kanban game simulation

team from the head of the team outdequeue () {return This. Items.shift (); } //get the first element in a teamFront () {return This. items[0]; } //determine if the team is emptyIsEmpty () {return This. Items.length = = 0; } //get the number of elements in a teamsize () {return This. Items.length; } }4. TestingHere, we can use the queue to simulate the game of ' drumming and passing flowers ':/** Name: Drum Pass * parameter: Namearr, an

2 3 4 5 6 X 8 9 Person 4 3 2 1 2 3 4 3 2 Reports 10 11 12 13 X 15 16 X 18 Person 1 2 3 4 3 2 1 2 3 Reports 19 20 X 22 23 24 25 26 X Person 4 3 2 1 2 3 4 3 2

\ Leq 10 \)For the previous \ (40 \ % \) data, \ (n \ Leq 500 \)For \ (100 \ % \) data, \ (5 \ Leq n \ Leq 5000, CI \ Leq 10 ^ 9 \)Question First, let's talk about the background of the question, and talk about the author who moved the question and made the magic change. Simple game theory \ (DP \) involves almost no knowledge of game theory.Obviously, the two are equivalent. \ (F [I] [J] \) indicates the

Link: click hereTest instructions description input input contains multiple sets of data, and the first line of a number T represents the number of input samples. Each set of samples starts with a number M output Each set of test data outputs a pair of digits, as described above. Two numbers are separated by a space. Sample Input

The problem defines a new sort algorithm, which is to put a number in a sequence of sequences, if its right number is smaller than itYou can move to the right of the past until the number on the right is larger than it.Easy to get, if the simulation is O (n^2) efficiency, certainly notThinking about it, this problem can be translated intoFor each element in this sequence, the right side of the string is less than its number and, if present, ++ansI did

; k2; k++) A for(l=0; l6; l++) the if(Map[i][j] = =Dir[k][l]) + { -dp[k][l][a[k][l]].x =i; $Dp[k][l][a[k][l] ++].y =J; $ } - - the for(k=0; k2; k++//output to the statistical results of the pieces - {Wuyi intFlag =0; the if(k = =0) -printf ("White :"); Wu Else -printf ("Black:"); About for(i=0; i6; i++) $ { - if(k = =1) -Qsort (Dp[k][i], a[k][i],sizeof(dp[k

Title Link: ZOJ 3879 Capture the FlagTest instructions: Given n teams attacking each other server, there are the following rules:1. If team A's server 1 is successfully attacked by Team B, Team A will lose n-1 points, and this n-1 point score will be split equally to other teams that successfully attack team A's server 1.egA total of 4 teamsA B 1C B 1At this point team A and team C both got 1.5 (3/2) points and Team B lost 3 points.2. If a team fails to maintain their servers, the team will lose

of the interval, but when we enumerate to these two positions, we can return to change the point to the answer of the contribution, that is, a difference.So find these two points, by looking at the current point from these two points to complete the processing of information (can also become the implementation of the tag).Eolv the big guy to maintain a list at each point, but after the teachings of the GMPOTLC, I was in this part through the chain forward to the star to achieve.The above diffic

output in the simplest fractional form, the first is the numerator and the second is the denominator."Input Sample"32 3 5"Output Example"22 3the sample Tips "There are 6 possible travel examples:[2, 3, 5]: The distance to travel: |2–0| + |3–2| + |5–3| = 5;[2, 5, 3]: |2–0| + |5–2| + |3–5| = 7;[3, 2, 5]: |3–0| + |2–3| + |5–2| = 7;[3, 5, 2]: |3–0| + |5–3| + |2–5| = 8;[5, 2, 3]: |5–0| + |2–5| + |3–2| = 9;[5, 3, 2]: |5–0| + |3–5| + |2–3| = 8.The answer is 1/6 * (5+7+7+8+9+8) =44/6=22/3"Data Range"30

