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Codeforces Beta Round #34 (Div. 2) C. Page Numbers C. Page Numbers time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output «Bersoft» company is working on a new version of its most popular text editor-Bord 2010. bord, like export other text editors, shocould be able to print out multipage documents. A user key

Title Link: Http://codeforces.com/contest/810/problem/BTest instructions: Given the number of days and the number of days the goods can be doubled, every day there is a certain amount of goods and the number of customers, ask how the goods can sell the most (the day before the goods will not be left to the next, each customer can only buy one goods).Simple greedy question, greedy strategy is: if two times the volume of goods sold out more, choose twice times, or choose one times.The amount of go

(infinite length) carries n operations each time the number of positions A and B is exchanged, asking the final sequence how many pairs of reverse order number.First use map to make the final sequence (only need to get the number of exchange positions in the topic). Then follow the tree-like array to reverse the way to operate. But it needs to be discretized.Consider a sequence a[l],l+1,l+2,...., r-1,a[r] where the numbers on L and R are exchanged (n

up to a[i], because only from the point of entry to the point, so that the soldiers can only walk one side.The code uses templates directly, and there are two implementations of the Dinic EK algorithm. #define Inf9000000000#define EPS (double) 1e-9#define mod1000000007#define pi3.14159265358979//********************* /#endif # define N 405#define M 100005#define maxn 450# Define MOD 1000000000000000007int n,a[n],b[n],s,e,sum,maxf,m,ans[n][n],s1;struct edge{int from,to,cap,flow; Edge (int

)) - #defineAll (n) n.begin (), N.end () - #defineLson (x) ((x - #defineRson (x) ((x - #defineINF 0x3f3f3f3f intypedefLong Longll; -typedef unsignedLong Longull; to using namespacestd; + Const intMAXN = 1e6 +5; - ll SAVEPOW[MAXN]; the intCNT[MAXN]; * ll F[MAXN]; $ intMain ()Panax Notoginseng { -Ios::sync_with_stdio (false); theCin.tie (0); Cout.tie (0); + intI, J, K, M, N; ACIN >>N; the for(inti =1; I i) + { -CIN >>K; $cnt[k]++; $ } -savepow[0] =1; - for(inti =1; I 1000000;

A. Vasya and socks I don't need to talk much about water issues. It's just a matter of brute force enumeration. #include B. Little Dima and Equation Needless to say, if you have a question, it is appropriate to enumerate S (x) directly, then obtain X, and then launch S (x) based on X. At that time, 81 was written as 72, and it was sad, sad. #include C. Present Binary + greedy. It is also a form of water question .... The result is a result of two points. Greedy to see if the current result

instructions A string of 1-8 to select a substring (in the original string can be discontinuous) the string of each number of the same must be adjacent and the number of each is not more than 1The game only thought of the DFS solution, and later proved to be wrongLater look at someone else's code a face mask =. =... Then a god said two arrays to understand the role of ... So refer to the wording of ClarisAnd....This also wrote for a long time and also less update status length can be 0 ...#incl

=1 case.#include D:chloe and Pleasant PrizesThe main idea: a tree with 1 roots, given the point right of each point, find two sub-trees that do not intersect and make their weights and as large as possible, if the output-1 is not found.Idea: The first thing that can not be found is the case of a chain (of course, 1 must be an endpoint of the chain), and then the answer directly on the tree DP, and trees DP for the longest chain of the basic hair.#include E:vladik and CardsTitle: In a given seque

instructionsGiven a series of numbers, requires a continuous two numbers and not less than K, ask each number in accordance with the requirements of the case, the minimum need to add how much, the output of the last string value.Official:If we don ' t make enough walks during daysIandI+ 1, it's better to make a additional walk on day I + 1 because it also counts as a walk during days I + 1 and I + 2 (and if we walk One more time on day I, it won '

Codeforces Round # FF (Div. 2): C. DZY Loves Sequences, C. DZY Loves Sequencestime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output DZY has a sequenceA, ConsistingNIntegers. We'll call a sequenceAI, Bytes,AIPipeline + pipeline 1, pipeline..., Pipeline ,...,AJ(1 digit ≤ DigitILimit ≤ limitJLimit ≤ limitN) A subsegment of the sequenceA. The value (JAccept-Encodi

Codeforces Round #257 (Div. 2/B)/Codeforces450B_Jzzhu and Sequences, B solution report It's a regular question., x y z-x-y-z. Note that if the number is less than 0, you must first modulo the negative number and then modulo it. You cannot directly add mod, but it may still be a negative number. The stamp code for me is kneeling ,,, #include Jzzhu and Sequencestime limit per test1 secondmemory limit per t

