[暖手][学习阶段-各路杂题][HDU-1028]Ignatius and the Princess III

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

41020

 

Sample Output

542627

 

最早用递归做的，闭着眼都知道超时了。

然后用dp做 A了一次 但是感觉很丑

在网上找到大神的办法 贴在下面

题解：

画出解答树，每个结点有两个值（暂时叫左值和右值）不断划分，原则是将大数化为小数，左值大于右值，结点左值的子树的右值大于等于结点的右值。

n = 5、6、7、8时解答树如下：

由n=8的解答树，可知该解答树有4个子树记作A[8][1]、A[8][2]、A[8][3]、A[8][4]

A[8][1] = A[7][1] + A[7][2] + A[7][3] + 1.

A[8][2] = A[6][2] + A[6][3] + 1

A[8][3] = A[5][3] + 1

A[8][4] = A[4][4] + 1

记A[8][0]为n=8的总结点，则A[8][0] = A[8][1] + A[8][2] + A[8][3] + A[8][4] + 1

ps：也可以用DFS

import java.util.Scanner;public class Main {public static void main(String[] args) {// TODO Auto-generated method stubScanner in = new Scanner(System.in);int n, i, j, k;  int[][] A = new int[121][70];     for (k=1; k<=120; k++)      {          A[k][0] = 1;          for (i=1; i<=k/2; i++)          {              for (j=i; j<=(k-i)/2; j++)                  A[k][i] += A[k-i][j];              A[k][i] += 1;              A[k][0] += A[k][i];          }      }  while(in.hasNext()){n= in.nextInt();System.out.println(A[n][0]);}}}