[2011山东ACM省赛] Mathman Bank(模拟题)
版权声明:本文为博主原创文章,未经博主同意不得转载。
https://blog.csdn.net/sr19930829/article/details/24187925
Mathman Bank
nid=24#time" rel="nofollow" style="color:rgb(83,113,197);text-decoration:none;">
Time Limit: 1000ms Memory limit: 65536K 有疑问?点这里^_^
题目描写叙述
computers. Therefore, they want to use computers to manage their banks. However, mathmen's programming skills are not as good as their math-
ematical skills. So they need your help to write a bank management system software.
The
system must support the following operations:
1.
Open account: given the customer's name and password, and an initial deposit,
open an bank account for the customer.
2.
Deposit: given the amount of money and the name of the customer, deposit
money in the customer's account.
3.
Withdraw: given the amount of money, the customer's name and password,
withdraw money from the customer's account.
4.
Transfer: given the amount of money, sender's name, sender's password and
receiver's name, transfer money from the sender's account to the
receiver's
account.
5.
Check: given the customer's name and password, print the balance of the
customer's account.
6.
Change password: given the customer's name and old password, and the
new password, replace the customer's old password with the new one.
Initially,
no account exists in the bank management system .
输入
the number of commands. Each of the following n lines contains one of the following commands, in the
following format:
1.
O "customer" "password" "initial deposit" - open an account for "customer",
set the password to "password" and deposit "inital deposit" money
in the account. "customer" and "password" are strings, and "initial
deposit" is an integer.
2.
D "customer" "amount" - deposit "amount" money in "customer"'sac-
count. "customer" is a string, and "amount" is an integer.
3.
W"customer" "password" "amount" - withdraw "amount" money from "customer"'s
account. "customer" and "password" are strings, and amount
4.
T "sender" "password" "receiver" "amount" - transfer "amount" money from
"sender"'s acount to "receiver"'s account. "sender", "pasword" and
"receiver" are strings, and "amount" is an integer.
5.
C "customer" "password" - check "customer"'s balance. "customer" and
"password" are strings.
6.
X "customer" "old password" "new password" - replace "customer"'s old
password with "new password". "customer", "old password" and "new
password" are strings.
All
of the strings appearing in the input consist of only alphebetical letters and
digits, and has length at most 10. All the integers in the input are nonnegative,
and at most 1000000.
Refer
to the sample input for more details.
输出
1.
Open account: If "customer" already has an account, output "Account
exists." (without quotation marks); otherwise, output "Successfully
opened an account." (without quotation marks).
2.
Deposite: If "customer" doesn't have an account ,output "Account does not
exist."; otherwise, output "Successfully deposited money."(without quotation
marks).
3.
Withdraw: If "customer" doesn't have an account, output "Account does
not exist."; otherwise, if "customer"'s password doesn't match "password",
output "Wrong password."; otherwise, if there are less money
than "amount" in "customer"'s account, output "Money not enough.";
otherwise, output "Successfully withdrew money." quotation marks).
4.
Transfer: If "sender"'s account or "receiver"'s account doesn't exist,
output "Account does not exist."; otherwise, if "sender"'s password
doesn't match "password", output "Wrong password."; otherwise,
if there is less money than 'amount" in "sender"'s account, output
"Money not enough."; otherwise, output "Successfully transfered money."
5.
Check: If "customer"'s account doesn't exist, output "Account does not
exist."; otherwise, if "customer"'s password doesn't match "password",
output "Wrong password."; otherwise, output the balance of "customer"'s
account.
6.
Change password: If "customer" doesn't have an account, output "Account
does not exist."; otherwise, if "customer"'s password doesn't match
"old password", output "Wrong password."; otherwise, output 'Successfully
changed password.".
演示样例输入
25
W Alice alice 10
O Alice alice 10
C Alice alice
D Bob 10000
O Bob bob 100
D Alice 50
C Alice alice
X Bob bob BOB
C Bob bob
O Bob bob 10
T Bob bob Alice 100000
W Alice alice 10
T Bob BOB Alice 100000
T Bob BOB Alice 100
C Alice alice
C Bob BOB
T Alice alice BOB 10
T ALICE alice Bob 10
X Jack jack JACE
X Alice ALICE alice
W Alice Alice 10
W Alice alice 200
T Alice alice Bob 80
C Alice alice
C Bob BOB
演示样例输出
Account does not exist.
