标签:algo gets lse each 表示 序列 star pac nes
A new e-mail service "Berlandesk" is going to be opened in Berland in the near future.
The site administration wants to launch their project as soon as possible, that‘s why they
ask you to help. You‘re suggested to implement the prototype of site registration system.
The system should work on the following principle.
Each time a new user wants to register, he sends to the system a request with his name.
If such a name does not exist in the system database, it is inserted into the database, and
the user gets the response OK, confirming the successful registration. If the name already
exists in the system database, the system makes up a new user name, sends it to the user
as a prompt and also inserts the prompt into the database. The new name is formed by the
following rule. Numbers, starting with 1, are appended one after another to name (name1,
name2, ...), among these numbers the least i is found so that namei does not yet exist in
the database.
4 abacaba acaba abacaba acab
OK OK abacaba1 OK
/** 分析:该题是,判断一个新输入字符串是否属于原先字符串序列 ①"不属于" : 打印"OK"表示插入成功 ②"属于" : 判断是第几次出现该串(序列号) --> 并打印:"该串" + 序列号 方法:map <string, int> 模板: map <string, int> my_map; pair <map <string, int> :: iterator, bool> pr; pr = my_map.insert (map <string, int> (str, 0)); if (!pr.second) { ++ my_map [str]; cout <<str <<my_map [str] <<endl; } else { cout <<"OK" <<endl; } **/
C/C++代码实现:
#include <iostream> #include <algorithm> #include <cstring> #include <cmath> #include <cstdio> #include <stack> #include <queue> #include <map> using namespace std; int N; string str; int main () { ~scanf ("%d", &N); map <string, int> my_map; pair <map <string, int> :: iterator, bool> pr; while (N --) { cin >>str; pr = my_map.insert (pair <string, int> (str, 0)); if (!pr.second) { // 插入失败 my_map [str] ++; cout <<str <<my_map [str] <<endl; } else { printf ("OK\n"); } } return 0; }
nyoj 991 Registration system (map)
标签:algo gets lse each 表示 序列 star pac nes
原文地址:https://www.cnblogs.com/GetcharZp/p/8974000.html