at t columbia la

widen the line segment (the Code allows the line segment to be extended by 1e-6 on both sides) and change it to a standard intersection. Another scenario is: When the line segments of the two endpoints are collocated, the problem also occurs. For example, it should have been closed, but it may be converted into a path because it determines whether to move forward based on the standard intersection of a line segment. The solution is that the points on other line segments at the endpoint after t

#includeView code Why does La 4254 time out?

LA 3635, LA3635 Regionals 2006> Europe-Northwestern3635-Pie Time limit: 3.000 secondsMy birthday is coming up and traditionally I'mServing pie. Not just one pie, no, I have a numberN of them, of various tastes and of various sizes. FOf my friends are coming to my party and eachThem gets a piece of pie. This shocould be one pieceOf one pie, not several small pieces since that looksMessy. This piece can be one whole pie though.My friends are very annoy

LA 6802 Turtle Graphics (water question) 6802-Turtle Graphics Question: Similar to snake games. Ask the coordinates of the end point after the given path (F-forward, L-left, R-right) and after more than two points. Solution: Simulate it. Reference code: #include #include #include #include using namespace std;const int MAXN = 100;const int dx[4] = {0, 1, 0, -1};const int dy[4] = {1, 0, -1, 0};int

used as the unit test framework, and Moq is used as the Mock test framework. We use Jenkins (Hudson) as a continuous integration tool, Thoughtworks's Twist as a regression and integration testing tool, and Powershell as a build script, git is used as the source code control tool. The first week of LA. In addition to taking part in the Meeting several times on the first day and understanding the overall project situation, especially the business logic

LA 4513 (Stammering Aliens-Hash for LCP) [Template: hash for LCP], 4513lcp 4513-Stammering AliensDr. ellie Arroway has established contact with an extraterrestrial civilization. however, all efforts to decode their messages have failed so far because, as luck wowould have it, they have stumbled upon a race of stuttering aliens! Her team has found out that, in every long enough message, the most important words appear repeated a certain number of times

tree), then try to find s[i in tire. L]. Specific reference code.The code is as follows:#include #include#includeusing namespacestd;Const intMAXN =4000* -+Ten;//4,000 words, each word the longest is 100, at most there are so manyConst intMAXM =300010;Const intMoD =20071027;intD[MAXM];CharSS[MAXM], t[ the];structtire{intch[maxn][ -]; intVAL[MAXN]; intsz; voidInit () {sz =1; Memset (Val,0,sizeof(Val)); memset (ch[0],0,sizeof(ch[0])); }//Initialize intIdxCharc) {returnA O'a'; }//Get number v

Topic PortalTest Instructions: Training guide P250Analysis: DFS memory search, the range or the graph is a known string composition of the automaton diagram, then by | (1 #include 　　DP (Memory Search) + AC automaton LA 4126 Password Suspects

is still a two-point search1#include 2#include 3#include 4 using namespacestd;5 6 Const DoublePi=acos (-1.0);7 intn,f;8 Doubler[10001];9 Ten BOOLOkDoubleArea ) One { A intsum=0; - for(intI=0; i) -Sum+=floor (r[i]/Area ); the returnsum>=f+1; - } - - intMain () + { - intT; +Cin>>T; A while(t--) at { -Cin>>n>>F; - Doublemaxn=-1; - for(intI=0; i) - { - intA; inCin>>A; -r[i]=pi*a*A; tomaxn=Max (maxn,r[i]); + } - DoubleL=0

Test instructionsIn a sequence of length n, find the shortest length sequence so that it is greater than or equal to s;Ideas:pointer, water problem;AC Code:#include /*#include */using namespacestd;#definefor (i,j,n) for (int i=j;i#defineRiep (n) for (int i=1;i#defineRIOP (n) for (int i=0;i#defineRJEP (n) for (int j=1;j#defineRJOP (n) for (int j=0;j#defineMST (SS,B) memset (ss,b,sizeof (ss));typedefLong Longll;templateclassT>voidRead (tnum) { CharCH;BOOLf=false; for(Ch=getchar (); ch'0'|| Ch>

Test instructions: Given n balls, each time you take one from each basket and put it in a new basket, and remove the same, sort by the number of balls, ask you to complete the loop with a few balls.Analysis: Mathematical problems, it is easy to find the top N and is the largest number of balls, so we just need to find the largest n.The code is as follows:#include LA 7500 Boxes and Balls (math)

