1 条题解

  • 0
    @ 2024-12-24 9:54:36

    C :

    #include<stdio.h>
    #include<math.h>
    #define PI 3.14159265;
    int main()
    {   
    	double r,s;
    	int n,x1,y1,z1,x2,y2,z2;
        while(scanf("%d",&n)!=EOF)
    	{
    		while(n--)
    		{
    			scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);
    			r=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)+(z1-z2)*(z1-z2));
    			s=4*r*r*r*PI;
    			printf("%.2lf %.2lf\n",r,s/3);
    		}
    	}   
    	return 0;
    }
    
    
    

    C++ :

    #include <math.h>
    #include <stdio.h>
    using namespace std;
    
    const double pi = acos(-1.0);
    
    int main()
    {
        //freopen("test.in", "r", stdin);
        //freopen("test.out", "w", stdout);
        int t;
        scanf("%d", &t);
        while (t--) {
            double r;
            double ax, ay, az, x, y, z;
            scanf("%lf%lf%lf", &ax, &ay, &az);
            scanf("%lf%lf%lf", &x, &y, &z);
            r=sqrt((x-ax)*(x-ax)+(y-ay)*(y-ay)+(z-az)*(z-az));
            printf("%.2f %.2f\n", r, 4*pi*r*r*r/3);
        }
        return 0;
    }
    
    

    Java :

    
    
    import java.text.DecimalFormat;
    import java.util.Scanner;
    
    public class Main {
        private static Scanner s = new Scanner (System.in) ;
        static DecimalFormat df = new DecimalFormat("0.00") ;
        public static void main(String[] args) {
        	int t = s.nextInt() ;
        	
        	for (int i = 0; i < t; i++) {
    			int x1 = s.nextInt() ;
    			int y1 = s.nextInt() ;
    			int z1 = s.nextInt() ;
    			int x2 = s.nextInt() ;
    			int y2 = s.nextInt() ;
    			int z2 = s.nextInt() ;
    			
    			int r = (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)+(z1-z2)*(z1-z2) ;
    			
    			double R = Math.sqrt(r) ;
    			
    			double V = 4*Math.PI*R*R*R/3 ;
    			
    			System.out.println(df.format(R)+" "+df.format(V));
    		}
        }
    }
        
    
    
    • 1

    信息

    ID
    1375
    时间
    1000ms
    内存
    32MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者