1 条题解

  • 0
    @ 2024-12-24 9:59:30

    C :

    #include<stdio.h>
    
    int main()
    {
    	int n;
    	double x1,x2,x3,y1,y2,y3;
    	while(scanf("%d",&n)!=EOF,n)
    	{
    		while(n--)
    		{
    			scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);
    			printf("%.1lf %.1lf\n",(x1+x2+x3)/3,(y1+y2+y3)/3);
    		}
    	}
    	return 0;
    }
    

    C++ :

    #include<stdio.h>
    
    int main()
    {
    	int n;
    	double x1,x2,x3,y1,y2,y3;
    	while(scanf("%d",&n)!=EOF,n)
    	{
    		while(n--)
    		{
    			scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);
    			printf("%.1lf %.1lf\n",(x1+x2+x3)/3,(y1+y2+y3)/3);
    		}
    	}
    	return 0;
    }
    

    Java :

    import java.util.*;
    public class Main{
        public static void main(String[] args){
            Scanner in=new Scanner(System.in);
            while(in.hasNextInt()){
            int n=in.nextInt();
            if(n==0) break;
            double[] x=new double[3];
             double[] y=new double[3];
            for(int i=1;i<=n;i++){
                double ave1=0,ave2=0;
              for(int j=0;j<3;j++){
                   x[j]=in.nextDouble();
                   y[j]=in.nextDouble();
                }
                for(int k=0;k<3;k++){
                    ave1+=x[k];
                     ave2+=y[k];
                    }
                    ave1=ave1/3.0;
                     ave2=ave2/3.0;
                     System.out.printf("%.1f",ave1);
                     System.out.print(" ");
                     System.out.printf("%.1f\n",ave2);
                    }
                }
            }
        }
    
    • 1

    信息

    ID
    1664
    时间
    1000ms
    内存
    32MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者