HDU 1028 Ignatius and the Princess III (母函数，或者DP）

Ignatius and the Princess III

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 2   Accepted Submission(s) : 2

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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

 

 

前面已经写了一篇用母函数做的：

http://www.cnblogs.com/kuangbin/archive/2011/07/30/2122192.html

母函数的意思就是把n 用1,2,3,4,5`````来表示的种数，构造母函数：(1+x+x^2+x^3+`````)*(1+x^2+x^4+````)*(1+x^3+x^6+````)````

 

其实发现不用母函数也可以做出来。

就是用递归也可以做，只不过递归的时候要DP的方法记录下来。

程序如下：

#include<stdio.h>

#include<string.h>

const int MAXN=130;

int dp[MAXN][MAXN];

//dp[i][j]表示 i 表示成最大的数不超过 j 的方法数

int calc(int n,int m)

{

    

    if(dp[n][m]!=-1) return dp[n][m];

    

    

    if(n<1||m<1) return dp[n][m]=0;

    if(n==1||m==1) return dp[n][m]=1;

    if(n<m) return dp[n][m]=calc(n,n);

    if(n==m) return dp[n][m]=calc(n,m-1)+1;

    return dp[n][m]=calc(n,m-1)+calc(n-m,m);

    

}     

int main()

{

    int n;

    memset(dp,-1,sizeof(dp));

    

    while(scanf("%d",&n)!=EOF)

      printf("%d\n",calc(n,n));

    return 0;

}    

这个的速度比母函数做要快。

而且想对好理解一点。

dp[n][m]的意思是把n,用最大不超过m的数来表示的方法数。

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