Solution C

#include <stdio.h>
#include <math.h>

int main()
{
	int a;
	double answer[5];
	double p[3];
	double E[4];

	//E1,2,3分别对应x1,2,3的期望
	E[1]=5*(1.0/2);
	E[2]=E[3]=6*(1.0/3);
	
	while(scanf("%d",&a) != EOF)
	{
		if(a==1)
		{
			p[0]=10*pow(0.5,5);
			p[1]=15*pow(1.0/3,2)*pow(2.0/3,4);
			p[2]=6*pow(1.0/3,5)*(2.0/3);

			answer[1]=p[0]*p[1]*p[2];
			printf("%0.6f\n",answer[1]);//P{X1=2,X2=2,X3=5}，
		}
		else if(a==2)
		{
			answer[2]=E[1]*E[2]*E[3];
			printf("%0.6f\n",answer[2]);//E（X1，X2，X3）
		}
		else if(a==3)
		{
			answer[3]=E[1]-E[2];
			printf("%0.6f\n",answer[3]);//E(X1-X2)
		}
		else if(a==4)
		{
			answer[4]=E[1]-2*E[2];
			printf("%0.6f\n",answer[4]);//E（X1-2*X2）
		}
		else if(a==0)
		{
			answer[0] = 0.000001;
			printf("%0.6f\n",answer[0]);
		}		
	}
	return 0;
}

Solution C++

#include <iostream>
using namespace std;
int main()
{
    int N;
    while(scanf("%d",&N) != EOF)
    {
	if (N==1)
 cout<<"0.001694"<<endl;
	else if(N==2)
       cout<<"10.000000"<<endl;
      else if(N==3)
      cout<<"0.500000"<<endl;
      else if(N==4)
      cout<<"-1.500000"<<endl;
      else
  cout<<"0.000001"<<endl;
	}
    return 0;
} 

Solution Java

import java.util.*;
public class Main {
	public static void main(String args[]) {
		Scanner cin = new Scanner(System.in);
		int a;
		 double b;
		while (cin.hasNext()) {
			a = cin.nextInt(); 
			if (a==1){
				 b=0.001694; 
System.out.printf("%.6f\n",b);
   
}
else if (a==2){
	  b=10.000000;
System.out.printf("%.6f\n",b);}
else if (a==3){
	b=0.500000;
System.out.printf("%.6f\n",b);}
else if (a==4){
	 b=-1.500000;
System.out.printf("%.6f\n",b);}
else if (a==0){
	 b=0.000001;
System.out.printf("%.6f\n",b);}

}
	}	}