sgu 130. Circle 卡特蘭數

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

#include <bits/stdc++.h>

using namespace std;

long long v[35];
void solve(){
    v[0]=1;
    v[1]=1;
    for(int i=2;i<=30;i++){
        for(int j=0;j<i;j++){
            v[i]+=v[j]*v[i-j-1];
        }
    }
    int n;
    scanf("%d",&n);
    printf("%lld %d",v[n],n+1);

}

int main(){
    solve();
    return 0;
}

#include <bits/stdc++.h>

using namespace std;

long long v[35];
void solve(){
    v[1]=1;
    for(int i=2;i<=30;i++){
        v[i]=v[i-1]*(4*i-2)/(i+1);
    }
    int n;
    scanf("%d",&n);
    printf("%lld %d",v[n],n+1);

}

int main(){
    solve();
    return 0;
}

import java.math.BigInteger;
import java.util.Scanner;

public class Main{

    public static void main(String[] arg){
        Scanner cin= new Scanner(System.in);
        BigInteger a[] = new BigInteger[35];
        a[1]=BigInteger.ONE;
        for(int i=2;i<=30;i++){
            a[i]=a[i-1].multiply(BigInteger.valueOf(4*i-2)).divide(BigInteger.valueOf(i+1));
        }

        int x;
        x=cin.nextInt();
        System.out.println(a[x]+" "+(x+1));

    }
}