Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3088 Accepted Submission(s): 2133
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
Author
Ignatius.L
1
//
动归是自下而上
2
#include
<
iostream
>
3
using
namespace
std;
4
int
f[
121
][
121
];
5
int
dp(
int
n,
int
k)
//
f[5][1]=f[4][1]+f[3][2]+f[2][3]+f[1][4]+f[0][5]
6
{
7
if
(n
==
0
)
return
1
;
//
所以f[0][5]=1,就是代表的是5=5
8
if
(n
<
k)
return
0
;
//
就是f[1][4]一不能分了=0
9
if
(n
==
k)
return
1
;
//
相当与f[2][2]=1,就是2+2一种
10
if
(f[n][k]
!=-
1
)
return
f[n][k];
//
不用重复计算
11
f[n][k]
=
0
;
//
没有初始化的就先赋值为0;
12
for
(
int
i
=
k;i
<=
n;i
++
)
//
以k为最小开始将n分开
13
f[n][k]
+=
dp(n
-
i,i);
//
f[5][1]=f[4][1]+f[3][2]+f[2][3]+f[1][4]+f[0][5]
14
return
f[n][k];
15
}
16
int
main()
17
{
18
int
n;
19
memset(f,
-
1
,
sizeof
(f));
//
初始化将f[][]数组的值全部赋值为-1
20
while
(cin
>>
n)
21
cout
<<
dp(n,
1
)
<<
endl;
//
假设n=5,则就是f(5,1)
22
return
0
;
23
}