puzzle choice

; Nan has other amazing features. For any floating point operation, if one or more of its operands is Nan, the result is Nan. This rule is very reasonable, but it has strange results. For example, the following program prints false: class Test { public static void main(String[] args) { double i = 0.0 / 0.0; System.out.println(i - i == 0); }} The principle behind this Nan calculation rule is that once a computation produces Nan, it will be damaged, and no further computation can f

Transmission DoorAnalysisUsing the segment tree to maintain the interval GCD, each query finds the first point that is not a multiple of x, and if there is an interval that gcd cannot be divisible by x, the interval is illegal.Code#include #include#include#includestring>#include#include#include#include#include#include#include#includeSet>#include#includeusing namespacestd;intd[2000100],cnt,a[500100];inlineintgcdintXintY) {returny==0? X:GCD (y,x%y);} InlinevoidBuildintLeintRiintwh) { if(le==R

Online Demo http://img.jb51.net/online/pintu/pintu.htm CopyCode The Code is as follows: Original JS works: JS puzzle game Complete annotations, object-oriented Reprinted please indicate from Reprinted please indicate from

, which is based on the number of attacks and the number of monsters, 1 on each line (each attack) represents the monsters that the attack can destroy, the original problem is converted to the minimum number of rows (several attacks) in the new matrix constructed ), each column in the new matrix consisting of these rows has at least one (because all monsters should be killed, and the range of the two attacks may overlap, A monster can be overwritten by two attacks, that is, each column can have

Although HTML was not born along with the Internet, their close relationships have almost ignored this history. HTML is so powerful and widely used. Since W3C declared that H is dead, the development of HTML 5 is reversed, forced W3C to accept and continue to develop.However, since the HTML design, it mainly targets the performance of static content, which is also doomed to its inherent defect. The Internet has evolved from its initial content performance to its application platform. In terms of

^ = expr is compiled, many C and C ++ compilers extract the value of X after calculating the expr, which makes the above usage work properly. Although it works properly, it still violates the C/C ++ rule that cannot repeatedly modify variables between two consecutive sequence points. Therefore, the behavior of this idiom is not clearly defined in C and C ++. To value its value, we can still write a Java expression that swaps the content of two variables without using temporary variables. But it

A number puzzle Time Limit: 3000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/others) total submission (s): 938 accepted submission (s): 276Problem descriptionlele was boring during his recent classes, so he invented a digital game to pass the time. This game is like this. First, he took out a few pieces of paper and wrote any number between 0 and 9 (a number can be re-rewritten). Then, he asked his classmates to write two numbers, X and K.

The examples in this paper describe the simple puzzle game implemented by Python plus pygame. Share to everyone for your reference. The implementation method is as follows: Import Pygame, sys, randomfrom pygame.locals import *# Some constants windowwidth = 500WINDOWHEIGHT = 500BACKGROUNDCOLOR = (255, 255, 255) BLUE = (0, 0, 255) BLACK = (0, 0, 0) FPS = 40VHNUMS = 3CELLNUMS = Vhnums*vhnumsmaxrandtime = 100# exit def terminate (): P Ygame.quit () Sys.

If the answer is within a fragment, then the direct suspension method solves the time complexity of $o (n\sum) $.If the $n$ is larger, then the $\sum$ is smaller.To find out the length of each point up to extend, enumerate each point upward of this segment as a short board.Figure out the sum of the length of the completely optional fragment and not the full selection, the largest extension distance to the right of the left, update the answer.Time complexity $o (n\sum^2) $.If the $n$ is relativel

); - if(b -) F+=dfs (a,b+1, c,d,e,f) * ( --B)/( Wu-sum); + if(c -) F+=dfs (a,b,c+1, d,e,f) * ( --c)/( Wu-sum); - if(d -) F+=dfs (a,b,c,d+1, e,f) * ( --D)/( Wu-sum); + DoubleMd= -;if(e==4) { for(intI=0;i4; ++i) Md=min (DFS (A,B,C,D,I,F)/( Wu-sum), MD); f+=MD;} AMd= -;if(f==4) { for(intI=0;i4; ++i) Md=min (DFS (a,b,c,d,e,i)/( Wu-sum), MD); f+=MD;} atvis[a][b][c][d][e][f]=true; - returndp[a][b][c][d][e][f]=F; - } - intMain () - { -memset (Vis,false,sizeofvis); in inta,b,c,d;

company purchased a server, then you can choose to host the server to the Hong Kong computer room; As the server needs 24 hours uninterrupted power and network to ensure the smooth, and need the relevant professional and technical personnel to maintain, self-management and maintenance of high cost, once the problem is not timely and may affect the entire network of the group, resulting in a very significant loss; therefore, Hong Kong computer room hosting is a good solution, Hong Kong has built

