HDOJ 1028 Ignatius and the Princess III（遞推）

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）代表的含義是i為n，j為劃分的最大的數字。

邊界：a(i,0) = a(i, 1) = a(0, i) = a(1, i) = 1;

i|j==0時，無論如何劃分，結果為1；

當（i>=j）時，

劃分為{j,{x1,x2…xi}},{x1,x2,…xi}的和為i-j,

{x1,x2,…xi}可能再次出現j，是以是(i-j)的j劃分，是以劃分個數為a(i-j,j);

劃分個數還需要加上a(i,j-1)（累加前面的）;

當（i < j）時，

a[i][j]就等于a[i][i]；