HDOJ 1028 Ignatius and the Princess III（递推）

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

思路：
（i，j）（i>=j）代表的含义是i为n，j为划分的最大的数字。
边界：a(i,0) = a(i, 1) = a(0, i) = a(1, i) = 1;
i|j==0时，无论如何划分，结果为1；

当（i>=j）时，
划分为{j,{x1,x2…xi}},{x1,x2,…xi}的和为i-j,
{x1,x2,…xi}可能再次出现j，所以是(i-j)的j划分，所以划分个数为a(i-j,j);
划分个数还需要加上a(i,j-1)（累加前面的）;

当（i < j）时，
a[i][j]就等于a[i][i]；

import java.util.Scanner;

public class Main{ static int a[][] = new int[125][125]; public static void main(String[] args) { dabiao(); Scanner sc = new Scanner(System.in); while(sc.hasNext()){ int n = sc.nextInt(); System.out.println(a[n][n]); } } private static void dabiao() { for(int i=0;i<121;i++){ a[i][0]=1; a[i][1]=1; a[0][i]=1; a[1][i]=1; } for(int i=2;i<121;i++){ for(int j=2;j<121;j++){ if(j<=i){ a[i][j]=a[i][j-1]+a[i-j][j]; }else{ a[i][j]=a[i][i]; } } } }
}

  

 

  
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文章来源: chenhx.blog.csdn.net，作者：谙忆，版权归原作者所有，如需转载，请联系作者。

原文链接：chenhx.blog.csdn.net/article/details/50982710