5 条题解

  • 0
    @ 2025-12-3 18:46:04
    #include <iostream>
    #include <cmath>
    #include <cstdio>
    using namespace std;
    int main (){
        int n;
        cin>>n;
        while(n--){
            double x1 ,y1,x2,y2;
            cin>>x1>>y1>>x2>>y2;
            double num=(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
            printf("%.2lf\n",sqrt(num));
        }
        return 0;
    }
    
    • 0
      @ 2025-11-12 20:20:27
      #include <stdio.h>
      #include <math.h>
      int main(){
          int n;
          scanf("%d",&n);
          while(n--){
              double x1,y1,x2,y2,x,y;
              double ans;
              scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
              x=x1-x2;
              y=y1-y2;
              ans=sqrt(x*x+y*y);
              printf("%.2f\n",ans);
          }
      }
      
      • 0
        @ 2025-10-4 16:51:39

        #include<stdio.h>

        #include<math.h>

        int main(){

        double x1,y1,x2,y2,m;

        int n;

        scanf("%d",&n);

        while(n--){

        scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);

        m=(x1-x2)(x1-x2)+(y1-y2)(y1-y2);

        m=sqrt(m);

        printf("%.2f\n",m);

        }

        return 0;

        }

        • 0
          @ 2024-10-15 8:41:13

          #include <stdio.h>

          #include <math.h>

          int main()

          {

          int n;

          scanf("%d",&n);

          for(int i=0;i<n;i++){

          double x1,y1,x2,y2;

          scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);

          double a=sqrt((x1-x2)(x1-x2)+(y1-y2)(y1-y2));

          printf("%.2lf\n",a);

          }

          return 0;

          }

          • 0
            @ 2023-10-9 1:42:32
            #include<stdio.h>
            #include<math.h>
            int main()
            {
                int n;
                scanf("%d",&n);
                while(n--)
                {
                    double x1,y1,x2,y2;
                    double dist;
                    scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
                    dist=sqrt(pow((x1-x2),2)+pow((y1-y2),2));
                     printf("%.2lf\n",dist);
                }
                return 0;
            }
            
            • 1

            信息

            ID
            164
            时间
            3000ms
            内存
            128MiB
            难度
            6
            标签
            (无)
            递交数
            1210
            已通过
            335
            上传者