; }}intMain () {intt,n,m; scanf ("%d%d",t,MoD); Init (); while(t--) {n=read (); m=read (); printf ("%d\n", (LL) ans1[n]*ans2[m]%MoD); } return 0;}Expand Euclid to seek inverse yuan#include #includestring.h>Const intn=1e7+ the; typedefLong Longll;intPr[n],p[n],cnt,mod;intInv[n],ans1[n],ans2[n];intRead () {intx=0;CharCh=GetChar (); while(ch'0'|| Ch>'9') ch=GetChar (); while(ch>='0'ch'9') {x=x*Ten+ch-'0'; ch=GetChar ();} returnx;}intEX_GCD (intAintBintx,inty) { if(!b) {x=1, y=0; returnA

() {///Expand nodes (Add new nodes to the tree) A for(; ii) { thech[cur][s[i]-'a']=alloc++; +cur=ch[cur][s[i]-'a']; -++Val[cur]; $ } $ } - voidIns () {///the value of the curried node (expands the value of the existing node) -len=strlen (s); theCur=0; - for(i=0; ii) {Wuyi if(ch[cur][s[i]-'a']==-1) Add (); the Else{ -cur=ch[cur][s[i]-'a']; Wu++Val[cur]; - } About } $ } - intQuery () {///Ask -len=strlen (s); -Cu

on the bit. 1 #pragmaComment (linker, "/stack:1024000000,1024000000")2#include 3#include 4#include 5#include 6#include 7#include 8#include 9#include Set>Ten#include One#include A#include -#include -#include the using namespacestd; - intdirx[]={0,0,-1,1}; - intdiry[]={-1,1,0,0}; - #definePI ACOs (-1.0) + #defineMax (a) (a) > (b)? (a): (b) - #defineMin (a) (a) + #definell Long Long A #defineEPS 1e-10 at #defineMOD 1000000007 - #defineN 506 - #defineINF 1e12 - intn,m,q; - intMp[n][n]; - int

1#include 2 3 intMain ()4 {5 Long Longi,j;6 intN//The number of prime numbers used to record the input to be multiplied7 Long Longproduct=1;//used to record the product of the first n prime numbers and initialize them to 18 intnum=0;//used to record the number of prime numbers currently found9 Ten //input Onescanf"%d", n);//Enter the number of prime numbers to be multiplied A - for(i=2;; i++)//traverse a positive integer of i>=2 to find the first n prime numbers

) { the if(Vis[s])return ; *Vis[s] =true; $ if(Bitcount (s) = =1){Panax NotoginsengNode[s].push_back (node{0,0}); - return ; the } + for(intL = (s1) s; L >0; L = (l1) s) { A intR = S ^l; theDFS (L); DFS (R);//all scenarios for a two subset of a root node. Node[l].size () left scheme number node[r].size () to the right of the scheme number + for(inti =0; I ){ - for(intj =0; J ){ $ Doublell = min (-sumw[r]/(Sumw[l] + sumw[r]) +

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=1251The power of map, but it runs too long.Code:1#include 2#include string.h>3#include 4#include 5#include 6#include 7#include 8#include 9#include Ten#include One using namespacestd; A -mapstring,int>M; - the intMain () { - stringx; - CharA; - while(true){ +scanf"%c",a); - if(a=='\ n'){ +scanf"%c",a); Ax="";//string x Empty at } - if(a=='\ n') Break; -x+=A; -m[x]+=1; - } - while(cin>>x) inprint

maximum weight of these edges is minimized, and the point pairs that satisfy any hostile relationship end up in the same connected block.Summary: This is a good question, I used O (n^2) to record all the distance to meet the point pair, and then O (M log m) with and check the merger and lookup points between the relationship between, apparently too violent.The most recent point pair of plane. This is the core of the subject, whether you are violent or non-violent, using O (n log n) to find the

Original refer:http://www.cnblogs.com/happyfreelife/p/4240100.htmlWhen a form with username and password is submitted, the viewer intelligently asks the user if they want to save the password. If developers don't want this "smart" thing, they can use Ajax to submit the form so that the viewer is not "smart"."When the browser is allowed to save the password for the site, the next time any page of the site is opened, the browser automatically detects that the page has a password element , and if t

if(C[i] (1k)) {111Addedge (II, jj+1); theAddedge (ii+1, JJ);113Addedge (JJ, ii+1); theAddedge (jj+1, ii); the}Else { the Addedge (ii, JJ);117Addedge (ii+1, jj+1);118 Addedge (JJ, ii);119Addedge (jj+1, ii+1); - }121 }122 }123 for(i=0; ii)124 if(!Pre[i]) the DFS (i);126 for(i=0; flagi2)127 if(Bn[i] = = bn[i+1]) -Flag =false;129 } the if(flag)131Puts"YES");