class 2 div 2 enclosure

Codeforces Round #287 (Div. 2) (question A, question B ), PS: makeup. I forgot to send it in the past few days. At that time, SLB in the dormitory cried. He didn't squeeze in for the first 20 minutes. Finally, he couldn't move the computer to the fourth floor, sad. I made two questions and understood them later. However, due to poor learning skills, sad became a problem. A. Amr and Musictime limit per te

knives. M/(k + 1. However, the problem occurs. Based on the WA result returned by test10, I come up with pai_^. It is possible that the K-knife is not enough for full or vertical cutting! You need to divide the remaining number of knives into vertical cutting (corresponding to the previous full cross cutting) or cross cutting (corresponding to the previous full vertical cutting. That is, the else part of the code. In fact, the complete A1 Expression is like this: N/(n-1) * m/(k-(n-1) + 1), mean

Codeforces Round #302 (Div. 2) (ABCD ), Match link: http://codeforces.com/contest/544 A. Set of Stringstime limit per test: 1 secondmemory limit per test: 256 megabytes You are given a stringQ. A sequenceKStringsS1, bytes,S2, middle..., middle ,...,SKIs calledbeautiful, if the concatenation of these strings is stringQ(Formally,S1 worker + workerS2 cores + cores... cores + CoresSKSignature = SignatureQ) An

Codeforces Round #313 (Div. 2) E. Gerald and Giant Chess theorem Lucas calculates the big composite number,E. Gerald and Giant Chesstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output Giant chess is quite common in Geraldion. We will not delve into the rules of the game, we'll just say that the game takes place onHLimit × limitWField, and it is painted in t

Codeforces Round #282 (Div. 2) --- A. Digital Counter, Digital Counter Time limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output Malek lives in an apartment block with 100 floors numbered from 0 to 99. the apartment has an elevator with a digital counter showing the floor that the elevator is currently on. the elevator shows each digit of a number with 7 light sti

Codeforces Round #297 (Div. 2) --- E. Anya and Cubes, Anya loves to fold and stick. Today she decided to do just that. Anya has n cubes lying in a line and numbered from 1 to n from left to right, with natural numbers written on them. she also has k stickers with exclamation marks. we know that the number of stickers does not exceed the number of cubes. Anya can stick an exclamation mark on the cube and ge

Codeforces Round #296 (Div. 2) A B C D A: record the answer when simulating the moving phase division B: Three cases: can reduce 2, can reduce 1, cannot reduce respectively consider clearly C: Use a set and a multiset to separate the rows and columns. Use the sorting and query provided by the set to split the corresponding blocks into two blocks and insert them b

?000,1?≤? a,? b,? c,? d? ≤? N )-maximum possible value of an integer in the cell and four integers that Vasya remembers.OutputPrint one integer-the number of distinct valid squares.Examplesinput2 1 1) 1 2Output2Input3 3 1) 2 3Output6NoteBelow is all the possible paintings for the first sample.In the second sample, only paintings displayed below satisfy all the rules.SourceB. Restoring PaintingMy SolutionDraw a 3*3 square chart, then mark A, B, C, D an

to.This image describes second the sample case:It's easy -to-see- summary price was equal to.Test instructions: Give you some cities, some cities are provincial, and other cities will be connected. The General city is 1-2-...- n -1 This connection method, seeking to connect the city multiplied product and. Solution: In order to prevent repeated calculation, the first consideration is the connection of the provincial capital, we first save all the con

'#') { intt = (g[s.x][s.y]=='.'?1:0); if(a[s.x][s.y].step[k]==-1|| A[q.x][q.y].step[k]+t A[s.x][s.y].step[k]) {///because there is no need to build roads between kingdoms, all may be searchedA[S.X][S.Y].STEP[K] = A[q.x][q.y].step[k] +T; Q.push (s); } } } }}intMain () {intM, N; scanf ("%d%d", m, N); for(intI=0; i) scanf ("%s", G[i]); Memset (A,-1,sizeof(a)); for(intI=0; i3; i++) BFS (M, N, i); intAns =Oo; for(intI=0; i) for(intj

equal to 8 squares in total. Test Instructions : give you x, ask you how many kinds of n*m circumstances make, in the current rectangle within the number of small squares is XExercisesColumn push: Sigma (K=1,k=min (n,m)) (n-k+1) * (m-k+1) =x;We enumerate N and get M///1085422276#include using namespaceStd;typedefLong Longll;#defineMem (a) memset (A,0,sizeof (a))#definePB Push_backinline ll read () {ll x=0, f=1;CharCh=GetChar (); while(ch'0'|| Ch>'9'){ if(ch=='-') f=-1; ch=Get

]+ = Size[u];//size is how many mark nodes (including yourself) are in the subtree } }}intMain () {//freopen ("In2.txt", "R", stdin); intN, M; CIN>> N >>m; for(inti =0; I 1; i++) { intu, v; CIN>> u >>v; U--; V--; Adj[u].push_back (v); Adj[v].push_back (U); } for(inti =0; I ) { intv; CIN>>v; V--; MARK[V]=true; } DFS (-1,0); intv =-1; for(inti =0; I ) if(Mark[i] (v = =-1|| D[V] D[i])) V=i; memset (d,0,sizeof(d)); DFS (-1, V); intsum =0; intM

)? ≠? (aJ,?bJ,?CJ).The next line contains a integer- q (1?≤? Q≤?100), denoting the number of the queries.Then followsQLines, containing space-separated, integers- ui and vi (1?≤? u i,? v i? ≤? N ). It's Guaranteed that ui? ≠? v I .OutputFor each query, print the answer-a separate line.Sample Test (s) input4 51 2 11 2 22 3 12 3 32 4 331 23 41 4Output210Input5 71 5 12 5 13 5 14 5 11

single of these pairs are suitable for the second query. In the second, the sample XOR equals 1 for any subarrays of an odd length.Test instructions: Find out how many pairs of xor and the logarithm of k are in the interval;Idea: Mo team algorithm, mainly how O (1) update, a[l]^a[l+1].....^a[r]=pre[r]^pre[l-1]=k;pre[i] means a[1]^a[2]^....^a[i];Pos[r]^pre[l-1]=k; pre[r]=pre[l-1]^k; update the pre;#include using namespacestd;#definell Long Lo

Codeforces Round #277.5 (Div. 2) Question 1: Question: The minimum number of exchange requests does not decrease the sequence. The number of exchanges cannot exceed n. Idea: Since n is not exceeded, select sorting is used. Although the complexity is very high, the number of exchanges is indeed at least. Find the minimum value each time, and then exchange it with the current value. Question: A. SwapSorttime

Codeforces Round #306 (Div. 2) What happened recently? It's always dropping points. sad = ....... I will do it clearly, but I have no idea when I arrive at the competition... It is probably too sleepy... A. Two Substrings time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given stringS. Your task