1 solutions

  • 0
    @ 2025-11-5 15:07:39

    C :

    #include <stdio.h>
     
    int main()
     
    {
        int a;
    float b;
        while(scanf("%d",&a) != EOF)
     {
     if (a==1) 
       b=28.0/45;
      else if(a==2)
        b=1.0/45;
        else if(a==3)
        b=16.0/45;
        else if(a==4)
        b=1.0/5;
      else if (a==0)
    		b=0.0001;
            printf("%.4f\n",b);
    		
    }
        return 0;
    }
    

    C++ :

    #include <stdio.h>
     
    int main()
     
    {
        int a;
        float b;
        while(scanf("%d",&a) != EOF)
     {if (a==1) 
        b=28.0/45; //将“作业答案”替换为该题答案。若有多行,用多行重复输出。
      else if (a==2)
        b=1.0/45;
      else if(a==3)
        b=16.0/45;
      else if(a==4)
        b=1.0/5;
      else 
    	b=0.0001;
            printf("%.4f\n",b);
    }
        return 0;
    }
    

    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;
    		DecimalFormat df=new DecimalFormat("#0.0000");
    		while(reader.hasNext()){
    			a = reader.nextInt();
    			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
    16427
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
    # Submissions
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