1 solutions

  • 0
    @ 2025-11-5 15:07:50

    C :

    #include <stdio.h>
     
    int main()
     
    {
        int a;
       float b,p1=0.4,p2=0.8,p3=0.9,p4=0.7,p=0.85;
        while(scanf("%d",&a) != EOF)
     {
     if (a==1) 
       b=p1*p2*p3+p1*p4-p1*p2*p3*p4;
      else if(a==2)
        b=2*p*p+2*p*p*p-5*p*p*p*p+2*p*p*p*p*p;
        
      else if (a==0)
    		b=0.000001;
            printf("%.6f\n",b);
    		
    }
        return 0;
    }
    

    C++ :

    #include<string>
    int main() 
    { 
    
        int a; 
    	float p1=0.4; float p2=0.8;  float p3=0.9; float p4=0.7; float p=0.85;
     
    
        while(scanf("%d",&a) != EOF) 
    
     {if (a==1){
     	
    printf("%.6f\n",p1*p2*p3+p1*p4-p1*p2*p3*p4);}
    	     
     
    else if(a==2)
    	
    {printf("%.6f\n",2*p*p+2*p*p*p-5*p*p*p*p+2*p*p*p*p*p);}  
    else  {
    	printf("%.6f\n",0.000001);
    }
     
    }
    }
    
    
    

    Java :

    import java.text.DecimalFormat;
    import java.util.Scanner;
    
    public class Main {
    	public static void main(String[] args){
    		
    		float f[] = new float[2];
    		Scanner reader = new Scanner(System.in); 
    		int a;
    		double p =0.85 ,
    			  p1 =0.4 ,
    			  p2 = 0.8,
    			  p3 = 0.9,
    			  p4 = 0.7;
    		DecimalFormat df=new DecimalFormat("#0.000000");
    		while(reader.hasNext()){
    			a = reader.nextInt();
    			if(a==1){
    				f = null;
    				f =new float[1];
    				f[0]=(float)(p1*p4+p1*p2*p3-p1*p2*p3*p4);	
    			}
    			else if(a==2){
    				f = null;
    				f =new float[1];
    				f[0]=(float)(2*p*p*p*p*p-5*p*p*p*p+2*p*p*p+2*p*p);
    			}
    			else {
    				f = null;
    				f =new float[1];
    				f[0] = (float) 0.000001;
    			}
    			/*switch(a){
    			case 1:f = null;
    				   f =new float[1];
    				   f[0]=(float)28/45;
    				   break;
    			case 2:f = null;
    			       f =new float[1];
    			       f[0]=(float)1/45;
    			       break;
    			case 3:f = null;
    		           f =new float[1];
    		           f[0]=(float)16/45;
    		           break;
    			case 4:f = null;
    		            f =new float[1];
    		            f[0]=(float)1/5;
    		             break;
    		    default:  f = null;
                          f =new float[1];
                          f[0]=(float)0.0001;
                           break;
    			}*/
    			for(int i=0;i<f.length;i++){
    				
    				System.out.println(df.format(f[i]));
    			}
    		}
    	}
    }
    
    • 1

    Information

    ID
    16431
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
    # Submissions
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