hdu1028 dp+母函数

这题是参考前辈的成果得到的，说来惭愧，居然花了一个下午才理解

记录一下。

dp是如此的强大，但要有很强的逻辑才能胜任，要找好不同状态之前的联系

/*
 * =====================================================================================
 *
 *       Filename:  hdu1028.c
 *
 *    Description:  
 *
 *        Version:  1.0
 *        Created:  2013年11月20日 11时22分16秒
 *       Revision:  none
 *       Compiler:  gcc
 *
 *         Author:  Wenxian Ni (Hello World~), niwenxianq@qq.com
 *   Organization:  AMS/ICT
 *
 * Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10631    Accepted Submission(s): 7537


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

1 1
2 2
3 3
dp[i][j] = dp[i-1][j-1] + dp[i-j][j]
数字i中，最大的是j

 * =====================================================================================
 */
 #include <stdio.h>
 #include <math.h>


 int main()
 {
    int n;
    int i, j;
    int dp[121][122];
    //dp[1][1] = 1;
    //dp[1][0] = 1;
    //dp[2][1] = 1;
    //dp[2][2] = 1;
    //dp[2][0] = 2;
    for(i=0;i<121;i++) 
     {
         dp[i][i] = 1;
     }

    for(i=1;i<121;i++) 
     {
        dp[i][122] = 0;
        for(j=1;j<=i;j++)
         {
             if(i != j)
                 dp[i][j]  = dp[i-j][j] + dp[i-1][j-1];

             dp[i][121] += dp[i][j];
         }
     }
    while(~scanf("%d",&n))
     {
         printf("%d
",dp[n][121]);
     }
     return 0;
 }

这里面，用最后一位存储最后的结果，之前老是计算错误，才发现数组越界，哎，估计晚上脑子不好使了

最精华的是母函数理论和实现

f(x) = (1+x^1+x^2+...)(1+x^2+x^4+...)...

#include<iostream>
using namespace std;

#define N 121

int main()
{
    unsigned long a[N],b[N];
    int i,j,k;
    int n;

    while(cin>>n)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));

        a[0]=1;
        for(i=1;i<=n;i++) //第i个多项式
        {
            for(j=0;j<=n;j++) //第i个多项式与之前的多项式相乘
            {
                for(k=0;k*i+j<=n;k++) //具体添加到某一项中
                    b[k*i+j]+=a[j];
            }

            for(k=0;k<=n;k++)
            {
                a[k]=b[k];
                b[k]=0;
            }
        }

        cout<<a[n]<<endl;
    }

    return 0;
}