1 条题解

  • 0
    @ 2025-4-7 21:41:58

    C :

    #include"stdio.h"
    #include"math.h"
    float bianchang(float x1,float x2,float y1,float y2)
    {
        return (sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)));
    }
    int main()
    {
        float x1,x2,x3,y1,y2,y3;
        while(1)
        {
            scanf("%f%f%f%f%f%f",&x1,&y1,&x2,&y2,&x3,&y3);
            float a,b,c,p;
            a=bianchang(x1,x2,y1,y2);b=bianchang(x1,x3,y1,y3);
            c=bianchang(x2,x3,y2,y3);
            if(a==0&&b==0&&c==0)
                break;
            p=(a+b+c)/2;
            float s;
            s=sqrt(p*(p-a)*(p-b)*(p-c));
            printf("%.1f\n",s);
        }
        return 0;
    }
    

    C++ :

    #include<stdio.h>
    #include<math.h>
    int main()
    {
      float x1,y1,x2,y2,x3,y3,a,b,c,p,s;
      while(scanf("%f%f%f%f%f%f",&x1,&y1,&x2,&y2,&x3,&y3))
      { if(x1==0&&y1==0&&x2==0&&y2==0&&x3==0&&y3==0) break;
      a=sqrt(1.0*(x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));
      b=sqrt((x3-x2)*(x3-x2)+(y3-y2)*(y3-y2));
      c=sqrt((x3-x1)*(x3-x1)+(y3-y1)*(y3-y1));
      p=(a+b+c)/2.0;
      s=sqrt(p*(p-a)*(p-b)*(p-c));
      printf("%0.1f\n",s);
      }
      return 0;
    }
    

    Java :

    import java.util.Scanner;
    
    public class Main
    {
    	public static void main(String[] args)
    	{
    		Scanner sc=new Scanner(System.in);
    		while(sc.hasNext())
    		{
    			int x1=sc.nextInt();
    			int y1=sc.nextInt();
    			int x2=sc.nextInt();
    			int y2=sc.nextInt();
    			int x3=sc.nextInt();
    			int y3=sc.nextInt();
    			if(x1==0&&y1==0&&x2==0&&y2==0&&x3==0&y3==0)
    				break;
    			getAreas(x1,y1,x2,y2,x3,y3);
    		}
    	}
    	public static void getAreas(int x1,int y1,int x2,int y2,int x3,int y3)
    	{
    		double ares=Math.abs((double)((x1*y2+x3*y1+x2*y3-x3*y2-x1*y3-x2*y1)/2.0));
    		System.out.printf("%.1f",ares);
    		System.out.println();
    	}
    }
    
    • 1

    信息

    ID
    2132
    时间
    3000ms
    内存
    128MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者