("%d\n", bcc_cnt); if(u==xv==y) Break; } } } Else if(Dfn[y]//Important } if(fa0child==1) iscut[x]=0;}intMain () {scanf ("%d",m); while(m!=0) {memset (First,0,sizeof(first)); N=0; tot=0; bcc_cnt=0; for(intI=1; i) { intu,v; scanf ("%d%d",u,v); N=Max (N,max (u,v)); Insert (U,V); Insert (V,U); } memset (DFN,0,sizeof(DFN)); memset (Low,0,sizeof(low)); memset (Iscut,0,sizeof(Iscut)); memset

Topic PortalTest instructions: The small island of the convex polygon in the sea, ask the island's point to the sea farthest distance.Analysis: Training Guide P279, two-point answer, and then the entire polygon to internal contraction, if the half-plane intersection non-empty, then these points constitute a half-plane, there are satisfied points./************************************************* author:running_time* Created time:2015/11/10 Tuesday 14:16:17* Fi Le Name:LA_3890.cpp ***************

output for three test cases.Sample Input3 15 4 10 2 7 7 5 12 2 7 12 4 9 13 6 3 2 1 3 4 0 2 5 4 6 10 4 8 2 5 4 2 5 6 5 8 2 5 8Sample OutputYES NOYESThe main idea: to give you a lot of targets, each target is an interval and has its own height, ask you can in a given interval to find a straight line to all these targets strung together.Analysis: Look at the online solution to know the practice, two points archery position, and then maintain a section of elevation, if a target in the current inter

Interval update + statistical update length Just a little bit of attention, T. #include #defineLson L, M, rt#defineRson m+1, R, rtusing namespacestd;Const intMAXN =100000+131;intma[maxn2], mi[maxn2], lazy[maxn2];intCnt;voidPushup (intRT) {Ma[rt]= Max (ma[rt1], ma[rt1|1]); MI[RT]= Min (mi[rt1], mi[rt1|1]);}voidPushdown (intRT) { if(LAZY[RT]! =-1) {Lazy[rt1] = lazy[rt1|1] =Lazy[rt]; Ma[rt1] = ma[rt1|1] =Lazy[rt]; Mi[rt1] = mi[rt1|1] =Lazy[rt]; LAZY[RT]= -1; }}voidBuild (intLintRintRT) {Ma[rt]=

first instruction that gives the coordinates of the starting position. You may assume there is no more than instructions in each test case, and all the integer coordinates is in the range (-300, 300). The input is terminated if N is 0.OutputFor each test case there'll be one output line in the formatCase X:there is w pieces.,where x is the serial number starting from 1.Note:the figures below illustrate the and the sample input cases. Sample Input50 0 0 1 1 1 1 0 0 071 1 1 5 2 1 2 5 5 1 3

Test instructions: Given a sequence of DNA of length n of M, a shortest sequence of DNA is obtained, which minimizes the total hamming distance.Hamming the distance is equal to the number of different positions of characters.Analysis: See this problem, my first feeling is to calculate the time complexity, good small, nothing, completely can violence, as long as the same position on each string,The choice appears most, if has the same choice ASIIC code small (because requires the dictionary order

-Rujia's white paper describes the topic./*problem:status:by wf,*/#include "algorithm" #include "iostream" #include "CString" #include "Cstdio" #include "string" #include "stack" #include "cmath" #include "queue" #include "set" #include "map" #define Lson l, M, RT 　　LA 3942 Remember The Word (trie tree)

*q =NewLine[n]; Aboutq[first=last=0] = l[0]; the for(intI=1; I) the { the while(First1])) last--; + while(First; -Q[++last] =L[i]; the if(Fabs (Cross (Q[LAST].V, q[last-1].V)) EPS)Bayi { thelast--; the if(Onleft (Q[last], l[i]. P)) q[last]=L[i]; - } - if(first1] = Getintersection (q[last-1], q[last]); the } the while(First 1])) last--; the //Delete a useless plane the if(last-first1)return 0; -P[last] =getin

#include #include#includeusing namespacestd;Const intMAXN =4000* -+Ten;Const intSigma = -;Const intMaxnn =300010;Const intMoD =20071027;Charwords[Maxnn];intdp[Maxnn];structtrie{intCh[maxn][sigma]; intVALUE[MAXN]; intsz; Trie () {sz=1; memset (ch[0] ,0,sizeof(ch[0]) ); } voidInit () {sz =1; memset (ch[0] ,0,sizeof(ch[0]) ); } intDixCharc) {returnC'a' ; } voidInsert (Char*s) {intLen =strlen (s); intU =0; intI, cur; for(i =0; i ) {cur=dix (s[i]); if( !Ch[u][cur]) {memset (Ch[sz],0,sizeof(Ch[