cryptocurrency slack

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Hdu4847 wow! Such Doge! (This question is simple + the input from the pitfalls)

series. In August 2013, images of the meme were spammed on Reddit's r/murica subreddit by 4chan's random imageboard,/B /. A search of the term Doge on Google Trends shows an explosion of popularity occurring in October 2013, and more so in the following month. by November 2013, the meme had become widespread on the Internet. google later created a Doge Easter Egg: When Doge meme was entered into the YouTube search bar, all of the site's text wocould be displayed in colorful Comic Sans, similar

50 lines of Python code to build a small blockchain

This article describes how to use Python to build a small blockchain technology that is implemented with Python2 with less than 50 lines of code. Although some think blockchain is a solution waiting for problems, there's no doubt that this novel technology is a Mar Vel of computing. But, what's exactly is a blockchain? While some people think the blockchain itself still has a lot of problems to solve, there is no doubt that this novel technology is a marvel of the computer world. But what exact

Blockchain technology learning-simple crypto currency-goofy coin

This article explains a very simple cryptocurrency, the goofy currency, before the advent of Bitcoin. It aims to understand the operation mechanism of Bitcoin gradually. The regular goofy coin (goofycoin), the creator is goofy. Is the simplest crypto currency known at the moment. There are only two rules: (1) Goofy can create a new currency at any time, and the newly created coin belongs to him; (2) The person who owns the currency can transfer it to

Use the command line to view the price of encrypted currency such as Bitcoin | Linux China __linux

Coinmon will check the price of the encrypted currency on the terminal. It will get all the details from the coinmarketcap.com API. is very useful for cryptographic money investors and engineers. --Sk This article navigates the ◈ installation Coinmon 22%◈ View the encrypted currency price from the command line 30% compiled from | https://www.ostechnix.com/coinmon-check-cryptocurrency-prices-commandline/ Author | Sk Translator | Geekpi A while ago, we

POJ 2400 KM algorithm minimum weight matching backtracking output all optimal matching schemes

A tough question First, the input is very painful. According to the online experts, the question matrix is reversed. Then I gave it back. If it is a match between n and n, there is no pressure to directly obtain the negative value of KM. However, if the points on both sides are different, it is said that there will be problems. [Cpp]# Include # Include # Include # Include # Include # Include # Define maxn505# Define MAXM 555555# Define INF 1000000000Using namespace std;Int n, m, ny, nx;Int w [

POJ 3565 Ants (Computational ry, KM)

20005Using namespace std;Struct Point {Double x, y;} Ant [105], tree [105];Double path [1, 101] [2, 101];Int cnt;Double lx [101], ly [101];Int match [101];Double slack;Bool v_x [101], v_y [101];Double dist (Point p1, Point p2 ){Return sqrt (p1.x-p2.x) * (p1.x-p2.x) + (p1.y-p2.y) * (p1.y-p2.y ));}Bool dfs (int k ){V_x [k] = true;Double temp;For (int I = 0; I If (! V_y [I]) {Temp = lx [k] + ly [I]-path [k] [I];If (zero (temp )){V_y [I] = true;If (match

Cycle issues for setup time and hold time

why calculate Setup time of the Slack need to consider adding cycles, Hold Time when you don't need it? Summary one:Because the data delay,launch edge and capture edge do not correspond to the same clock edge of the clock signal source clocks, due to the existence of the time of Setup, the addition cycle needs to be considered. When a single clock cycle is checked, the tool defaults to capture Edge-launch edge=1 cycles.Hold time because the two corre

bzoj3571: [Hnoi2014] frame minimum product match + minimum product XX summary,

easily solved. (It seems that a template like this, basic all similar problems can be resolved)1#include 2#include 3#include 4#include 5#include 6 using namespacestd;7 #defineMAXN 1008 #defineINF 1000000009 Ten intCases,n; One intA[MAXN][MAXN],B[MAXN][MAXN],VAL[MAXN][MAXN],SLACK[MAXN],VALX[MAXN],VALY[MAXN],LINKY[MAXN]; A BOOLVISX[MAXN],VISY[MAXN]; - - structpoint{ the intx, y; - }ans; - -Pointoperator-(point A,point b) {return(point) {a.x-b.

KM matching template

1 intG[n][n];2 intLx[n], ly[n];3 intSlack[n];4 intMatch[n];5 BOOLVisitx[n], visity[n];6 intN;7 8 BOOLHungary (intu)9 {TenVisitx[u] =true; One for(inti =0; I i) A { - if(Visity[i]) - Continue; the Else - { - if(Lx[u] + ly[i] = =G[u][i]) - { +Visity[i] =true; - if(Match[i] = =-1||Hungary (Match[i])) + { AMatch[i] =u; at return true; - } - } -

HDU 3488 Tour

HDU_3488 It is worth mentioning that if we use the KM algorithm to find the perfect matching of minimum weights, we need to initialize the edge weight to the MAX-G [I] [j] and then find the perfect matching of the maximum right, and then convert the result back. If you initialize G [I] [j] to-G [I] [j] to find the perfect matching of the maximum permission, the program I write Will time out, at the moment, I have no idea whether this idea is a problem or my writing is a problem. #include#inclu

