grandpa rabbit

"Kingdoms Wei Zhi Bu" said: "Cloth has horse name Yue Red Rabbit" "Three Kingdoms" description is more clear. The Red Rabbit Horse is the treasure Ma Liang which Dong Zhuo brings from the West cool. Dong Zhuo to win the young Generals Lui BU, he gave the horse to him. Lui Bo after the horse really killed the original master Dingyuan, to the door of Dong Zhuo, when his own. Later, the horse followed Lui bu's

At home during the Spring Festival, grandpa can still take care of his own lives, but less than two weeks after work, he received a phone number from his home, saying grandpa was seriously ill and asked me to go back quickly. Surprised me! As a result, when I got home with my girlfriend and met my grandfather, I shed tears. In this short time, my grandfather became like this. I was lying in bed, with my eye

spread on the network, and finally the site's boss also make a "this story is a fictitious, if there is the same, pure coincidence!" There is also a little funny color, but think about it, so the site to create a need for how much cost, what kind of technology? The answer to believe that I do not have to say that we all understand it, can only say that the creation of this site is really very clever, with the help of Baidu this tree, short-term between can have so high flow, can also use the bi

Javascript animation Grandpa Mao is full of days, javascript is full of daysVar minSize = 50;Var maxSize = 100;Var newOn = 200;Var flakeColor = "# fff ";Var flak = $ ("$ (Function (){Var dheight = $ (document). height (); // get the height of the screenVar dwidth = $ (document). width (); // obtain the screen widthSetInterval (function (){Var startPostion = Math. random () * dwidth; // locationVar startOpacity = 0.7 + Math. random () * 0.3; // transpa

practice, so I feel great progress. This study lasted for more than a semester, some videos watched several times.Big Four semester began to play know, in fact, this thing a long time ago downloaded, but it seems like the beginning of the invitation to register and then did not play, and then a discovery without invitation can also register, only to start playing up. One time to see the cosine of the skill sheet, found inside a language called Python, happen to be 51cto on the article said that

Question link: http://poj.org/problem? Id = 2092 Ideas: First, statistical data is sorted by count in descending order. If count is equal, data is sorted by num in ascending order. Then, all the num with the second largest count is output;Code: #include Poj_2092 grandpa is famous

var minsize=50;var maxsize=100;var newon=200;var flakecolor= "#fff";var flak=$ ("$ (function () {var dheight=$ (document). Height ();//Gets the height of the screenvar dwidth=$ (document). width ();//Gets the width of the screenSetInterval (function () {var startpostion=math.random () *dwidth;//positionvar startopacity=0.7+math.random () *0.3;//Transparencyvar endpositiontop=dheight;//heightvar endpositionleft=math.random () *dwidth;//generates a left random numbervar durationfall=2000+math.rand

Title: The second person in the ranking. Analysis: Count sorting + statistics. Note: rhubarb says there is a problem with the topic. How can he put it in the DP category. ) #include Zoj 2250-grandpa is famous

After finding the convex hull, the violent enumeration edge is $check$, note that the convex hull is a line (or two lines) of the case to output $no$#include That's it."POJ 1228" Grandpa ' s Estate convex bag

) { if(T > EPS)return 1; if(Fabs (T) return 0; return-1;}DoubleDist (Point A, point B) {returnsqrt (A-B) * (Ab));}BOOLCMP (Point A, point B) {intt = sign (a-p[0]) ^ (b-p[0])); if(T = =0) returnDist (A, p[0]) 0]); returnT >0;}voidGraham (intN) { intk=0; for(intI=0; i) { if(P[k].y>p[i].y | | (sign (p[k].y-p[i].y) = =0 p[k].x >p[i].x)) K=i; } Swap (p[0], p[k]); Sort (P+1, p+N, CMP); sta[0]=0, sta[1]=1, top=1; if(N 2) {Top= N1; return ; } for(intI=2; i) { while

;}BOOL operator== (Constpoint A,Constpoint b) {returnDCMP (a.x-b.x) = =0 dcmp (a.y-b.y) = =0; }DoubleDot (vector A, vector B) {returna.x*b.x + a.y*b.y;}DoubleLength (Vector A) {returnsqrt (Dot (A, a));}DoubleAngle (vector A, vector B) {returnACOs (Dot (A, B)/Length (a)/Length (B)); }DoubleCross (vector A, vector B) {returna.x*b.y-a.y*b.x;}DoubleAngle (Vector v) {returnatan2 (V.Y, v.x);}BOOLOnsegment (Point P, point A, point B) {//endpoint does not count returnDCMP (Cross (a-p,b-p)) = =0 dcmp

, the new convex hull given this pile of points, and then determine whether there are at least 3 points on each side of the convex hull, if one edge does not meet the conditions, then output No. otherwise yes.Their own do not know where the wrong change, attached to the Big God Code:#include #include #include #include #include #define EPS 1e-8using namespace STD;structpoint{Doublex, y;}; Point p[1010],Stack[1010];intN,top;Doublemulti (point P1, point P2, point p3) {return(p2.x-p1.x) * (P3.Y-P1.Y