1 条题解

  • 0
    @ 2024-12-24 9:49:24

    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;
    }
    

    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;
    } 
    
    

    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);}
    
    }
    	}	}
    
    • 1

    信息

    ID
    865
    时间
    1000ms
    内存
    128MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
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