1 solutions

  • 0
    @ 2024-12-10 22:16:30

    C :

    #include<stdio.h>
    double f(int n)
    {
        int i;
        double a=1;
        for(i=1;i<=n;i++)
        a=a*i;
        return a;
    }
    void main()
    {
        int n;
        while(scanf("%d",&n)==1)
        {
           printf("%.lf\n",f(2*n)/(f(n)*f(n)*(n+1)));
        }
    }
    
    

    C++ :

    #include <iostream>
    #include <cstring>
    using namespace std;
    
    long long f[25][25];
    
    int main(){
    	int n;
    	while (cin>>n){
    		memset(f,0,sizeof(f));
    		for (int i=1;i<=n;i++)
    			f[i][0]=1;
            for (int i=1;i<=n;i++){
    			for (int j=1;j<=i-1;j++)
    				f[i][j]=f[i-1][j]+f[i][j-1];
    			f[i][i]=f[i][i-1];
    		}
    		cout<<f[n][n]<<endl;
    	}
    	return 0;
    }
    

    Pascal :

    program acm1095;
    var a:array[0..20] of int64;
        i,j,n:longint;
    begin
      a[0]:=1; a[1]:=1;
      for i:=2 to 20 do
        for j:=i-1 downto 0 do
         a[i]:=a[i]+a[i-1-j]*a[j];
      while not eof do
       begin
         readln(n);
         writeln(a[n]);
       end;
    end.
         
    

    Java :

    import java.util.*;
    import java.math.*;
    
    public class Main{
        public static void main(String[] args){
        Scanner in = new Scanner(System.in);
        while(in.hasNextInt()){
            int n = in.nextInt();
            int i = n;
            int j = n*2;
            BigInteger up = BigInteger.valueOf(i);
            BigInteger down = BigInteger.valueOf(j);
            BigInteger a = BigInteger.valueOf(0);
            for(int k = i, l = 1;l<i;k--,l++)
                up = up.multiply(BigInteger.valueOf(k - 1));
            for(int k = j, l = 1;l<i;k--,l++)
                down = down.multiply(BigInteger.valueOf(k - 1));
                a = down.divide(up).divide(BigInteger.valueOf(n+1));
            System.out.println(a);
        }
    }
    }
    
    • 1

    Information

    ID
    534
    Time
    1000ms
    Memory
    32MiB
    Difficulty
    (None)
    Tags
    # Submissions
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