目录

SGU 130 - Circle

Description

On a circle border there are $2k$ different points $A_1, A_2, \cdots , A_{2k}$, located contiguously. These points connect $k$ chords so that each of points $A_1, A_2, \cdots, A_{2k}$ is the end point of one chord. Chords divide the circle into parts. You have to find $N$ - the number of different ways to connect the points so that the circle is broken into minimal possible amount of parts $P$.

Input

The first line contains the integer $k$ ($1\leq k\leq 30$).

Output

The first line should contain two numbers $N$ and $P$ delimited by space.

Sample Input

1
2

Sample Output

1
23

Analysis

我们可以采用分治的方法,固定某个点,从其上引一条弦,将圆分成左右两部分。我们可以将这两部分看成新的圆,那么方案数就是这两个圆的方案数相乘。即:$$f[N] = \sum{\left(f[i - 1] \cdot f[N - i]\right)}$$ 其中 $1\leq i\leq N$。$f[i-1]$ 表示左边的圆,为 $k = i - 1$ 时的情况,$f[N - i]$ 为右边的圆,表示 $k = N - i$ 时的情况。这样,我们只要递推一下就可以了。

Solution

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
#include <iostream>
 
using namespace std;
 
const int MAX = 32;
 
long long f[MAX];
 
int main()
{
    int N;
    f[0] = 1; f[1] = 1; f[2] = 2;
    for(int i = 3; i < MAX; i++)
    {
        for(int j = 1; j <= i; j++)
        { f[i] += f[j - 1] * f[i - j]; }
    }
    while(cin >> N)
    { cout << f[N] << " " << N + 1 << endl; }
    return 0;
}

这也算是一道数学题,然而想到分治这一点还是有些难度的。