class 2 div 2 enclosure

diamonds.AC Code:1#include 2 #defineINF 0x3f3f3f3f3 #definell Long Long int4 using namespacestd;5 6 Const intMAXN = 1e5+Ten;7 8 intT_1[MAXN], T_2[MAXN];9 intN, Num_c, num_d;Ten One intLowbit (intx) A { - returnx (-x); - } the voidAddintNointStintvalue) - { - for(inti = st; I lowbit (i)) - { + if(no) t_1[i] =Max (T_1[i], value); - ElseT_2[i] =Max (T_2[i], value); + } A } at - intQueryintNointSt) - { - intres =0;

Div 2 A simulation. Simulation. Simulation. I have read a wrong sentence. I have always done this kind of thing recently... According to his experience, he thought that Freda always said "Lala." at the end of her sentences, while rainbow always said "Miao ."At thebeginning of his sentences. Cause wawawa ---- find problem --- solve ----- AC #include B proportional sequence infinite summation formula A + AP ^

". ExamplesinputAbacabaOutputYESinputJinotegaOutputNONoteIn the first sample case, one possible list of identifiers would is "numberstring number character number string numb Er ". Here are Kostya would obfuscate the program: Replace all occurences of number with A, the result would is "a string a character a string a", Replace all occurences of string with B, the result would is "a B a character a b a", Replace all occurences of character with C, the result would is "a B

Pure div+css made beautiful click button and Close button, click the Click button also has effect. These are not pictures.Three points worth noting:1. Among them, the main use of rotate. It allows the text to rotate the angle2. There are radius, do the round corner dedicated, very useful properties. For people like me save time for drawing.Transition of CSS3 in the 3.W3C standard this is kind of description: "CSS transition allows CSS property values

Div2 A: Water question Div2 B: change each number to 1. the required number of steps is the F value of this number (the odd number minus one and the even number is divided by 2) Div1 A: record the highest current height. Div1 B: Sb question, there will be many points with the same X coordinate, but there may be some key points, so multiply the CNT! /(2 ^ c). Because the total CNT size cannot exceed 10000

){ +scanf"%lld%lld%lld",x,y,z); - if(y>=z) { theA.push_back ({x,y-z}); *sum1+=x*1LL; $sum+= (x*y) *1LL;Panax Notoginseng } - Else { theB.push_back ({x,z-y}); +sum2+=x*1LL; Asum+= (1LL) *x*Z; the } + } - if(check ()) { $coutreturn 0; $ } - sort (A.begin (), A.end (), CMP); - sort (B.begin (), B.end (), CMP); the //cout -ll ans1=sum,ans2=sum;Wuyill xx=sum1%S; the for(intI=0; I){ -ll min=min (xx,a[i].s); Wuans1-=min*a[i].x; -xx-=Min; About

Test instructions: Minus the previous number, plus the number of the back, to ensure that the last number, plus and minus the same number of times;Puzzle: Emmmmm, see is a greedy, first on the value sort, the same on the subscript sort, the rule is to find the first one, and then from the back to the next table is not used the first, subtract, and then until found,But this time complexity is O (n^2), think for a long time or do not know how to use pri

Codeforces Round #221 (Div. 2) D. Maximum Submatrix 2 (thinking question ), Question address: codeforces 221 DThis is the first question of D in CF in our life. (At that time, only question A was made ..) When I reviewed the Chinese New Year, I took a few D questions by the way. Currently, the question "D" of CF cannot be done during the competition. However, yo

0x3f3f3f3ftypedefLong Longll; typedef unsignedLong Longull;using namespacestd;Const intMAXN = 3e5+5; Priority_queueA;intHAS[MAXN];intMain () {Ios::sync_with_stdio (false); Cin.tie (0); Cout.tie (0); intI, J, K, M, N; CIN>>N; ll ans=0; for(intI=1; ii) {cin>>K; A.push (MP (K,i)); if(A.top (). first>=k)Continue; Ans+=k-A.top (). First; if(!Has[a.top (). Second]) {A.pop (); Has[i]=1; } Else{has[a.top (). Second]=0; Has[i]=1; }} coutEndl; return 0; } Codeforces Round #43

Tags: there is a start personal Force str msql ASE Post key pointASignBTest instructionsAnalysisCTest instructionsThere are two kinds of pizza, each pizza can be divided into S block, there are n individuals, respectively, give the number of blocks that n people need, eat the first 1 pieces of the value obtained, eat the value of the second 1, ask in the need for the minimum number of pizza can get the maximum value of how muchAnalysisKey points: Everyone to the best, the two pizza the rest will

