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Hdu 1524 A Chess Game (SG) Question: There are n vertices in a directed acyclic graph. Each vertex has only one pawn and two people. Each time based on this graph, you can only move any pawn one step. If a player cannot move at a certain point, the player will lose. Analysis: this topic is the most typical directed acyclic graph game. We use dfs to calculate the SG values of all vertices and then perform a

Test instructionsThere are n heaps of stones, Alice takes first, each can choose to take a heap of stones in the 1~x (the total number of stones), you can also choose to divide the heap of stones into any two piles. Alice and Bob take turns and take the last stone to victory.Ideas:Because the range of the number is relatively large, it is best to find the regular solution by the results of the SG table.SG (4k+1) =4K+1;

*7) and 5Third Type 35 (5*7) and 2On the whole, according to the unique decomposition theorem of positive integers, any positive integer x must have x= (P1^R1) * (P2^R2) *......* (PN^RN)Define SUM=R1+R2+...+RN, the sum value is the total number of these stones, then sg=sg[sum1]^sg[susm2]^ ....Is that the problem again? How should we ask for this sum?We can get th

the back, that is, its transfer state and its number.Each stone can be regarded as a heap of stones, if it is the first heap of stones, the number of stones it represents is n-1-p.Because the gravel heap is not interfering with each other, the game can be seen as consisting of several games with only a dozen stones. (consider it separately)It can reach the sub-state of NIM and update its own SG values. Like sweeping the staircase, even if the number

Official: A Simple NimSg[0]=0When X=8k+7 is Sg[x]=8k+8,When X=8k+8 is Sg[x]=8k+7,The rest of the time sg[x]=x; (k>=0)It can be found by the rules of the table, the mathematical induction method can be proven.The specific table is placed in the code, see the INIT function/* by*/#include#include#include#includeusing namespaceStd;typedefLong LongLL;ConstLL n=1e6+Ten

Reprint please indicate the source, thank you http://blog.csdn.net/ACM_cxlove? Viewmode = contents by --- cxloveThis is a game against Directed Acyclic graphs.Question: There are N locations in which there is a topological relationship, which must be observed when moving. Finally, the mobile app wins and asks if there is a winning strategyHttp://acm.hdu.edu.cn/showproblem.php? Pid = 1, 1524A directed acyclic graph. The SG function of a vertex is the s

Stain on a 2*n grid every time a lattice or 2*2 lattice that can dye 1*1 is the most full of all the squares. M is the number of stained squares Alice initiatorThe 1*1 point SG value of 1 SG "I" means the SG value of continuous 2*i SG "3" is a continuous 2*3 blank latticeIf n = 3The case of a 1*1 lattice, then the subs

each position:if the described position is a winning position print a ' W '. If the described position is a losing position print a ' L '. Print a newline after all test case.Sample Input2 2 532 5 123 2 4 74 2 3 7 125 1 2 3 4 532 5 123 2 4 74 2 3 7 120Sample OUTPUTLWWWWLSource Norgesmesterskapet 2004 Test instructions: M heap stone players can remove Si stone from a heap each time cannot take the puzzle: Preliminary learning sg function

Question link: Https://icpcarchive.ecs.baylor.edu/index.php? Option = com_onlinejudge Itemid = 8 category = 242 page = show_problem problem = 1669 Root: regionals 2006: Asia-Beijing Regionals 2006> Asia-Beijing3668-a funny stone gametime limit: 3.000 seconds Question meaning: There are n piles of stones. You can choose three stacks of I, J, K each time. require (I Solution: Expansion of Nim game First, we only consider the transfer of one stone. It is a stone

Test instructions and NIM game almost, is to take the stone when the most can only take the original half, such as a pile of 5 stones up to two.First use a table to see the previous part of the SG value, and then find the rules to do.It is the most important procedure to make a table to find the value of SG.#include #include#definell Long LongintMain () {intvis[ -]; intsg[ -];

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=1536Test instructions: First enter K to indicate the size of a collection after which the input set represents the number of elements that can only go to this collection for this pair of stonesThen enter an m to indicate the next m-query for this setAfter m lines each row input an n means there are n heaps each heap has n1 a stone ask this line to indicate whether the state is win or lose if win input W otherwise lThe definition of the

