HDU 1028 Ignatius and the Princess III （母函数或者dp,找规律，）

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15942    Accepted Submission(s): 11245

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

 

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一开始自己想了一种解法，类似dp，但是应该不是dp，应该算找规律，速度没dp快，因为多了一层循环，虽然最里面一层循环很小，

#include<queue>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 130
int n,d[N][N];//d[i][j]表示组成不超过j的数组成i有多少种方法

int main()
{
    for(int i=1;i<=120;i++)d[i][1]=1;
    d[0][0]=1;
    for(int i=2;i<=120;i++)
    {
        for(int j=2;j<=i;j++)
        {
            for(int k=j;k>=1;k--)
            {
                d[i][j]+=d[i-k][min(i-k,k)];
            }
        }
    }
    while(~scanf("%d",&n))
    {
        cout<<d[n][n]<<endl;
    }
    return 0;
}

看了网上的正规dp解法，稍加改进

#include<queue>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 130
int n,d[N][N];

int main()
{
    for(int i=1;i<=120;i++)d[i][1]=1;
    d[0][0]=1;
    for(int i=2;i<=120;i++)
    {
        for(int j=2;j<=i;j++)
        {
            d[i][j]=d[i][j-1]+d[i-j][min(j,i-j)];
        }
    }
    while(~scanf("%d",&n))
    {
        cout<<d[n][n]<<endl;
    }
    return 0;
}

还有一种母函数的做法

以后再学习