1 条题解

  • 0
    @ 2025-4-12 21:41:02

    C :

    #include<stdio.h>
    #include<math.h>
    int main ()
    {
        double a,b,c,d;
        while(scanf("%lf%lf%lf%lf",&a,&b,&c,&d)!=EOF)
        {
            printf("%.2lf\n",sqrt((a-c)*(a-c)+(b-d)*(b-d)));
        }
        return 0;
    }
    

    C++ :

    #include<iostream>
    #include <string>
    using namespace std;
    #include<cmath>
    #include<iomanip>
    
    
    int main(){
    
    	double x1, x2, y1, y2;
    	double l;
    
    
    
    	//int n = 10;
    	while (cin >> x1 >> y1 >> x2 >> y2){
    		
    		l = sqrt(pow((x1 - x2),2) + pow((y1 - y2),2));
    		cout << fixed << setprecision(2) << l << endl;
    		//n--;
    
    
    	}
    	return 0;
    }
    
    
    
    

    Java :

    import java.util.Scanner;
    public class Main
    {
    	public static void main(String[] argsStrings)
    	{
    		int x1,y1,x2,y2;
    		double distance;
    		Scanner inputScanner=new Scanner(System.in);
    		while(inputScanner.hasNextInt())
    		{
    			x1=inputScanner.nextInt();
    			y1=inputScanner.nextInt();
    			x2=inputScanner.nextInt();
    			y2=inputScanner.nextInt();
    			distance=Math.sqrt(Math.pow(x1-x2, 2)+Math.pow(y1-y2, 2));
    			System.out.printf("%.2f\n", distance);
    		}
    	}
    }
    
    
    • 1

    信息

    ID
    1085
    时间
    1000ms
    内存
    128MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者