HDU 3718 Similarity

HDU_3718 First, we need to convert the characters in the string into the number of [0, k-1] with the same meaning, and then scan the array sequentially, we can get the maximum value of various matching between the two sets of numbers, this completes the graph creation. Then use the KM algorithm to find the optimal match. #include#include#define MAXD 30#define MAXN 10010#define INF 1000000000char a[MAXN][5], b[MAXN][5];int visa[MAXN], visb[MAXN], x[MAXN], y[MAXN];int G[MAXD][MAXD], yM[MAXD], N,

HDU 3395 Special Fish

HDU_3395 We can split a fish into two points, which respectively represent attack and attack, and then set the Edge Weight of G [I] [j] = 1 to value [I].^ Value [j], and then use the KM algorithm to find the optimal matching of the Bipartite Graph. #include#include#define MAXD 110#define INF 1000000000int yM[MAXD], G[MAXD][MAXD], N, value[MAXD];int A[MAXD], B[MAXD];int visx[MAXD], visy[MAXD], slack;char b[MAXD];int init(){int i, j; scanf("%d", N);

hdu2255 Ben-off make money, Max right match, km algorithm

tree, The Edge (I,J) may be added to the equal sub-graphIn order for the A[i]+b[j]>=w (I,J) to always be true, and at least one edge is added to the equal sub-graph, d=min{a[i]+b[j]-w (i,j)},i in the interlaced tree, J is not in the interlaced treetime complexity: need to find an O (n) time augmentation path. Each augmentation requires an O (n) sub-index change. Each time you change the top, enumerate the edges to find the D value, the complexity is O (N2), and the total complexity is O (N4). S

Bellman-ford algorithm--Solving the shortest path problem of single source point

distance s. When 1 times slack is made on each side, the branches that start from S and have a maximum level of 1 are generated. That is, the shortest path to the vertices with a maximum of 1 edges associated with S is found, and the 2nd-pass relaxation of each edge creates a 2nd-level branch, which means that the shortest path of those vertices connected by 2 edges is found .... Because the shortest path contains up to |v|-1 edges only, you only nee

"Turn" Bellman_ford algorithm

Original link: http://www.cnblogs.com/Jason-Damon/archive/2012/04/21/2460850.htmlExcerpt from Baidu EncyclopediaThe Bellman-ford algorithm is a single-source shortest path algorithm with negative weights, which is very inefficient, but the code is easy to write. That is, the continuous relaxation (relaxation), each slack to update each edge, if the n-1 can be updated after the relaxation, then the picture has a negative ring (that is, the negative pow

Laravel basic tutorial-SSH task

after the task is executed. This command receives an API token, the name of the room, and the user name of the sender displayed in the message: @servers(['web' => '192.168.1.1'])@task('foo', ['on' => 'web']) ls -al@endtask@after @hipchat('token', 'room', 'Envoy')@endafter If necessary, you can also send custom messages to the HipChat room. When building a message, the available variables of the task are also available in the message: @after @hipchat('token', 'room', 'Envoy', "$task ran in

Two-dimensional graph with weighted matching (thinking of pushing box problem)

staggered road X4, Y2, X3, Y0 X1, not found? there is a X4---------Y2, X3, X0--Y1-X2 The inverse of the path attribute becomes the upper (right) graph. At this point, all the vertices in set X already have corresponding matching, that is, complete matching! That is, the maximum weight of this binary graph match! X0-Y1X1-Y0X2-Y4X3-Y3X4-Y2Maximum power value is 30What about the minimum weight matching requirement? Very simple, before solving the ownership value of the opposite number, the results

HDU2426 interesting Housing problem (km matches)

Test instructions: N students arranged to M dormitory, each student to the dormitory has a rating, positive, 0, negative, now the evaluation is negative, can not let this student go to this room, ask how to arrange to let all the students live in the dormitory and the most evaluation.Idea: When building the weight of a graph, filter out negative edges.#include #include #include #include #include #include #include #include #include #include #include #include #define Lson (rt#define Rson (rt#defin

HDU 2255 Ben-off make money

KM algorithm binary graph maximum weighted value matching#include #include#include#includeConst intMAXN =356;Const intINF = (1 to)-1;intW[MAXN][MAXN];intLX[MAXN],LY[MAXN];intLINKY[MAXN];intVISX[MAXN],VISY[MAXN];intSLACK[MAXN];intNx,ny;BOOLFindintx) {Visx[x]=true; for(inty =0; Y ) { if(Visy[y])Continue; intt = lx[x] + ly[y]-W[x][y]; if(t==0) {Visy[y]=true; if(linky[y]==-1||find (Linky[y])) {Linky[y]=x; return true; } } Else if(Slack

"Hdoj" 2255 Ben-off make money

Maximum binary graph matching, O (n^3).1 /*2255*/2#include 3#include 4#include 5#include 6#include 7 using namespacestd;8 9 #defineMAXN 305Ten #defineINF 0XFFFFFFF One A intW[MAXN][MAXN]; - intLINK[MAXN]; - intLX[MAXN], LY[MAXN]; the intSlack; - BOOLS[MAXN], T[MAXN]; - intN; - + BOOLDfsinti) { -S[i] =true; + for(intj=1; jj) { A if(T[j]) at Continue; - intTMP = lx[i]+ly[j]-W[i][j]; - if(TMP = =0) { -T[J] =true; - if(!link[j] | |DFS (Link[j]))

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