Codeforces Round #269 (Div. 2) Solution, Another DIV2 attack was triggered overnight, and the problem was solved. Now the final test result is not available, so it is estimated that it will be ready. Passed the Pretest of the first four questions. Update: This time rank220, It's a pitfall! A: give you six lengths (four limbs, Head, and trunk) to determine the species. The judgment rules are very clear, but

-th (1?≤? I? ≤? n) position of the sequence contains the character "+", if on theI-th step from the Wall the "plus" wire runs above the "minus" wire, and the character "-" otherwise.OutputPrint either "Yes" (without the quotes) if the wires can be untangled or "No" (without the quotes) I f The wires cannot be untangled.Sample InputInput-++-OutputYesInput+-OutputNoInput++OutputYesInput-OutputNoHintThe first testcase corresponds to the picture in the statement. To untangle the wires, one can firs

And finally came back to the familiar round.Math a-pouring RainSet an unknown, solve the equation, fortunately no hack point#include Mathematics b-coat of AnticubismTest instructions: To increase the minimum length of a stick, so as to form a polygon.Analysis: Then the formation of the triangle, the original n sticks are divided into a, A and B close (a#include DP C-reberland LinguisticsTest instructions: The suffix is composed of several substrings of length

+--only, representing +1,-1The second string contains? may be + or--now ask you what the probability of the second string equals the first string is worth. the correct: for no, direct sentenceYes? We're going to have to do a lot of violence. Character length less than 11, duly violent///1085422276#include using namespacestd; typedefLong Longll;#defineMem (a) memset (A,0,sizeof (a))#defineMeminf (a) memset (A,127,sizeof (a));#defineTS printf ("111111\n");#definefor (I,A,B) for (int i=a;i#defineF

sticks.OutputIf you can make a bear from the given set, print string "Bear" (without the quotes). If you can make an elephant, print string "elephant" (w?thout the quotes). If you can make neither a bear nor an elephant, print string "Alien" (without the quotes).Sample Test (s) input4 2 5 4 4 4OutputBearInput4 4 5 4 4 5OutputElephantInput1 2 3 4 5 6OutputAlienNoteIf you ' re out of the creative ideas, see

, S;intMain () {#ifdef LOCAL freopen ("534b.in","R", stdin);#endifCin>> V0 >>VT; CIN>> T >>D; if(V0 >VT) Swap (V0, VT); T-=1; VH=V0; S=V0; while(VH 1) *d t) {VH+ = min (d, vt+ (t1) *d-VH);//cout s + =VH; T--; } while(VH > VT | |t) {s+=VT; VT+ = min (d, vh-VT);//cout t--; } coutEndl; return 0;}C. Polycarpus ' DiceMath problems, such as the first dice can take 1, 2 then a value range is (1+ left dice to take the smallest) ~ (The dice to take the la

#include 2#include 3#include 4#include 5 using namespacestd;6 7 Const intMAXN =200000+5;8 CharA[MAXN], B[MAXN];9 intC0[MAXN], C1[MAXN];Ten One Long LongRes; A - intMain () - { the #ifndef Online_judge -Freopen ("In.txt","R", stdin); - #endif //Online_judge - + while(SCANF ("%s%s", A +1, B +1) !=EOF) { - intLa = strlen (A +1); + intLB = strlen (b +1); A atmemset (C0,0,sizeof(C0)); -memset (C1,0,sizeof(C1)); - - fo

that can be reversed for a certain part, so that the sequence is incremented, note that then also print space-separated integers denoting start and end, the output is the position, not the currentThe value. And if the reversal has the reverse () function, use sort () WA.Idea: To find the subscript from left to right, which is smaller than the previous number, and then to the right-to-left to find the subscript larger than the next number. Find and break#include Copyright NOTICE: This article f

///#include #include #include#include#include#include#include#include#include#include#includeSet>#includeusing namespacestd;#defineMem (a) memset (A,0,sizeof (a))#definePB Push_back#defineFi first#defineSe Second#defineMP Make_pairtypedefLong Longll;Const intN =201000;Const intM =1000001;Const intINF =0x3f3f3f3f;Const intMOD =1000000007;Const DoubleEPS =0.000001; ll A[n],n;intMain () {scanf ("%i64d",N); ll ans=inf; for(intI=1; i) {scanf ("%i64d",A[i]); Ans=min (ans,a[i]);} for(intI=1; ians; ans