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Time: 2016-03-31-23:49:13 Thursday Title Number: [2016-03-31][codeforces][659c][tanya and Toys].md The main topic: There are $10^9$ species of goods, the value of the item I, has used n items, to M-yuan, ask the most can buy the number of items not owned Analysis: Greedy, from the lowest start to buy, assuming $m = 10^9$, then also buy items not more than $10^6$, because $\frac{(1+k) K}{k} Problems encountered: The answer could be 0

Sad story: Our own OJ data seems a bit of a problem. Standard Range WA 5%ExercisesCopy to Google TranslateTranslation ResultsFirst, the proof of VFK God Ben to prove that the problem is three points.{I come to tell you the truth of the world = =Because this problem can be the minimum cost maximum flowEvery time the shortest circuit length does not dropSo it's a single peak.The shortest-circuit length is the difference value.So the first-order guide does not fall.Is it simple, rude? You have to

To the users of the illustrator software to the detailed analysis of the puzzle to share the production of toys design drawings tutorial. Tutorial Sharing: Step 1 First open the Preferences (CTRL+K), and then set the keyboard increment to 1in. Step 2 Using the Line tool, click in any position on the page, pop-up dialog box, input length: 30in, Angle: 30 degrees. Step 3 Now

process using a Huawei glory, a ZTE, a sister with a cool pie, aFake Android, all test OK. Two air conditioning tests are normal. Perfect! A large number of test processes have been found to be in the receiving area. Generally there is a 5M above, if you add some driving circuit effect will be better, but if not the effect of the result is very good. so Easy to put, so that we can use the phone as our external energy remote control. It's the home appliance. .end!Welcome to Exchange, sharing the

New toys-comprehension-arrangements in the dark-DizzyDude-Linux general technology-Linux technology and application information. The following is a detailed description. Yesterday, I accidentally found a new toy, an idle wireless router. The reason for this is that I once saw an article about telnet to the SOHO Wireless Router. I think it is very interesting. In addition, our teacher said that he logged on to his router to perform operations, he said

Also, let's take a look at my Linux toys list. By: Wu Yin Date: 2008-01-31 Homepage: Http://blog.csdn.net/wooin Email: Lazy_fox # msn.com Copyright information: This article is copyrighted by Wu Yin and his lovely wife. It can be freely transmitted and replicated for non-commercial purposes. For commercial purposes, any behavior in this article shall be approved by the author.Contact info: la

The computational geometry finally opened the hole ...Cross product + two points.#include #include#include#include#defineMAXN 5050using namespacestd;structpoint{intx, y; Point (intXinty): x (x), Y (y) {} point () {} friend pointoperator-(point X,point y) {returnPoint (x.x-y.x,x.y-y.y); }}P[MAXN];structline{Point X,dt; Line (point X,point DT): X (x), dt (DT) {} line () {} friendint operator*(line X,line y) {returnx.dt.x*y.dt.y-y.dt.x*X.dt.y; }}L[MAXN];intn,m,cnt[maxn],x1,y1,x2,y2,x,y;BOOLCheckint

The following content is reproduced from mobileside.cn-your mobility! Our passion! Those who have heard of my teched 2007 courses in Beijing or Guangzhou may still remember the power toys for. NET Compact Framework 3.5 ctp I used. The powerful remote Performance Monitor and. Net cf clr profiler may be impressive. Now, the good news is that this tool has been officially released. You can download it from the Microsoft website. In:Http://www.mic

Given a string consisting of 'w', 'I', 'n', and 'G', these rules can combine two characters into one, for example, "II" can be synthesized into 'w', "ww" can be synthesized into 'I' or 'N' Which types of characters can this string be finally synthesized? Seeing this question, I think of broad search... Actually, it is not necessary. It can be completely solved by action. If f [I] [J] [k] is the J character starting from I, can it be merged into a character [k] Then J loops from the outer layer,

title link : Http://codeforces.com/problemset/problem/659/CTest Instructions :N is an already existing number, M is the maximum number available andAsked to choose the numbers that they did not have, and the numbers of those numbers could not exceed mand require the number of optional numbers the more the betterOutput an AnswerProblem Solving Ideas :Just started to open a bool type of 1e9 array, and then judge canBut hand it over to find out the memory limitThe 1e9 array is then converted to an