Successfully opened an account.
10
Account does not exist.
Successfully opened an account.
Successfully deposited money.
60
Successfully changed password.
Wrong password.
Account exists.
Wrong password.
Successfully withdrew money.
Money not enough.
Successfully transfered money.
150
0
Account does not exist.
Account does not exist.
Account does not exist.
Wrong password.
Wrong password.
Money not enough.
Successfully transfered money.
70
80
提示
来源
解题思路:
模拟银行开设账户,存款,取款。转账等业务,题目没难度,依照题意模拟,写代码时要细致。
watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvc3JfMTk5MzA4Mjk=/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/Center" />
代码:
#include <iostream>
#include <string.h>
using namespace std;
struct Node//为每一个账户开设一个结构体
{
string Name;//姓名
string Code;//密码
int Money;//存款金额
}node[1002];
int find(Node x[],int m,string name)//两个功能,推断name是否已经开了账户,假设已经开了返回在数组中的位置
{
for(int i=0;i<m;i++)
{
if(x[i].Name==name)
{
return i;
}
}
return -1;
}
int main()
{
int n;char cm;
int m=0;
cin>>n;
while(n--)
{
cin>>cm;
if(cm=='O')//开设账户
{
string name,code;
int money;
cin>>name>>code>>money;
if(find(node,m,name)==-1||m==0)//该姓名没有开账户
{
node[m].Name=name;
node[m].Code=code;
node[m].Money=money;
m++;
cout<<"Successfully opened an account."<<endl;
}
else
cout<<"Account exists."<<endl;
}
if(cm=='D')
{
string name;int money;
cin>>name>>money;
if(find(node,m,name)==-1)
cout<<"Account does not exist."<<endl;
else
{
node[find(node,m,name)].Money+=money;
cout<<"Successfully deposited money."<<endl;
}
}
if(cm=='W')
{
string name,code;int money;
cin>>name>>code>>money;
if(find(node,m,name)==-1)
{
cout<<"Account does not exist."<<endl;
}
else
{
if(node[find(node,m,name)].Code!=code)
{
cout<<"Wrong password."<<endl;
}
else if(node[find(node,m,name)].Money<money)
{
cout<<"Money not enough."<<endl;
}
else
{
node[find(node,m,name)].Money-=money;
cout<<"Successfully withdrew money."<<endl;
}
}
}
if(cm=='T')
{
string name1,code,name2;
int money;
cin>>name1>>code>>name2>>money;
if(find(node,m,name1)==-1||find(node,m,name2)==-1)
cout<<"Account does not exist."<<endl;
else if(node[find(node,m,name1)].Code!=code)
cout<<"Wrong password."<<endl;
else if(node[find(node,m,name1)].Money<money)
cout<<"Money not enough."<<endl;
else
{
node[find(node,m,name1)].Money-=money;
node[find(node,m,name2)].Money+=money;
cout<<"Successfully transfered money."<<endl;
}
}
if(cm=='C')
{
string name,code;
cin>>name>>code;
if(find(node,m,name)==-1)
{
cout<<"Account does not exist."<<endl;
}
else if(node[find(node,m,name)].Code!=code)
{
cout<<"Wrong password."<<endl;
}
else
{
cout<<node[find(node,m,name)].Money<<endl;
}
}
if(cm=='X')
{
string name,code1,code2;
cin>>name>>code1>>code2;
int len=find(node,m,name);
if(len==-1)
{
cout<<"Account does not exist."<<endl;
}
else if(node[len].Code!=code1)
{
cout<<"Wrong password."<<endl;
}
else
{
node[len].Code=code2;
cout<<"Successfully changed password."<<endl;
}
}
}
return 0;
}
[2011山东ACM省赛] Mathman Bank(模拟题)的更多相关文章
- [2011山东ACM省赛] Identifiers(模拟)
Identifiers Time Limit: 1000ms Memory limit: 65536K 有疑问?点这里^_^ 题目描写叙述 Identifier is an important ...
