HDU--1028--Ignatius and the Princess III--完全背包Ignatius and the Princess III

time limit: 2000/1000 ms (java/others)    memory limit: 65536/32768 k (java/others)

total submission(s): 11896    accepted submission(s): 8424

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

sample output

题意：把一个数因式分解，点到顺序的不算，比如5=1+4和5=4+1就算是一种方式

题解：等于在式子的‘+’号后面一次加上一个数字形成新的，而且要求加上的这个数字是当前式子里面最大的，这样就不会重复

如：（去想象这个表中的数据是一行一行从左向右刷出来的“一行的定义是每个+n算一行”，每次刷的位置的各种情况都是根据先前已经算出来了的数据得来的）

分解的那个整数 1

2 3

4 5 6

+1 1  

1+1

1+1+1 1+1+1+1

1+1+1+1+1

1+1+1+1+1+1

————————————————————————————————————————

+2 2

1+2

1+1+2 1+1+1+2

1+1+1+1+2

2+2

1+2+2

1+1+2+2

2+2+2

+3 3

1+3

1+1+3

1+1+1+3

2+3

1+2+3

3+3

+4 4

1+4

1+1+4

2+4

+5 5

1+5

+6 6

因式分解的方式数：1

2 3 5

7 11

比如+1的时候，6就从5里面找，因为5+1=6，+2的时候6就从4里面找，+3的时候就从3里面找，这么算下去，到+6的时候，6本身也算进来，所以就等于从0里面加一个