coursera game theory

0.618033988749894848204586834 ... Split the integer into the array, split it into three parts, and then multiply to reduce the loss of precision. See the code specifically: m y Span style= "Display:inline-block; width:0px; Height:2.616em; " > Code #include #include #include #include #include using namespace STD;typedef Long LongLL; LL tmp[3] = {618033988,749894848,204586834}; LL MOD =1000000000;intMain () {intT LL m,

HDU 2147Based on the initial theory, Kiki's game can draw a picture to find out the rule of victory and defeat. # Include HDU 1850Being a good boy in Spring Festival is more of a theory. Anyway, it seems that it is recorded by YY +. # Include

Enumerates the possible states of the first step and determines whether they are required to fail.#include "Game theory" "SG function" "Enumeration" bzoj1874 [BeiJing2009 Wintercamp] Take a stone game

Because all the beans in the first bottle are equivalent, set SG (i) to indicate the SG value of the bottle I, which can be transferred to the state of SG (j) ^sg (k) (IJust consider the number of beans is an odd bottle, because if the number of beans is even, repeated or meaningless.For the number of programs or something ... The enumeration is good.#include "Game theory" "SG function" "Enumeration" bzoj1

Two people, if N is a multiple of m+1, then the first person no matter what number, the second person only need to make two people get a number equal to m+1 on the line, so the second personI'm sure to win.If n is not a multiple of m+1, then the first person only needs to take n% (m+1), and then the following situation is the same as above#include #include #include using namespace Std;int main (){int n, m;int T;scanf ("%d", t);while (t--){scanf ("%d%d", n, m);if (n% (m+1)) printf ("first\n");els

States, so as to obtain the rule about general mandatory states. However, this article intends to discuss the topic without involving a deep game theory. Consider n = 1 first. At this time, James can take the pile of stones, so that he can win. Assume n = 2. If (X, Y) is used to represent the number of two stones in a round (x N = 3. If James can adjust the state to (0, x, x) by taking away and moving

First, this is basically a series of articles about ACM game theory. Today, let's talk about the simplest game-bashgame. Its game rules are as follows: There are a pile of N stones, two smart people to play, each person can go to 1 ~ M stones, the last stone wins. For example, n = 7, M = 3 The procedure is as follows,

analytical: a typical symmetric game . When n>=3, no matter how a chooses, B chooses a special position (one or 2 consecutive) to divide the remaining loop into two links, the same size. Next, regardless of which of the two links the A chooses, b just needs to select the other link and do the same. The rest of the system is still symmetrical. Until finally, B must be the last to operate. #include inlineintRead () {CharC=getchar ();intx=0, f=

, the number of matches in each heap.Output format:Outputs the minimum number of matches taken in the first round. If there is no guarantee to win, output-1.Input/Output sampleInput Sample # #:65 5 6 6 5 5Sample # # of output:21stDescriptionkNim Game of the initiator win: a[1] ^ a[2] ^ ... ^ a[n]!=0So we now need to take a number of stones, so that the opponent regardless of how to take the stone or and not 0.XOR + is not 0???Linear Base!!!The very gr

Address: HDU 5011 During the competition, so many people were dumbfounded .. Helpless, this question is really not done, game theory is not at all, you have to make up. I didn't expect the code to be like this .. At that time, I thought about a lot of methods of water and I was not confused .. The Code is as follows: # Include HDU 5011 game (

contains multiple test cases, each with 2 rows, and the first line contains an integer m (1Output if the initiator can win, please export his first feasible number of scenarios, otherwise please output 0, the output of each instance takes up one row.Sample Input35 7 90Sample Output1AuthorlcyCalculate the total number of scenarios, you can see Nimbo-Baidu EncyclopediaThe way to convert from a non-singular situation to a singular situation can be: 1) make a = C (+) b2) Make B = A (+) c3) Make C =

("%d",n);Wuyi for(intI=1; i){ thescanf"%d%d%d",c[i].x,c[i].y,C[I].R); -st[2*i-1]=point (c[i].x-c[i].r,i,-1); Wust[2*i]=point (C[i].x+c[i].r,i,1); - } AboutSort (st+1, st+2*n+1, CMP); $ for(intI=1; i2*n;i++){ -Point x=st[i];tmp=x.x; - if(x.tp==-1){ -It=s.upper_bound (Point (0, X.id,1)); A if(It==s.end ()) Addedge (fa[x.id]=0, x.id); + Else{ thePoint y=*it; - if(y.tp==1) $Addedge (fa[x.id]=y.id,x.id); the

OutputStan Winsollie winsTest instructions: There are two numbers, a, a, a large number minus a multiple of decimals, first a person who becomes 0 wins.Train of thought: There are three cases (assuming a is a large number):1, if a%b==0, that must be the initiator win.2, if a3, if a>2*b, then certainly is the initiator wins, because can always turn into (a,a%b) and (A,a%b+b) the state, because these two states and the second situation is adjacent, therefore certainly has one kind to be in the lo

The main topic: Anti-Nim game, that is, take the last one to loseFirst State 1: Assuming that all of the heap is 1, then the heap number for even win, otherwise the initiator will be defeatedThen state 2: Assuming that there are two heaps of the same number and not 1, then the FLAC has the ability to control the field, namely:If the initiator takes a pile, then he can choose to leave a pile of 1 or all to take, so that the two piles are finally only 1

Recently I started the game theory, which is in the initial entry stage. I found a few questions to do this. This is a deformation of the NIM game. The distance between two coordinates is considered as the number of stones in a pile of stones, and N is the number of stones. Therefore, according to the conclusion of the NIM ga