atm program c

, K, l, q = 1;int a1 = 123456, a2 = 123456, a3 = 1000;printf ("Welcome to ATM machine \ n");while (q){printf ("Please enter your account \ n");scanf ("%d", k);printf ("Please enter password \ n");scanf ("%d", l);if (A1 = = Ka2 = = L){printf ("Your account balance is%d", A3);q = 0;}Else{printf ("Your password is wrong, please re-enter \ n");Q = 1;}}do{printf ("Please select \ n");printf ("1 query \ n");printf ("2 withdrawals \ n");printf ("3 Change pas

#include void Main (){int i=0,j=0,many=100000,many1=100000,num,num1,num2,num4=0,s=0,s1=0;int a=123;int b;printf ("");for (i;i{printf ("*");}printf ("\ n");printf ("");printf ("*");for (i=0;i{printf ("");}printf ("*");printf ("\ n");printf ("");printf ("*");for (i=0;i{printf ("");} printf ("*");printf ("\ n");printf ("");printf ("*");for (i=0;i{printf ("");}printf ("Welcome to the use of love to deposit ATM");for (i=0;i{printf (""); }printf ("*"); prin

[BZOJ1179] [Apio2009] AtmQuestion DescriptionInputThe 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, these two integers are between 1 and N, and the two integers of line i+1 represent the intersection number of the start and end point of the road. Next n lines, one integer per line, in order to represent the amount of money in the ATM ma

DescriptionInputThe 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, these two integers are between 1 and N, and the two integers of line i+1 represent the intersection number of the start and end point of the road. Next n lines, one integer per line, in order to represent the amount of money in the ATM machine at each intersection. The ne

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

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

Import OS‘‘‘This module is Db_handler, as the database connection engine(This is primarily a JSON file)‘‘‘def file_db_handler (conn_params): Db_path = '%s\%s '% (conn_params["path"],conn_params["name"]) Return Db_pathdef db_handler (conn_params): If conn_params["engine"] = = "File_storage": Return File_db_handler (Conn_params)def file_handler (conn_params): If conn_params["file"] = = "Mall": Return File_db_handler (Conn_params) elif conn_params["file"] = = "Trolley":

The final work of the Software Institute of Yunnan University The Final Course the 2006 Spring TermSchool of Software, Yunnan University The object-oriented programming with Java The ATM and Pizza Sale ProjectsLecturer:michael Zheng team members Id Name Role Phone e-Mail Work Description Score 20051120217 Ding Liang Project Leader, programme

; 4 , distinct must be placed at the beginning:when distinct when applied to multiple fields, the scope of its application is all fields following it, not just one field next to it, and distinct can only be placed in front of all fields,Select ID, distinct name from A; -- The error will be indicated because the distinct must be placed at the beginning 5 , for NULL the processing: distinct the NULL is not filtered, that is, the returned result contains NULL value of

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