1 条题解

  • 0
    @ 2024-12-24 9:59:26

    C :

    #include<stdio.h>
    #include<math.h>
    int main()
    {
    	int  n,i;
    	int x1,y1,x2,y2;
    	double h;
    	while (scanf("%d",&n)!=EOF){
    		for (i=1;i<=n;i++){
    		scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
    		h=sqrt( ( ( x1-x2 )*( x1-x2 ) ) + ( (y1-y2)*(y1-y2) ) );
    	printf("%.2f\n",h);
    	if (i==n){	
    	break;
    }
    	}
    }
    	return 0;
    }
    

    C++ :

    #include <stdio.h>
    #include <math.h>
    
    class CPoint {
        int x,y;
    public:
        CPoint(int xx,int yy)
        {
            x=xx;
            y=yy;
        }
        double operator- (CPoint c);
    };
    
    double CPoint::operator- (CPoint c) {
        return sqrt((x-c.x)*(x-c.x)+(y-c.y)*(y-c.y));
    }
    
    int main() {
        //freopen("test.in", "r", stdin);
        //freopen("test.out", "w", stdout);
        int t;
        int ax, ay, bx, by;
        scanf("%d", &t);
        while (t--) {
            scanf("%d %d %d %d", &ax, &ay, &bx, &by);
            CPoint a(ax, ay), b(bx, by);
            printf("%.2f\n", a - b);
        }
        return 0;
    }
    
    

    Java :

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

    信息

    ID
    1467
    时间
    1000ms
    内存
    32MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者