C语言解答

#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++解答

#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解答

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));
		}
    }
}