LongLl;typedef pairint,int>PII;structdata{intA,b,c;} P[n];BOOL operator(Data a,data b) {returna.c*b.ba.b;} InlinevoidUpdintx,inty) { if(xy;}intDp[n];intMain () {intt,n,t,j,fn,ans=0, I; //freopen ("score.in", "R", stdin); //freopen ("Score.out", "w", stdout); //int T1=clock ();Cin>>u; while(t--) {ans=0; CIN>>n>>T; for(i=1; i) scanf ("%d%d%d",p[i].a,p[i].b,p[i].c); Sort (P+1, p+1+N); Memset (DP,-1,sizeof(DP)); dp[0]=0; for(i=1; i) { //Debug (P[I].A);fn=p[i].a-p[i].b*T; //D

{static final int finish_line = 75;Private listPrivate Executorservice exec = Executors.newcachedthreadpool ();Private Cyclicbarrier barrier;Public horserace (int nhorses,final int pause) {Barrier = new Cyclicbarrier (nhorses,new Runnable () {@Overridepublic void Run () {StringBuilder s = new StringBuilder ();for (int i = 0; i S.append ("=");}System.out.println (s);for (Horse horse:horses) {System.out.println (Horse.tracks ());}for (Horse horse:horses) {if (Horse.getstrides () >= finish_line) {

//source code from Laekov for c0x17#definePrid "BXJL"#include#include#includeusing namespaceStd;typedefLong Longdint;Const intMAXN =100003;Const intMoD =998244353;CharA[MAXN];intN, M, NE[MAXN], SZ[MAXN];voidPrenext () {ne[1] =0; for(inti =2, j =0; I i) { for(; J A[i]! = a[j +1]; j =Ne[j]); if(A[i] = = A[j +1] J +1i) {Ne[i]= ++J; } Else{Ne[i]=0; }} memset (SZ,0,sizeof(SZ)); for(inti = n; I --i) {++Sz[i]; Sz[ne[i]]+=Sz[i]; }}intMainintargcChar*args[]) { if(ARGC 2|| strcmp (args[1],"-NF") {f

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5112Problem Solving report: Sweep again1#include 2#include 3#include 4#include 5#include 6 using namespacestd;7 Const intMAXN = 1e6+5;8 9 structnodeTen { One Doublet,x; A }REC[MAXN]; - DoubleMax (DoubleADoubleb) - { the returna > B?a:b; - } - BOOLCMP (node A,node b) - { + returnA.T b.t; - } + DoubleFabs (Doubled) A { at returnD >0? d:-1.0*D; - } - intMain () - { - intT,n,kase =1; -scanf"%d",T); in while(t--) - { t

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5122Problem-Solving report: Define a sorting algorithm, each round can be randomly find a number, the number and the subsequent smaller than this number of exchanges, has been judged, until there is no smaller than this number, so called a round, now given a sequence of length n, you ask, at least through how many rounds, you can make this sequence into order.Because we can only compare with the number of the following, so I just need to coun

have IDs that are smaller than L, so change the directionBayinow=T[now][j]; the } the returnans; - } - the intMain () the the { the -Freopen ("Input.txt","R", stdin); the theFreopen ("output.txt","W", stdout); the 94N=read (); q=read (); theFor1 (i,n) a[i]=a[i-1]^read (); theid[0]=-1; theInsert (rt[0],a[0],0);//Insertion 098For1 (i,n) Insert (rt[i-1],a[i],i);//Insert Prefix XOR value About intLEN=SQRT (n); m=n/len+ (n%len!=0); -For0 (i,m-1)101For2 (j,i*len+1, N)102B[i][j]=max (b

Problem descriptionafter repeated attempts, ll finds the greedy strategy is very awful in practice. even there is no apparent evidence to proof it is better than a random one. so he has to drop this strategy and try to discover a better one. Inputthere are 100 test cases. each test case begins with two integers n, m (5 Outputfor each test case, first output a single line containing the number of steps S. then s lines follow, each contains two integers indicating the position of an arbitrary blo