hack cf

CF 518D (Ilya and Escalator-Take Dp directly if the number of combinations is too large), 518 dilya D. Ilya and Escalatortime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output Ilya got tired of sports programming, left university and got a job in the subway. He was given the task to determine the escalator load factor. Let's assume thatNPeople stand in the queue for the escalator. At each second one of t

CF 508D (Tanya and Password-Euler path, freelai algorithm), 508dpassword- D. Tanya and Passwordtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output While dad was at work, a little girl Tanya decided to play with dad's password to his secret database. Dad's password is a string consistingNKeys + keys 2 characters. She has written all the possibleNThree-letter continuous substrings of the password on pie

CF 269 E Flawed Flow, 269 flawed Emuskald considers himself a master of flow algorithms. Now he has completed his most ingenious program yet-it calculates the maximum flow in an undirected graph. The graph consistsNVertices andMEdges. Vertices are numbered from 1N. Vertices 1 andNBeing the source and the sink respectively. However, his max-flow algorithm seems to have a little flaw-it only finds the flow volume for each edge, but not its direction. he

// ------------------------------------------------------ variables ----------------------------------------------------------- 51 int T, I; 52 // ----------------------------------------------------------- work ------------------------------------------------------------- 53 getint (t ); 54 rep (I, 1, t) Work (I ); 55 56 // bytes //---------------------------------------------------------------------------------------------------------------------- 57 # ifdef debug 58 cout 59 # endif 60 return

XML (. net cf) Recently, I was working on a smart device project and needed to save the data. (Dataset. readxml is too convenient .) the trouble is that it takes four seconds to load an XML file containing only two rows of data. it seems that the "small" stuff on a PC like XML is still a dinosaur for smart devices! The mainstream configuration of Pocket PC is only 400 MHz! It is best to use whiteelephant. White Elephant Expensive gifts that ar

The weekend suddenly hurt, bored to look at TP, and then there is only CF such a TX game on hand, so there is a little record. Disable dual-machine debugging. It is basically the kernel API kddisabledebugger. First, let's take a look at the function definition. Ntstatus kddisabledebugger (void ); Because the parameters are available, you can directly change ret. Tessafe + 0x59dd:B15479dd 75b0 JNE tessafe + 0x598f (b154798f) Direct NOP Tessafe + 0x

Give you a tree with a permission $ A _ {I} $, with an edge $ W _ {e} $, defines a path named $2-path $. Each edge can pass through a maximum of 2 times and its weight is $ \ sum _ {x} A _ {x }\; -\; \ sum _ {e} W _ {e} \ cdot K _ {e} $, $ K _ {e} $ indicates the number of edge passes, for a total of $ q $ queries, each query goes through $2-path $ with the maximum weight value of $ X, Y $. Before reading the question, I felt that I could not do it... I thought it was a bit difficult, but I felt

Http://codeforces.com/problemset/problem/339/D Returns a sequence. Each time you change one of the values and then ask the F () of the sequence, the F () of the sequence is defined as: every two adjacent elements are halved by bit or length, in the sequence where two adjacent elements are split by bit or the length is halved again .... Replacement by bit or/different or, Until the sequence length is 1, output this number. OP maintains the operator symbol of the current node, and g maintains the

Every DP question in CF can give me a long review. It's so classic !!!!!!!! Give two strings A and B. Each operation divides a into two parts: C and D to form a DC. For example, a is abcdefg, ABC | defg -----> defg | ABC, then specify the number of operations K and ask the number of methods that convert A to B in step K. Link: http://codeforces.com/problemset/problem/176/ B Resolution: first obtain the number of cyclic points (Cyclic points: Generate

Tag: getch, But I'm ATI ext Clu pre direct perm OSI A very rewarding question (? ° ?)?? The limit is m, which seems to be not well handled. In general, what I know is exactly the K condition. Instead of using the combination number/dp state transfer/slope to divide the two points, there will be only room for rejection. We can first process the num [I], indicating that there are at least I perfect number of solutions, and then we can get the ANS [m] (exactly m ). How to obtain the num array? Set

1 #include View code Http://codeforces.com/contest/359/problem/ B 1 #include View code Cf B. Permutation

Topic linksTest instructionsA and b take turns in constructing a word, adding one character at a time, asking for the prefix of a given n string, the person who cannot add the character loses the game, and the person who loses is the first round of the game. Ask a initiator, after the K-wheel game, the last person who wins.Ideas:It is very obvious that n strings are inserted into the dictionary tree first, because there is a fork in the dictionary tree, it is not possible to judge the string len