Codeforces Round #513 by Barcelona bootcamp (rated, div. 1 + div. 2) Solved:2 out of 8 ... rank:2730 Unrated A. Phone Numbers Difficulty: Universal group. Simulation can be: Summary: Do not waste too much time on simple questions, to improve the accuracy of the problem. #include

Tag: INF pair scanf has root tree oid test i++ form ClassF-uniformly branched Trees#include #defineLL Long Long#defineFi first#defineSe Second#defineMk Make_pair#definePII Pair#definePLI Pair#defineull unsigned long Longusing namespacestd;Const intN = 1e3 +7;Const intINF =0x3f3f3f3f;ConstLL INF =0x3f3f3f3f3f3f3f3f; LL dp[n][ One][n], Inv[n], f[n], finv[n], g[n][ One];intN, D, mod;voidinit () {inv[1] = f[0] = finv[0] =1; for(inti =2; i MoD; for(inti

such then All Stars lie on that Line.OutputPrint three distinct integers on a single line-the indices of the three points so form a triangle that satisfies the C Onditions stated in the problem.If There is multiple possible answers, you could print any of the them.Sample Test (s) input30 11 01 1Output1 2 3Input50 00 22 02 21 1Output1 3 5NoteIn the first sample, we can print the three indices on any order.In the second sample, we had the following pic

print mnumbers h(1),?...,? H(m). If There is several correct answers, you could output any of the them. It is guaranteed that if a valid answer exists and then there was an answer satisfying the above restrictions.Examplesinput31 2 3Output31 2 31 2 3input32 2 2Output11 1 12input22 1Output-1Idea: The person with

", u, v, W); Edge[u].push_back (Mk (W, v)); Edge[v].push_back (Mk (W, u)); } LL ans=0; for(inti =1; I ) { if(!Vis[i]) { Base. Init (); Id.clear (); DFS (i,0); for(intj =0; J +; J + +) {num[0] =0, num[1] =0; for(Auto x:id) {num[(d[x]GT;GT;J) 1]++; } BOOLFlag =false; for(intK =0; K the; k++) { if((Base. A[k]>>j) 1) {flag=true; Break; }} LL tmp= num[0]* (num[0]-1)/2+ num[1]*

　　A question, water problem, judge all points is not into degrees and out of the same can be.Question B, test instructions understand is water problem.Question C. The other party must win at least one of the points after the score mod K. This method can be used to judge.Problem d, construction problem, not = =.E, test instructions is the choice of a node each time, the two sons of the same length of the chain to merge into a new chain of the same length. Ask if the end can be a straight chain, i

it is borrow (0,1), current digital and for nowShould pay attention to the borrow and not borrow the case of the decision (originally to consider less than borrow 9 can not borrow, was hack)1#include 2 using namespacestd;3 intans=0, a[ the]={0};4 intmark[ the][5][255]={0};5 intDfsintIdintBorintNow )6 {7 if(id==a[0]+1)8 {9ans=Max (ans,now);Ten returnans; One } A - if(mark[id][bor][now]!=-1)returnMark[id][bor][now]; -

[f[i]]=F[i]; G[i]=F[i]; } intCNT =1; for(inti =1; I ) if(H[i]) {Id[h[i]]=CNT; H[CNT]=H[i]; CNT++; } for(inti =1; I ) G[i]=Id[g[i]]; for(inti =1; I ) if(G[h[i]]! =i) {puts ("-1"); return 0; } for(inti =1; I ) if(H[g[i]]! =F[i]) {Puts ("-1"); return 0; } printf ("%d\n", CNT-1); for(inti =1; I ) printf ("%d", G[i]); Puts (""); for(inti =1; I ) printf ("%d", H[i]); Puts (""); return 0;}View CodeThis is the most pit of C, the first step on the pit I thought in time

To implement arbitrary drag of Div, we might as well analyze the whole process. When the mouse clicks on the DIV, triggers an event, lets the div position attribute (left,top) changes with the mouse position change, when the mouse is released, the div position attribute uses the mouse to release the position. When the

General website will do click Logo, will return to the homepage, how to do it, we first thought, to the picture plus links on it What do you do with the big exercises in the first four verses? There is no difficulty, if you feel that there is no relationship between the difficulty, this class, I take you to do this exercise!"First step overall layout and public CSS definition"Let's analyze this page first.The page is mainly divided into 5 large chunks