wincor atm

Learning C language for one weeks, the teacher assigned to work, using C to write an ATM simulation program, customer interface has, withdrawals, 2 deposits, 3 transfers, 4 exit.And after the input error is required, there are three chances to re-lose, define the input password hellowold, with Switch branch.#include #includestring.h>#defineKey "Hellowold"//Define Password#defineZhanghu "123456789"#defineLine "==================\n"intNum//Select TypeCh

Recharge function Multiple purchases, each purchase of multiple pieces Insufficient balance tip recharge View shopping History The ATM implementation functions are as follows: Specify maximum overdraft limit Withdrawals available Regular repayment (fixed monthly repayment, such as number 15th) can be deposited Regular billing Support multi-user login, transfer between users Multi-user support Admi

first element out of the team, repeat 2, until the queue has no elements.Forgive bloggers lazy to stick pseudo-code, I will directly talk about the problem, anyway, there are templates # manual funnySecond, APIO2009 ATMOriginal title link (from Bzoj): http://www.lydsy.com/JudgeOnline/problem.php?id=1179Title Description:Input:The first line consists of two integers n, M. n indicates the number of intersections, and M represents the number of road strips. Next m line, two integers per line, thes

and safer than e-Bank (?) Disadvantages: In the event of an accident... 3. Use a mobile phone to get money from UnionPay ATM Scenario: A mobile phone is bound to a bank savings account. Each Bank can only be bound to one savings account. Each UnionPay ATM has a mobile operator assigned a number, such as 13808881818 Send the bank code, account password, and amount to the

First, the demand1. Analog implementation of an ATM + shopping mall Program1). realization function1. Quota 15000 or Custom2. Implement the shopping mall, buy things into the shopping cart, call the credit card interface checkout3. Can withdraw cash, handling fee 5%4. Monthly billing 22nd, 10th monthly repayment day, overdue, according to the total amount owed 5 daily interest5. Support Multi-Account login6. Support transfer between accounts7. Record

Label: Package com.bank.unionpay; Interface of the bank card public interface I_yinhangka { //abstract method //public Abstract default adornment abstraction public abstract void Cxyue ();//check balance void Cunkuan ();//Deposit void Qukuan ();//withdrawal void Accound ();//Account } 　　 Package com.bank.unionpay; Import java.util.ArrayList; Import java.util.List; public class Zhujiemian {public void Zhujiemian () { System.out.println ("Bank self-Service terminal s

This system all uses the Python basic knowledge realityProgram StructureAtm|--readme|--ATM #ATM主程序目录| |--__init__.py| |--atm.py #ATM执行程序| |__manage.py #ATM管理端--management portal for adding users, modifying quotas, and freezing accounts| || |--conf #配置文件| | | |--__init__.py| | | |--setting.py| || |--core #主要程序逻辑| | | |-

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