省赛2D题

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2167

因为所有客户资料是与账户名挂钩的，所以以一个map做映射，做账户名和一个结构体的映射，结构体是cipher（string）和balance（int）。

#include <iostream>#include <map>#include <string>#include <stdio.h>using namespace std;typedef struct pwd_bal{string password;int balance;}p;int main(){int t;cin>>t;map<string, p> a;while(t--){char c;cin>>c;if(c == 'O'){string name;p temp;cin>>name>>temp.password>>temp.balance;if(a.insert(make_pair(name,temp)).second){cout<<"Successfully opened an account."<<endl;}else{cout<<"Account exists."<<endl;}}else if(c == 'D'){string name;int money;cin>>name>>money;map<string,p>::iterator it = a.find(name);if(it != a.end()){(it->second).balance += money;cout<<"Successfully deposited money."<<endl;}else{cout<<"Account does not exist."<<endl;}}else if(c == 'W'){string name, password;int money;cin>>name>>password>>money;map<string,p>::iterator it = a.find(name);if(it != a.end()){if((it->second).password == password){if((it->second).balance >= money){(it->second).balance -= money;cout<<"Successfully withdrew money."<<endl;}else{cout<<"Money not enough."<<endl;}}else{cout<<"Wrong password."<<endl;}}else{cout<<"Account does not exist."<<endl;}}else if(c == 'T'){string name1, password, name2;int money;cin>>name1>>password>>name2>>money;map<string,p>::iterator it = a.find(name1);map<string,p>::iterator itt = a.find(name2);if(it != a.end() && itt != a.end()){if((it->second).password == password){if((it->second).balance >= money){(it->second).balance -= money;(itt->second).balance += money;cout<<"Successfully transfered money."<<endl;}else{cout<<"Money not enough."<<endl;}}else{cout<<"Wrong password."<<endl;}}else{cout<<"Account does not exist."<<endl;}}else if(c == 'C'){string name, password;cin>>name>>password;map<string,p>::iterator it = a.find(name);if(it != a.end()){if((it->second).password == password){cout<<(it->second).balance<<endl;}else{cout<<"Wrong password."<<endl;}}else{cout<<"Account does not exist."<<endl;}}else if(c == 'X'){string name, password1, password2;cin>>name>>password1>>password2;map<string,p>::iterator it = a.find(name);if(it != a.end()){if((it->second).password == password1){(it->second).password = password2;cout<<"Successfully changed password."<<endl;}else{cout<<"Wrong password."<<endl;}}else{cout<<"Account does not exist."<<endl;}}}}

另外，今天在帮胖胖和一宁查错的时候，学习了一个新插件，Notepad++的compare，直接在plugin安装那里选择最新的稳定的版本即可，查错十分方便。