1 solutions

  • 0
    @ 2025-11-5 15:06:53

    C :

    #include <stdio.h>
     
    int main()
    {
        int a;
        float b[2];
        while(scanf("%d",&a) != EOF)
    {
    if (a==1)
      {
      b[0]=0.0833;
      b[1]=0.05;
      for(int i=0;i<2;i++)
      printf("%.4f\n",b[i]);
      }
     else
         {
       b[0]=0.0001; 
       printf("%.4f\n",b[0]);
          }
     }
        return 0;
    }
    

    C++ :

    #include <stdio.h>
     
    int main()
    {
        int a;
        float b;
        while(scanf("%d",&a) != EOF)
        {if (a==1){
          b=1.0/12; printf("%.4f\n",b);
          b=1.0/20; printf("%.4f\n",b);
        }
         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[9];
    		Scanner reader = new Scanner(System.in); 
    		int a;
    		DecimalFormat df=new DecimalFormat("#0.0000");
    		while(reader.hasNext()){
    			a = reader.nextInt();
    			if(a==1){
    				f = null;
    				f =new float[2];
    				f[0]=(float) 1/12;
    				f[1]=(float)1/20;
    				
    			}
    			else {
    				f = null;
    				f =new float[1];
    				f[0] = (float) 0.0001;
    			}
    			for(int i=0;i<f.length;i++){
    				
    				System.out.println(df.format(f[i]));
    			}
    		}
    	}
    }
    
    • 1

    Information

    ID
    16411
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
    # Submissions
    0
    Accepted
    0
    Uploaded By