q1 smartwatch

Test instructions: A 4-bit prime number changes one digit at a time, and the middle process also takes the prime number, asking to be the smallest step of the other.After a linear sieve, the BFS will be fine again. I write the two-way, actually not necessary.#include #include#includestring>#include#include#include#include#include#include#includeSet>#include#include//#include using namespaceStd;typedefLong Longll;Const intMAXN =1e5;BOOLISNOT[MAXN];voidSeive (intn = maxn-1){ intm = sqrt (n+0.5)

Note: My English is very slag, the main idea is mostly from Baidu ~=0= the topic of farmers John needs to seize his cattle, he and his cattle in a straight line (estimated to be one-dimensional creatures), John in N (0≤n≤100,000), his cattle in K (0≤k≤100,000), John Next can move from X to X+1 or x-1 or 2*x, ask John how many steps he needs at least to find his cow. With BFS can be solved ~ but need pruning or will be the MLE is worth mentioning is the number of steps do not use the structure of

Encryption and decryption of php code Php code encryption /* * @ Auther: wangyaofeng * @ Time: 2014.11.6 * @ Action: encrypt the php project. Note that if there is a framework directory in the project or there is no need to encrypt the Directory, remove it in advance. **/ Class Encryption { Private $ c = ''; // stores ciphertext Private $ s = '', $ q1, $ q2, $ q3, $ q4, $ q5, $ q6; // store

Given a number, the breadth overrides the output record. For example:Given binary Tree {3,9,20,#,#,15,7} , 3 / 9 / 7Return its level order traversal as:[ 3], [9,20], [15,7]]I thought about it. Use queues to store data for each layer. Hey.With two queues, one queue is used for output, one queue is used to record all the left and right nodes of the next layer, and then two queues are used for the loop./** Definition for binary tree * struct TreeNode {* int val; * TreeNode *left; * T

stack and updates the top value to the current incoming team element.Pop: The value of the non-empty stack, except for the last one (to be out of the queue), the rest is loaded into the empty stack. Note Update the top value to the last second incoming team element.Top: Returns the top value of the real-time update directlyEmpty: Determine if two stacks are emptyclassStack { Public: Queueint>Q1; Queueint>Q2; intN; //Push element x onto stack. void

same timeDaemon Thread: Service to non-daemon thread (Emperor's relationship with Eunuch, Emperor died, courtiers to immolation)The role of the queue:Decoupling of the implementation program (allows for loosely coupling between programs)Improve processing efficiency#!/usr/bin/python#author:seanimport queue# FIFO mode FIFOq1 = queue. Queue () q1.put (' D1 ') q1.put (' D2 ')

1000,000, only or steps are used. # Include # Include # Deprecision MAX 1000000Typedef _ int64 LL; Struct List{List * last;Int value;List * next;}; Bool isPrime [MAX + 1]; List * where [MAX + 1]; Int isResult [MAX + 1]; Int init (){Int I;LL temp, temp1;List * p;List * p1;List * q1;List * head;// Int step = 0; // number of steps used for testing Memset (isPrime, 0, sizeof (isPrime )); For (I = 2; I {P = new List;Where [I] = p;If (I = 2){P-> last = NUL

[] args) {Scanner in=NewScanner (system.in); intn =In.nextint (); Long[] q1 =New Long[n + 1]; Long[] q2 =New Long[n + 1]; q1[1] = 1; q2[1] = 1; q1[2] = 2; q2[2] = 6; for(inti = 3;i ) {Q1[i]= 2 * q1[i-1]; Q1[i]%=MOD; Q2[i]= 2 *

#-*-Coding:utf-8-*-Import timeImport RandomImport QueueImport threadingQ1 = Queue.queue ()def producer (name):Count = 0While Count Print "Making ..."Time.sleep (Random.randrange (3))Q1.put (count)Print "Procucer%s has produced%s Baozi ..."% (Name,count)Count + = 1Print "ok!"DEF consumer (name):Count = 0While Count Time.sleep (Random.randrange (4))If not Q1.empty ():data =

(NM) algorithm: Finding the solution with the smallest dictionary order"According to the definition of the 2-sat built in the diagram, it can be found that if I to J has a path, if I select, then J also to choose, or if J is not selected, I also can not choose;Therefore, a very intuitive algorithm is obtained:(1) Set a state for each point v,v=0 is not determined, v=1 indicates that the selection is determined, v=2 indicates that the selection is not selected. Called a dot ishave determinedIf a

estimate with Manhattan distance need to know coordinates, so not suitable for the two-tuple hash into a number, and can not use a number to compress the state, because to add value, code is very long, and write out after not know why have been hanging in ghost there are 3 cases, Program Run Results 78 ... Actual 77. POJ does not support #include //Rey#include using namespacestd;Const intMAXW = -; Const intMAXN = Max;//14*14*3/4+2 149 ints[3]; intt[3]; intW,h,n;Charmaze[maxw][maxw+1]; intdeg[

Title: There is a batch of software to install, ":" Before the software needs to be installed after the ":" Software Installation (":" There may be more than one software), "*" indicates that the software requires a reboot to complete the installation.The minimum number of reboots required for all software installations to complete.Problem Solving Ideas:Double queue top sort, with two queue q1,q2 respectively to save the software that does not need to

3 using namespacestd;4 structnode5 {6 intp;7 intnum;8}q[1000005],q1[1000005];9 intr,f,g,ff;Ten inta[1000005]; One voidInquene (inti) A { - while(q[--r].num>a[i]r>=f); - //printf ("r1=%d\n", r); theq[++r].p=i; - //printf ("r2=%d\n", r); -q[r].num=A[i]; -r++; + //printf ("r3=%d\n", r); - } + voidDequene (inti) A { at while(q1[--g].numff); -q1