It doesn't look like a single eye.And then someone says it's a network flow.Then I would like to build the map Ah, and then will not (is this konjac konjac too rubbish), there must be a network flow solutionAnd then went to the group to ask the Gdut giant, he said his teammates burst search + pruning (I was very impressed)Then I also wrote a 2^50 search pruning, actually really past (do not know the data is weak or a miracle)#include #include#include#include#include#include#include#include#inclu

intA[MAXN], B[MAXN], N, Len, H; + - intMain () { the //freopen ("Test.txt", "R", stdin);Bayi thescanf"%d%d%d", n, len, h); the for(inti =1; I "%d", b[i]); -Sort (b +1, B +1+len); -Tree.build (1,1, Len); the intAns =0; the for(inti =1; I i) { thescanf"%d", a[i]); A[i] = h-A[i]; theA[i] = Lower_bound (b +1, B +1+len, A[i])-b; - if(I Len) { theTree.Set(A[i], Len,1); theTree.insert (1); the }94 } the for(inti = len; I i) { theTree.Set(A[i], Len,1); theTree.i

Test instructions: The number of M operations for N is counted as the number of lucky numbers in the Count interval. Add a value to each number in the intervalDaniel on the internet is a tree-like array of line-segment trees can also be solved just a little hanging just learning line tree to strengthen the knowledge of the reserve and Code capability line Tree Single point update thank you for your advice#include#include#include#include#includeUsingNamespace Std;int nM;int ans;int a[1000000];int

him.InputThe first line of the "input contains four space-separated positive" integer numbers not exceeding 100-lengthes of the STI Cks.OutputOutput TRIANGLE If it is possible to construct a non-degenerate TRIANGLE. Output SEGMENT If the first case cannot take place and it was possible to construct a degenerate triangle. Output Impossible if it is impossible to construct any triangle. Remember that is to use three sticks. It is not a allowed to break the sticks or use their partial length.Sampl

,m,d;intA[MAXN],D1[MAXN],D2[MAXN];intVIT[MAXN];inthead[maxn],k;structedge{intv; intNext;} EDGE[MAXN1];voidinit () {k=0; memset (Head,-1,sizeof(head)); memset (D1,0,sizeof(D1)); memset (D2,0,sizeof(D2));}voidAddedge (intUintv) {EDGE[K].V=v; Edge[k].next=Head[u]; Head[u]=k++; EDGE[K].V=u; Edge[k].next=Head[v]; HEAD[V]=k++;}voidDFS1 (intUintt) { for(intI=head[u]; i!=-1; I=Edge[i].next) { intv=edge[i].v; if(vit[v]==0) {Vit[v]=1; D1[V]=t+1; DFS1 (V,t+1); } }}voidDFS2 (intUintt) { fo

-'0';returnx*F; + } - + Const intN =200005; A ConstLL mod =998244353; at - ll KSM (ll A, ll b) { -LL res =1; - while(b) { - if(B 1) Res = res * A%MoD; -A = a * a%MoD; inb >>=1; - } to returnRes; + } - the LL B[n]; * $ intMain () {Panax Notoginseng intn = Read (), M =read (); -LL A =read (); the for(intI=1; iread (); +LL ans =1, Inv2 = KSM (2, MoD-2); A for(intI=1; ii) { theLL L = b[i]-b[i-1]; +LL tmp =KSM (A, L); -Ans = ans * tmp% mod * (tmp +1) %MoD; $Ans

(root); } intMid = (L + R)/2; if(R 2, value); Else if(L > Mid) Query_1 (L, R, mid+1, R, root*2+1, value); Else{query_1 (L, Mid, L, Mid, Root*2, value); Query_1 (Mid+1, R, mid+1, R, root*2+1, value); } tree[root].sum= tree[root*2].sum + tree[root*2+1].sum;}intQuery_2 (intLintRintLintRintroot) { if(L = = L r = =R) {returntree[root].sum.ans[0][1]; } if(!tree[root].tempt. Judgeone ()) {down (root); } intRET =0; intMid = (L + R)/2; if(R 2); Else if(L > Mid) ret = Query_2 (l, R, mid+1

>using namespacestd;#defineCIN (x) scanf ("%d", x)#definefor (I,n) for (i=0;i#defineCLR (a,v) memset (A, (v), sizeof (a))inta[200005];intN;DoubleFDoublex) { Doubleans1=0, ans2=0; Doubleansans1=0, ansans2=0; for(intI=0; i) {ans1=max (ans1+ (A[i]-x), (a[i]-x)); Ansans1=Max (ANS1,ANSANS1); Ans2=min (ans2+ (A[i]-x), (a[i]-x)); Ansans2=min (ans2,ansans2); } returnMax (ansans1,-ansans2);}DoubleSanfen (DoubleLDoubleR) { DoubleMid,midmid,fmid,fmidmid; Mid= (l+r)/2; Midmid= (mid+r)/2; Fmid=f (m