每次都选最左边的点，然后以这个点为原点

统计和这个点构成的三角形面积和

不难想到极角排序然后由叉积很容易求出

1 const oo=1 shl 30; 2       eps=1e-8; 3 var i,j,k,m,n:longint; 4     x,y:array[0..6010] of longint; 5     z:array[0..6010] of double; 6     ans,xx,yy:int64; 7  8 procedure swap(var a,b:longint); 9   var c:longint;10   begin11     c:=a;12     a:=b;13     b:=c;14   end;15 16 procedure sort(l,r:longint);17   var i,j:longint;18       p,q:double;19   begin20     i:=l; j:=r;21     p:=z[(l+r) shr 1];22     repeat23       while z[i]
      
       p+eps do dec(j);25       if i<=j then26       begin27         swap(x[i],x[j]);28         swap(y[i],y[j]);29         q:=z[i]; z[i]:=z[j]; z[j]:=q;30         inc(i); dec(j);31       end;32     until i>j;33     if l
       
        y[i] then z[j]:=oo51         else z[j]:=-oo52       else z[j]:=(y[j]-y[i])/(x[j]-x[i]);53     sort(i+1,n);54     xx:=0; yy:=0;55     for j:=i+1 to n do56     begin57       ans:=ans+(x[j]-x[i])*yy-(y[j]-y[i])*xx;58       xx:=xx+x[j]-x[i]; yy:=yy+y[j]-y[i];59     end;60   end;61   writeln(abs(ans)/2:0:1);62 end.