robocoin atm

1 //Multi-School 5 1001 HDU5781 ATM Mechine2 // Http://acm.hdu.edu.cn/search.php?field=problemkey=2016+multi-university+training+contest+5source=1searchmode=source3 //Memory Search4 //two points each time, the decision is optimal, so the maximum query 11 times5 //Dp[i][j] Currently able to determine the I-dollar case, can also be warned J times6 //next fetch K, actual remaining T7 //dp[i][j]=min (1≤k≤i) ({(∑ (0≤t8 //k=0 no money to take, of course the

The depositor said that 2000 yuan from the ATM is all fake money (figure) 03:12:16 Source: wang.com (Guangzhou) netizens commented on 3548 items and click to view 　 Mr Meng showed at the bank's door that he had obtained the same counterfeit money from the bank. Information Times photo taken by Huang like Information Times reported on March 13, December 6Mr Meng reported that he had obtained 2000 yuan in cash from a bank teller machine at the t

"+password");}else{System.out.println ("Please enter a new password:");String newpass = Scanner.next ();System.out.println ("Please enter your new password again");String newpass2 = Scanner.next ();if (!newpass.equals (Newpass2)) {System.out.println ("Two times input new password inconsistent!");}else{Password = newpass;SYSTEM.OUT.PRINTLN ("Change password successfully!"); The current password is: "+password);}}Iscontinue ();}/***** Exit*/private static void layout () {System.out.println ("Exit

Samsung Pay can withdraw money without card The use of Samsung Pay and Apple Pay is basically the same, is to take the phone binding bank cards, and ultimately to achieve payment, and Apple Pay on the line can be implemented in a number of banks without card withdrawals business, and now have netizens exposed that Samsung Pay can actually also achieve without card withdrawals Samsung on the mobile phone on the smart pay function, the phone will be placed on the

Puzzle: Bare pinch point + shortest path (DP)#include 　　Bzoj 1179 [APIO2009]ATM

[V] = Dis[u] +Val[v]; the if(!Vis[v])98 { AboutVIS[V] =1; - Q.push (v);101 }102 }103 }104 } the }106 107 intMain ()108 {109 intI, x, y, u, v; then =read ();111m =read (); theMemset (Head,-1,sizeof(head));113memset (Head1,-1,sizeof(Head1)); the for(i =1; I ) the { thex =read ();117y =read ();118 Add (x, y);119 } - for(i =1; I read ();121s =read ();122p =read ();123 Dfs (s);124 for(i =1; I A[i]; the

)); thememset (Newmoney,0,sizeof(Newmoney));106memset (Isbar,false,sizeof(Isbar));107memset (DFN,0,sizeof(DFN));108CNT = time =0;109 for(inti =1; I if(Dfn[i] = =0) Tarjan (i); the if(CNT = =1) {printf ("%d", newmoney[cnt]);return 0;}111 for(intI=0; i) the {113scanf"%d", a); theIsbar[belong[a]] =true; the } theStart =Belong[s];117memset (head2,-1,sizeof(head2));118 intKKK =0;119 for(inti =1; I ) - {121 for(intU = head[i]; ~u; U =Edge1[u].next) {122

=Q.front (); + for(ITER it=g2[u].begin (); It!=g2[u].end (); it++) A if(dis[*it]it]) the { +dis[*it]=dis[u]+w[*it]; - if(!inq[*it]) $ { $Q.push (*it); -inq[*it]=1; - } the } -Q.pop (); inq[u]=0;Wuyi } the } - intMain () Wu { -scanf"%d%d",n,m); About for(intI=1; ii) $ { -scanf"%d%d",us[i],vs[i]); - G[us[i]].push_back (Vs[i]); - Rg[vs[i]].push_back (Us[i]); A } SCC (); + for(intI=1; i"%d"

/******************************************************************Auther:liuyongshui*****About:This is the BANK ATM applicantion******Date:20130511**************************************************************/#include Local graph:

# Include # Include Int main (void){Int password, one, two, money1 = 1000000, money2 = 999999, a = 0; // The value of money1 is the same as money2.Printf ("enter your password: \ n ");Scanf ("% d", password );If (Password = 123456 ){System ("CLS ");Printf ("data loading... please wait ");Sleep (1000); // wait for one secondPrintf (".");Sleep (1000 );Printf (".");Sleep (1000 );Printf (".");Sleep (1000 );Printf (".");Sleep (1000 );Printf (".");System ("CLS ");Printf ("select your operation: \ n 1

www.lydsy.com/JudgeOnline/problem.php?id=1179 (Topic link)Test instructions: Give a graph, each node has a bit of power. Mark some points to find a path that can repeat through an edge, making the total point right and maximum. Repetition of a point can not be repeated to calculate the right.SolutionToday's exam question, Dijkstra unfortunate GI rotten.　　Warning:dijkstra handle the longest road when there will be some bad situation, so don't use!! Now that we can repeat some of the edges, once w

; the intAdde_ (intUintv) { +e_[++k_]=v; -nxt_[k_]=head_[u],head_[u]=K_; $ } $ - intrebuild () { - for(intI=1; i){ the intu=scc[f[i]],v=Scc[e[i]]; - if(u!=v) adde_ (u,v);Wuyi } the for(intI=1; i) - if(Ok[i]) ok_[scc[i]]=1; Wu } - About intq[maxn],d[maxn],inque[maxn],ans,h; $ intSPFA () { -t=0; -s=Scc[s]; -q[++t]=s; Ainque[s]=1; +d[s]=Val[s]; theans=D[s]; - while(ht) { $ intu=q[++h]; the //printf ("%d\n", u); the for(intI=head_

It should be machine.Similar to the POJ3783 balls type.Now the upper bound is I, guess the wrong number is J, start guessing K-yuan, there are two cases:1 guessed: (i-k + 1) * Dp[i-k][j]2 Guess not k * Dp[k-1][j-1]The mean value of both cases is the expectation of the first guess as K, 1 In fact, M not to 2000, by the two ideas up to more than 10 times (beginning also did not think, has been unable to n^3 the complexity of lowering).#include 　　HDU5781 ATM