- [2011山东ACM省赛] Sequence (动态规划)
Sequence Time Limit: 1000ms Memory limit: 65536K 有疑问?点这里^_^ 题目描述 Given an integer number sequence ...
- [2011山东ACM省赛] Binomial Coeffcients(求组合数)
Binomial Coeffcients nid=24#time" style="padding-bottom:0px; margin:0px; padding-left:0px; ...
- [2012山东ACM省赛] Pick apples (贪心,完全背包,枚举)
Pick apples Time Limit: 1000MS Memory limit: 165536K 题目描述 Once ago, there is a mystery yard which on ...
- [2012山东ACM省赛] Pick apples (贪心,全然背包,枚举)
Pick apples Time Limit: 1000MS Memory limit: 165536K 题目描写叙述 Once ago, there is a mystery yard which ...
- 山东ACM省赛历届入口
山东省第一届ACM大学生程序设计竞赛 山东省第二届ACM大学生程序设计竞赛 山东省第三届ACM大学生程序设计竞赛 山东省第四届ACM大学生程序设计竞赛 山东省第五届ACM大学生程序设计竞赛 山东省第六 ...
- [2013山东ACM]省赛 The number of steps (可能DP,数学期望)
The number of steps nid=24#time" style="padding-bottom:0px; margin:0px; padding-left:0px; ...
- hdu 4023 2011上海赛区网络赛C 贪心+模拟
以为是贪心,结果不是,2333 贪心最后对自己绝对有利的情况 点我 #include<cstdio> #include<iostream> #include<algori ...
- 第八届山东ACM省赛F题-quadratic equation
这个题困扰了我长达1年多,终于在今天下午用两个小时理清楚啦 要注意的有以下几点: 1.a=b=c=0时 因为x有无穷种答案,所以不对 2.注意精度问题 3.b^2-4ac<0时也算对 Probl ...
随机推荐
- sqlhelper中事务的简单用法
sql1="INSERT INTO tablename(Id,col1,col2) VALUES(@Id,@col1,@col2) update tablename2 set col=@co ...
- Eclipse在当前行之上插入一行
在当前行之上插入一行快捷键: Ctrl + Shift + Enter 在当前行之下插入一行快捷键: Shift + Enter
- java.lang.ExceptionInInitializerError异常
今天在开发的过程中,遇到java.lang.ExceptionInInitializerError异常,百度查了一下,顺便学习学习,做个笔记 静态初始化程序中发生意外异常的信号,抛出Exception ...
- Python paramiko ssh 在同一个session里run多个命令
import threading, paramiko strdata='' fulldata='' class ssh: shell = None client = None transport = ...
- HDU1045(KB10-A 二分图最大匹配)
Fire Net Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Su ...
- MySQL两种存储引擎: MyISAM和InnoDB
MySQL两种存储引擎: MyISAM和InnoDB 简单总结 MyISAM是MySQL的默认数据库引擎(5.5版之前),由早期的ISAM(Indexed Sequential Access Me ...
- js-ES6学习笔记-Class(2)
1.与函数一样,类也可以使用表达式的形式定义. const MyClass = class Me { getClassName() { return Me.name; } }; 这个类的名字是MyCl ...
- 配置ArcGIS Server使用Windows AD Windows集成身份认证
1.配置 ArcGIS Server 以使用 Windows Active Directory 用户和角色. 2.填写Windows域账号凭证,对账号的要求如下: 需要能读取域中的用户和组,一般从属于 ...
- singleInstance和singleTask导致startActivityForResult回调失败
先来了解下这两种启动模式: 1.singleInstance,全局唯一,它的实例在全局(即在众多任务栈中)是唯一的,它单独地存在于属于自己的任务栈中,而且这个任务栈没有其他实例. 2.singleTa ...
- JNI使用方法
JNI可以让我们在java代码中调用本地库的功能. 下面记录一下JNI简单的使用方法 创建java端接口 public class JNIIterface { // 导入最终生成的dll文件 stat ...