class 2 div 2 enclosure

Http://stackoverflow.com/questions/3190706/nonlocal-keyword-in-python-2-x---answer----Python can read nonlocal variables in 2.x, just not the change them. This was annoying, but can work around it. Just Declare a dictionary, and store your variables as elements thereinpython2.x Natural support for Nolocal class variable access, as long as they do not change thei

, for example, the rule of "balance" (The winning player have to is at least, points ahead to win a set) have no power within the present problem.Test instructions: Test instructions good fan, do the problem by enumerating test instructions.The title says two people are playing, each game has a lot of rounds, each round of winning players can get a point. If someone's score is k in a match, the game ends and a game is re-opened.Now give the two people a total score, a, B, ask you whether the sco

This is my interview in a company encountered problems, then did not answer!!So see the small partners must pay attention to!!Change the browser width to see the effect:Right and left.Then we'll look at the code:First method: (floating)styletype= "Text/css">. Left,.right,.center{Border:1px solid;Height:100px;text-align:Center;Line-height:100px;font-size:50px; }. Left{float: Left;width:100px; }. Right{float: Right;width:100px; }. Center{margin:0 100px; }style>

favorite pizza, otherwise we can only buy a pizza, It means that some people can't eat their favorite pizza, and now we need to think about what kind of pizza to buy is better, in the greedy mind, we must be for two kinds of pizza like the difference is not obvious (that is, ABS (A-B) is small), can not eat their favorite pizza, This brings the loss is the smallest, and then judge let like eat type1 eat type2, or like to eat type2 eat type1 which loss more small lost on the good. Cod

. Instead He gave you an array a1, A2, ..., Aksuch that In other words, n was multiplication of all elements of the given array. Because of precision difficulties, Barney asked you to tell him the answer as an irreducible fraction. In all words need to find it as a fraction p/q such that, where is the greatest common divisor. Since p and Q can be extremely large, you have need to find the remainders of dividing each of the them by 109 + 7. Please note that we want the P and Q to is 1, not the th

2011 SourceHttp://acm.hust.edu.cn/vjudge/contest/view.action?cid=121703#problem/EMy SolutionDecomposition factorization, at least three different qualitative factors are lucky number, with the modified decomposition factorization template, is only a record of the quality because of the number of good, no tube is what and different quality because a number of more than equals 3 return 3The decomposition of the quality factor template do not know why WA, changed to

Tags: i++ ras span Art else add div start sumTest instructions: Some stock prices, we can choose to buy and sell, but only one operation a day, ask the maximum profit Idea: For the day, if the sale before he was smaller, we must be looking for a minimum day of buy, but do not know whether the sale is the most profitable, so we can use Multiset, this and set similar, but can store the same number, and sort So we delete that smallest, add

, add two B, (b-a) join ans, The first B is the equivalent of turning a into B and the second b being itself. 1#include 2 using namespacestd;3 Const intn=3*1e5+5; 4 intN,a[n];5 intMain ()6 { 7scanf"%d",N); 8 for(intI=1; i"%d",A[i]); 9priority_queueint,vectorint>,greaterint> >Q;Ten Long Longans=0; One for(intI=1; i) A { - if(Q.empty ()) - { the Q.push (A[i]); - Continue; - } - intt=q.t

A. Helpful mathstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputXenia The beginner mathematician is a third year student at elementary school. She's now learning the addition operation.The teacher have written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum is only contains numbers 1, 2 and 3. Still, that's isn ' t enough for Xenia. She is

of the coins. All denominations must is distinct:for any i? ≠? J we must have Di? ≠? DJ. If There is multiple tests, print any of them. You can print denominations in atbitrary order. Examplesinput 18 Output 30 4 1 5 10 25 Input 3 Output 20 2 5 2 Input 314 Output 183 4 6 5 2 139 I really don't have a clue ab

possible path. He needs at least strength = 2 initially.Case 2:note this to start from (a) he needs at least strength = 1.Hint Added by: Varun Jalan Date: 2011-12-15 Time limit: 0.336s Source limit: 50000B Memory limit: 1536MB Cluster: Cube (Intel G860) L

Tip: Run this code to download the jquery pluginCode principle: Through jquery set div css, get div's left, margin offset, margin offset algorithm is to use the width of the page window minus the div width, the resulting value divided by 2 is the left offset, the right offset algorithm is the same Div Vertical/Horizont

read his blog, and tell it in detail.1#include 2 3 using namespacestd;4 5 #defineMP Make_pair6 #definePB push_back7typedefLong LongLL;8typedef pairint,int>PII;9 Const Doubleeps=1e-8;Ten Const DoublePi=acos (-1.0); One Const intk=2e5+7; A Const intmod=1e9+7; - - intn,mx; thevectorint>Mp[k]; - - intDfsintXintf) - { + inta[3],num=0; - for(intI=0; I) + if(mp[x][i]!=f) A { at intv=Mp[x][i]; -a[

,sizeof(c)); for(i=1; i1); for(i=1; i) {scanf ("%d",x); X++; P2[i]= Getsum (X-1); Modify (x,-1); } memset (P3,0,sizeof(p3)); for(i=n;i>=1; i--) {P3[i]+ = p1[i]+P2[i]; if(P3[i] >= (n-i+1) ) {P3[i]= p3[i]-(n-i+1); if(I! =1) p3[i-1]++; }} memset (C,0,sizeof(c)); for(i=1; i1);//For (i=1;i//cout//cout for(i=1; i) { intLow =1, high =N; while(Low High ) { intMid = (Low+high)/2; if(Getsum (mid-1) >P3[i]) high= mid

case of the root of 1 and O (1) Move the root#include #includeusing namespaceStd;inlineintRead () {intXCharC; while((C=getchar ()) '0'|| C>'9'); for(x=c-'0';(C=getchar ()) >='0'c'9';) x= (x3) + (x1) +c-'0'; returnx;}#defineMN 200000structedge{intNx,t;} e[mn*2+5];inth[mn+5],en,d[mn+5],d1[mn+5],d2[mn+5],c[mn+5],g;inlinevoidInsintXinty) {e[++en]= (Edge) {h[x],y};h[x]=en; e[++en]= (Edge) {h[y],x};h[y]=en;}voidCalintXintd) { if(!d1[x]) D1[x]=d;Else i

. If There is multiple possible solutions, print any of them.Sample Test (s) input20 11 0Output2 1Input50 2 2) 1 22 0 4) 1 32 4 0) 1 31 1 1) 0 12 3 3) 1 0Output2 5 4) 1 3NoteIn the first case, the answer can is {1, 2} or {2, 1}. In the second case, another possible answer is {2