1 solutions
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0
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
- 0
- Accepted
- 0
- Uploaded By