(s): 6389

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

代码一：-----母函数解法：

构造母函数如下：
　　G(x)=(1+x+x^2+x^3+...+x^n)*(1+x^2+x^4+...)*(1+x^3+x^6+...)*...*(1+x^n);

#include <cstdio>
#include <iostream>

using namespace std;

int main()
{
    int n;
    int c1[125], c2[125];
    
    while(~scanf("%d", &n))
    {
        for(int i = 0; i <= n; ++i)
        {
            c1[i] = 1; 
            c2[i] = 0;
        }
        for(int i = 2; i <= n; ++i)
        {
            for(int j = 0; j <= n; ++j)
            {
                for(int k = 0; k <= n/i*i; k += i)
                    c2[j+k] += c1[j];
            }
            for(int j = 0; j <= n; ++j)
            {
                c1[j] = c2[j];
                c2[j] = 0;
            }
        }
        printf("%d\n", c1[n]);
    }
    return 0;
}

代码二：递归做法：

#include <cstdio>
#include <iostream>

using namespace std;
int f[125][125];

int fun(int a, int b) // fun(n, m)表示将整数 n 划分为最大数不超过 m 的划分
{
    if(f[a][b])
        return f[a][b];
    if(a == 1 || b == 1)
        return 1;
    if(a < b)
        return f[a][b] = fun(a, a);
    if(a == b)
        return f[a][b] = 1 + fun(a, b-1);
    if(a > b)
        return f[a][b] = fun(a-b, b) + fun(a, b-1); 
}

int main()
{
    int n;
    memset(f, 0, sizeof(f));
    while(~scanf("%d", &n))
    {
        printf("%d\n", fun(n, n));
    }
    return 0;
}