Ignatius and the Princess III （HDU 1028） ——母函数(另解DP)

        Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11999    Accepted Submission(s): 8494

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
   
   
   
   

这是一个典型的母函数问题（当然还有其他解法，比如说背包）

下面是这题的解题代码，也是母函数的模板型代码：

#include<stdio.h>
int main()
{
    int n,i,j,k;
    int c1[1000],c2[1000];

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

下面是别人对代码的一些解释：

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
    int N;
    int c1[125],c2[125];
    while(cin>>N)
    {
        int i,j,k;
        for(i=0;i<=N;i++)//初始化第一个表达式的系数
        {
            c1[i]=1;
            c2[i]=0;
        }
        for(i=2;i<=N;i++)
        {//从第二个表达式开始，因为有无限制个，所以有n个表达式
            for(j=0;j<=N;j++)
            {//从累乘的表达式后的一个表达式第一个到最后一个
                for(k=0;k+j<=N;k+=i)
                {//k为第j个变量的指数，第i个表达式每次累加i
                    c2[j+k]+=c1[j];
                }
            }
            for(j=0;j<=N;j++)
            {//滚动数组算完一个表达式后更新一次
                c1[j]=c2[j];
                c2[j]=0;
            }
        }
        printf("%d\n",c1[N]);
    }
    return 0;
}

母函数是一类解决数学中组合问题的方法，它有很多种，不懂可以百度一下“母函数”

另解:
 这是有关计数问题的DP中的划分数. 相当于n个无区别的物品,将它们划分成不超过m组(当然这里m=n),求划分方法数. 考虑n的m划分a1+a2+...+am = n, 如果对于每个i都有ai>0, 那么{ai-1}就对应了n-m的m划分. 另外,如果存在ai=0, 那么这就对应啦n的m-1划分. dp[i][j] : j的i划分的总数. 综上,我们可得出如下递推关系: dp[i][j] = dp[i][j-i] + dp[i-1][j].

/*************************************************************************
    > File Name: dp.cpp
    > Author: tj
    > Mail: [email protected]
    > Created Time: 2015年01月02日 星期五 14时24分28秒
 ************************************************************************/

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int main()
{
	int n, dp[130][130];

	while(~scanf("%d%d", &n,&m))
	{
		memset(dp, 0, sizeof(dp));
		dp[0][0] = 1;
		for(int i=1; i<=n; i++)
		{
			for(int j=0; j<=m; j++)
			{
				if(j >= i) dp[i][j] = dp[i][j-i]+dp[i-1][j];
				else dp[i][j] = dp[i-1][j];
			}
		}
		printf("%d\n", dp[m][n]);
	}
	return 0;
}