sgu 130Circle dp

130. Circle

time limit per test: 0.25 sec. 
memory limit per test: 4096 KB

 

On a circle border there are 2k different points A1, A2, ..., A2k, located contiguously. These points connect k chords so that each of points A1, A2, ..., A2k 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 <= k <= 30).

 

Output

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

 

Sample Input

2

Sample Output

2 3

思路:选取一个固定点,比如P0点,然后分别和列举其他点Pj的连线,左边被分出j-1个点,右边则是i-j个点,左右分别是子图

感想:考虑不周,以为只能和最左或者最下的连接,但是只要两边不互交就行了

#include <cstdio>

using namespace std;

long long f[31];

void calc(){

    f[0]=1;

    f[1]=1;

    f[2]=2;

    for(int i=3;i<31;i++){

        for(int j=1;j<=i;j++){

            f[i]+=f[j-1]*f[i-j];

        }

    }

}

int main(){

    int n;

    calc();

    scanf("%d",&n);

    printf("%I64d %d\n",f[n],n+1);

}