deutz 6206

) { returnR.compareto (l);//L>r } Else return(L.add (R)). CompareTo (Biginteger.valueof (0));//l+r>0 } Public Static voidMain (string[] args) {//TODO Auto-generated method stubsScanner cin=NewScanner (System.inch); Biginteger[][]p=Newbiginteger[4][2]; intt=Cin.nextint (); for(intcas=0; cas){ for(intI=0;i4; i++) {p[i][0]=Cin.nextbiginteger (); p[i][1]=Cin.nextbiginteger (); } BigInteger FA=fxy (p[1], p[2], p[0]), Fb=fxy (p[1], p[2], p[3]); BigI

(NewBufferedinputstream (system.in)); intNcase =Cin.nextint (); for(intT=1; t) {PO P1=NewPO (); PO P2 =NewPO (); PO P3=NewPO (); PO aim=NewPO (); p1.x= Cin.nextbiginteger (); P1.Y =Cin.nextbiginteger (); p2.x= Cin.nextbiginteger (); P2.Y =Cin.nextbiginteger (); p3.x= Cin.nextbiginteger (); P3.Y =Cin.nextbiginteger (); Aim.x= Cin.nextbiginteger (); AIM.Y =Cin.nextbiginteger (); Booleanres =Solve (P1, P2, p3, AIM); if(RES) System.out.println ("Accepted"); ElseSystem.out.println ("Rejected"); }

)); C=x3.multiply (x3). Add (Y3.multiply (y3)); G=bigdecimal.valueof (1); G=y3.subtract (y2). Multiply (x1). Add (Y1.subtract (y3). Multiply (x2)). Add (Y2.subtract (y1). Multiply (x3)); X=b.subtract (C). Multiply (y1). Add (C.subtract (A). Multiply (y2)). Add (A.subtract (B). Multiply (Y3)). Divide (bigdecimal.valueof (2)); Y=c.subtract (B). Multiply (x1). Add (A.subtract (C). Multiply (x2)). Add (B.subtract (A). Multiply (x3)). Divide ( Bigdecimal.va

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Check Count (2)Time limit:10000/5000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 6206 Accepted Submission (s): 1975Problem description gives you a m*n checkered checkerboard with a non-negative number in each lattice.Take out a number of the number, so that the arbitrary two number of the lattice does not have a common edge, that is, the number of the 2 lattice is not adjacent, and the number of the maximum.Input inc

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