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1536 Calculate the SG value. Note: 1. S is unordered. 2. You cannot create tables with N = 10000, because the number of tables that can be obtained is given and there will be points that cannot be obtained. Code:# Include # Include Int s [101], SG [10001], K; Int Mex (int n ){ If (SG [N]! =-1) return

output "Nacci", the output of each instance occupies one row.Sample INPUT1 1 11 4 10 0 0Sample Outputfibonacci#include #include#definell Long Longusing namespacestd;intf[ the]={1,2,3,5,8, -, +, the, -, the,144,233,377,610,987};intsg[1005];BOOLhas[1005];voidGETSG () {memset (SG,0,sizeof(SG)); for(intI=1; i +; i++) {memset (Has,false,sizeof(has)); for(intj=0;j the; j + +) { if(f[j]i) has[

Test instructionsThere are n heap of stones, two people take turns, each time can only take a pile of at least one to more than half stones, until can not be taken.Judge whether the initiator will win.Analysis:The key is to ask the SG function, but the direct analysis is not very good analysis, and then find the law.After some "ingenious" analysis, there is such a rule:If n is an even number, SG (n) = N/2;I

Cutting Game Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 3227 Accepted: 1195 DescriptionUrej loves to play various types of dull games. He usually asks other people to play with him. He says that playing those the can show his extraordinary wit. Recently Urej takes a great interest in a new game, and Erif Nezorf becomes the victim. To get away from suffering playing such a dull game, Erif Nezorf requests your help. Th

Problem Description Any college student should not be unfamiliar with the Fibonacci (Fibonacci numbers), which is defined as: F (1) = 1; F (2) = 2; F (n) =f (n-1) +f (n-2) (n>=3); So, 1,2,3,5,8,13 ... is the Fibonacci series. There are a lot of related topics in hdoj, such as 1005 Fibonacci again is once the Zhejiang province game problem. Today, another topic about Fibonacci appears, it is a small game, defined as follows: 1, this is a two-person game; 2, a total of 3 piles of stones, the numbe

number for the above formula when we make a sieve.#include #include#includeusing namespacestd;#defineMAXN 5000050intPrime[maxn],cnt=0;BOOLISPRIME[MAXN];intM[MAXN];intA[MAXN];intsg[ the];voidGetprime () {memset (IsPrime,0,sizeof(IsPrime)); memset (M,0,sizeof(m)); Memset (A,0,sizeof(a)); for(intI=2; i) { if(!Isprime[i]) {prime[cnt++]=i; for(intJ=i+i; ji) {isprime[j]=1; if(!m[j]) m[j]=i; } M[i]=i; } } for(intI=2; i) A[i]=a[i/m[i]]+1;}voidGETSG () {memset (

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=15241#include 2#include 3 using namespacestd;4 Const intM =10000000;//5 Const intN =1005;6 intSg[n], head[n];7 intCNT;8 9 structnode{Ten int from; One intto ; A intNext; - }edge[m]; - the voidAddedge (int from,intTo ) { -EDGE[CNT]. from= from; -Edge[cnt].to =to ; -Edge[cnt].next = head[ from];//edges from the same parent node +head[ from] = cnt++; - } + A voidinit () { atCNT =0;//total number of sides -memset (head,-1,sizeof(h

500030 0 1Sample Output0 2 31-1-1-10HintsourceIdeassg function, gameConstructs the SG function, takes the pawn position as the state, its successor is the J,K position. Enumerates the Ijk and accumulates tot according to the value of the SG function after operation.True Magic Code1#include 2#include 3 using namespacestd;4 5 Const intN = -;6 7 intN,a[n],sg[n],ans

sg function Great God: http://blog.csdn.net/luomingjun12315/article/details/45555495Template:intF[n],sg[n];BOOLS[m];voidGETSG (intN) {memset (SG,0,sizeof(SG)); for(intI=1; i) {memset (S),false,sizeof(S)); for(intj=1; f[j]) {S[sg[i-f[j]]]=true; } for(intj=0;; J + +)