[x];i!=-1;i=next[i])using namespaceStd;inlineintRead () {intx=0, f=1;CharC=GetChar (); for(;! IsDigit (c); C=getchar ())if(c=='-') f=-1; for(; IsDigit (c); C=getchar ()) x=x*Ten+c-'0'; returnx*F;}Const intmaxn= -;Const intMod=1000000007; typedefLong Longll;intn,m,es[11maxn];ll g[11MAXN],ANS[MAXN];intMain () {n=read (); m=read (); Rep (I,1, m) U[i]=read ()-1, V[i]=read ()-1; Rep (S,0,(11) {Rep (I,0, N-1)if(s (1; Rep (I,1, m)if((s (11; } xp[0]=1; Rep (I,1, m) xp[i]= (xp[i-1]1)%MoD; g[0]=1; Rep (

Use the cross product to determine whether the point is on the left or right side of the segment, then two points can beCode:#include POJ 2318 TOYS (computational geometry)

Simply calculate the geometry and determine whether the point is inside the quadrilateral with a cross product.#include #include#include#includeusing namespacestd;Const intmaxn= the+Ten;intN,m,x1,y1,x2,y2;intU[MAXN],L[MAXN];intANS[MAXN];BOOLFLAG[MAXN];structpoint{intx, y;} P[MAXN];Chars[10000];voidinit () {memset (flag,0,sizeofflag); memset (ans,0,sizeofans);}intFintA1,intB1,intA2,intB2) { //returns the cross product of a vector (A1,B1) and a vector (A2,B2) returna1*b2-a2*B1;}BOOLJudgeintA

Cross-product utilization inferred point segments left or right, and then you can 2 minutesCode:#include POJ 2318 TOYS (computational geometry)

After passing through the new employee training room in Java, I checked it out and saw that they were doing an exercise that the teacher gave them. After walking around, I felt like I was back on campus, students are practicing in the dormitory! Because they practice for the purpose of practice, training allows them to learn the Java language, HTML tags, JSP, struct, spring, and so on, but they write toys that make them intoxicated! Because they

State space Search? Use a map to judge weight. #include #include#include#include#includeusing namespacestd;intdx[]={1,0,-1,0},dy[]={0,1,0,-1};structStatus {inta[4][4]; Status Move (intXintYintdir) {Status res= * This; if(!a[x][y])return* This; intNx=x+dx[dir];intny=y+Dy[dir]; if(nx0|| Nx>3|| ny0|| Ny>3)return* This; if(A[nx][ny])return* This; Res.a[nx][ny]=1; res.a[x][y]=0; returnRes; } BOOL operatorConstStatus b)Const { for(intI=0;i4; i++) for(intj=0;j4; j + +)if(A[i]

toys:http://poj.org/problem?id=2318 To give you a box, there are n panels separated into n+1 parts, give you the coordinates of the M toy, ask each part has several toys. Train of thought: to each toy, the two-segment subscript, judge the toy in the left or right side of the line, enumerated after statistics. More Wonderful content: http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/ #include

The emphasis is on the production of fur textures. The authors introduce a very simple way to do this with a filter. There are fewer filters to use: Noise and radial blur filters, with only a few more styles to be added later, and the adjustment of

Description Analysis Is not a look like what network streamline planning and so on.However, there is no train of thought, a violent, plus pruning but run fastPlot, toy for edge, shop for PointSearch for the selected points, then one side of the

Poj_2318 You can use the cross product to determine the left or right of the online segment. # Include # Include String . H> # Define Maxd 5010 Int N, m, X1, Y1, X2, Y2, a [maxd], B [maxd], H [maxd]; Long Long Int Det ( Int X1, Int Y1,Int X2,