5 条题解

  • 0
    @ 2025-11-11 19:03:39
    #include <stdio.h>
    int main(){
        double x1,y1,x2,y2,x3,y3,s;
        while(1){
            scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);
            if(x1==0&&x2==0&&x3==0&&y1==0&&y2==0&&y3==0)break;
            s=0.5*(x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2));
            if(s<0)s=-s;
            printf("%.1f\n",s);
        }
    }
    
    • 0
      @ 2025-1-8 10:52:25
      #include<stdio.h>
      #include<iostream>
      #include <iomanip>
      using namespace std;
      const int N=100010;
      int f[N];
      int g[N];
      double S;
      int main()
      {
      	while(1)
      {
      		for(int i=1;i<=3;i++)
      	{
      		cin>>f[2*i-1]>>g[2*i];
      	}	
      	if(f[5]==0&&f[3]==0&&f[1]==0&&g[2]==0&&g[4]==0&&g[6]==0)
      	{
      		break;
      	}
      	S=0.5*(f[1]*(g[4]-g[6])+f[3]*(g[6]-g[2])+f[5]*(g[2]-g[4]));	
      	if(S<0)
      	{
      		S=S*-1;
      	}
          cout << fixed << setprecision(1) << S << endl;
      }
      	
      	return 0;
      }
      最近迷上了数组
      
      • 0
        @ 2023-11-21 20:13:16

        #include<stdio.h> #include<math.h>

        int main() { int a,b,c,d,e,f; while(scanf("%d %d %d %d %d %d",&a,&b,&c,&d,&e,&f)!=EOF) { if(a0&&b0&&c0&&d0&&e0&&f0) break; double s; s=(ad+cf+eb-af-cb-ed)*0.5; if(s<0) s=-s; printf("%.1lf\n",s); } }

        • 0
          @ 2023-10-8 21:19:04

          #include<stdio.h> #include<math.h> int main() { int x1,y1,x2,y2,x3,y3; double s; scanf("%d %d %d %d %d %d",&x1,&y1,&x2,&y2,&x3,&y3); while(x1!=0||y1!=0||x2!=0||y2!=0||x3!=0||y3!=0) { s=0.5abs((x1-x3)(y2-y3)-(x2-x3)*(y1-y3));

          printf("%.1lf\n",s); scanf("%d %d %d %d %d %d",&x1,&y1,&x2,&y2,&x3,&y3); } return 0; }

          • 0
            @ 2023-9-28 8:52:52
            #include<stdio.h>
            #include<math.h>
            int main()
            {
             int x1, y1, x2, y2, x3, y3;
             double p, p1, p2, p3, s;
             scanf("%d %d %d %d %d %d", &x1, &y1, &x2, &y2, &x3, &y3);
             while (x1!= 0||y1!= 0||x2!= 0||y2!= 0||x3!= 0||y3!= 0)
             {
             p1 = sqrt(pow(y2 - y1, 2) + pow(x2 - x1, 2));
             p2 = sqrt(pow(y3 - y2, 2) + pow(x3 - x2, 2));
             p3 = sqrt(pow(y1 - y3, 2) + pow(x1 - x3, 2));
             p = (p1 + p2 + p3) /2;
             s = sqrt(p*(p - p1)*(p - p2)*(p - p3));
             printf("%.1f\n", s);
             scanf("%d %d %d %d %d %d", &x1, &y1, &x2, &y2, &x3, &y3);
             }
             return 0;
            
            }
            //float貌似过不去
            
            • 1

            信息

            ID
            146
            时间
            3000ms
            内存
            128MiB
            难度
            6
            标签
            (无)
            递交数
            977
            已通过
            323
            上传者