因为已经看了一眼题解，知道是算中点和长度，相同时构成一个矩形，所以就把所有的线段算出来，然后排序，相同的就更新答案
为了避免误差，我们都用整数存，中点直接相加就行了，没必要除2，长度也只要平方就行了，不用开根，算面积就用叉积算，这样做就完全没有误差了
虽然复杂度本来是不行的但是出题人没想卡咱，就过了
1 const 2 maxn=1500; 3 type 4 point=record 5 x,y:longint; 6 end; 7 segment=record 8 aa,bb:point; 9 x,y:longint; 10 l:int64; 11 end; 12 var 13 a:array[0..maxn]of point; 14 seg:array[0..maxn*maxn]of segment; 15 n,tot:longint; 16 ans:int64; 17 18 procedure swap(var x,y:segment); 19 var 20 t:segment; 21 begin 22 t:=x;x:=y;y:=t; 23 end; 24 25 procedure sort(l,r:longint); 26 var 27 i,j:longint; 28 x:segment; 29 begin 30 i:=l; 31 j:=r; 32 x:=seg[(l+r)>>1]; 33 repeat 34 while (x.x<seg[i].x) or ((x.x=seg[i].x) and (x.y<seg[i].y)) or ((x.x=seg[i].x) and (x.y=seg[i].y) and (x.l<seg[i].l)) do 35 inc(i); 36 while (x.x>seg[j].x) or ((x.x=seg[j].x) and (x.y>seg[j].y)) or ((x.x=seg[j].x) and (x.y=seg[j].y) and (x.l>seg[j].l)) do 37 dec(j); 38 if i<=j then 39 begin 40 swap(seg[i],seg[j]); 41 inc(i); 42 dec(j); 43 end; 44 until i>j; 45 if i<r then sort(i,r); 46 if j>l then sort(l,j); 47 end; 48 49 procedure init; 50 var 51 i,j:longint; 52 begin 53 read(n); 54 for i:=1 to n do 55 with a[i] do 56 read(x,y); 57 for i:=1 to n-1 do 58 for j:=i+1 to n do 59 begin 60 inc(tot); 61 with seg[tot] do 62 begin 63 aa:=a[i]; 64 bb:=a[j]; 65 x:=a[i].x+a[j].x; 66 y:=a[i].y+a[j].y; 67 l:=sqr(int64(a[i].x-a[j].x))+sqr(int64(a[i].y-a[j].y)); 68 end; 69 end; 70 sort(1,tot); 71 end; 72 73 function cj(a,b,c:point):int64; 74 begin 75 exit(abs(int64(a.x-b.x)*int64(c.y-b.y)-int64(c.x-b.x)*int64(a.y-b.y))); 76 end; 77 78 procedure up(var x:int64;y:int64); 79 begin 80 if x<y then x:=y; 81 end; 82 83 procedure work; 84 var 85 l,r,i,j:longint; 86 begin 87 l:=1; 88 r:=1; 89 while l<=tot do 90 begin 91 while (r<tot) and (seg[r+1].x=seg[l].x) and (seg[r+1].y=seg[l].y) and (seg[r+1].l=seg[l].l) do 92 inc(r); 93 for i:=l to r-1 do 94 for j:=i+1 to r do 95 up(ans,cj(seg[i].aa,seg[i].bb,seg[j].aa)); 96 l:=r+1; 97 r:=l; 98 end; 99 write(ans); 100 end; 101 102 begin 103 init; 104 work; 105 end.