class 2 div 2 enclosure offers a wide variety of articles about class 2 div 2 enclosure, easily find your class 2 div 2 enclosure information here online.

Variable access & modification in Python 2.x closures (enclosure)

Http:// 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

Codeforces Round #397 by Kaspersky Lab and Barcelona bootcamp (div. 1 + div. 2 combined) c-table Tennis Game 2

, 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

"CSS face question" Three div requires horizontal alignment, left and right two div width fixed to 100px, the middle of the div filled with the remaining width (at least 2 ways)

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>

Codeforces Round #437 (Div. 2, based on Memsql start[c]up 3.0-round 2)

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

Codeforces Round 362 (Div 2) e formula derivation + fast Power + Fermat theorem Please a[i]= (1-a[i-1]) except 2 n times n is a continuous multiplier

. 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

UESTC Summer Training #2 Div.2 E decomposition factor (except for the remainder may be a prime number)

2011 SourceHttp:// 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

Codeforces Round #437 (Div. 2, based on Memsql start[c]up 3.0-round 2) E. Buy low Sell high

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

Codeforces Round #437 (Div. 2, based on Memsql start[c]up 3.0-round 2)

, 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

Codeforces Round #197 (Div. 2) A. Helpful Maths "string/to a single add-on calculation that contains only the numbers 1, 2, 3, which require reordering, so that the number of added numbers from small to large"

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

Codeforces Round #437 (Div. 2, based on Memsql start[c]up 3.0-round 2)

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

UESTC Summer Training #2 Div.2 A dp, recursive, multi-stage problem

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

div Vertical/Horizontal Center 2 (div width and height are dynamic)

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

Codeforces Round #397 by Kaspersky Lab and Barcelona bootcamp (div. 1 + Div. 2 combined) E. Tree folding

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[

Codeforces Round #285 (Div.1 B & Div.2 D) Misha and permutations summation--two points + Tree array

,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

Codeforces Round #397 by Kaspersky Lab and Barcelona bootcamp (div. 1 + div. 2 combined)

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

Codeforces Wunder Fund Round (div. 1 + div. 2 combined) B Guess the permutation

. 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

Wunder Fund Round (div. 1 + div. 2 combined)

Map[i][j]=min (A[i],a[j]), that is, the value is the smaller of the two, and array A is a sequence that allows you to restore the array A through this array of valuesIdea: In each row if the number of different elements equals n, then this sequence except 0 think the other value is the number of series in the array A, the remaining as long as the 0 is changed to N.Another way of thinking (being taught by others) is that the maximum value in each row is actually the value of the array a[i], but

Codeforces Round #397 by Kaspersky Lab and Barcelona bootcamp (div. 1 + div. 2 combined) A

home or contestSolution: Since the departure from home will return, then we just have to compare the number of times the odd-and-even sex is on the line, the numbers are at home, odd is not back1#include 2 using namespacestd;3 intN;4 Chars1[ $],s2[ $];5 intMain ()6 {7 intans=0;8Cin>>N;9scanf"%s", S1);Ten for(intI=0; i) One { Ascanf"%s", S2); - if(STRSTR (S2,S1)! =NULL) - { theans++; - } - } - if(ans%

Codeforces Round #513 by Barcelona bootcamp (rated, div. 1 + div. 2) c. Maximum Subrectangle

, longer lengths, i.e. larger areas After that, you can select another sequence in two, use a double loop to get the sub-columns, and then divide the limit by this sum to get a target value. After that, in the array of initial preprocessing to find the target value, return is greater than its first value, that is, with Upper_bound, to find the length minus one, so that the longest length. Finally, by multiplying the length of the lookup with the length of the current child column, the result is

Codeforces Round #413, rated, Div. 1 + div. 2 C. Fountains (greedy or tree-like array)

); - } About Else $ { - if(p>d)Continue; - Dd.push_back (Make_pair (p,b)); -dd_max=Max (dd_max,b); A } + } the if(cc_max!=0 dd_max!=0) Ans=cc_max+dd_max;//first Case -Ans=max (Ans,solve (cc,c));//Two of them are coins . $Ans=max (Ans,solve (dd,d));//Two of them are diamonds . theprintf"%d\n", ans); the } the return 0; the}In addition, the tree array can also be done, you can refer to

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