Codeforces Beta Round #4 (Div. 2 Only) C. Registration system hash

C. Registration system

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/problemset/problem/4/C

Description

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.

Input

The first line contains number n (1 ≤ n ≤ 105). The following n lines contain the requests to the system. Each request is a non-empty line, and consists of not more than 32 characters, which are all lowercase Latin letters.

Output

Print n lines, which are system responses to the requests: OK in case of successful registration, or a prompt with a new name, if the requested name is already taken.

Sample Input

4
abacaba
acaba
abacaba
acab

Sample Output

OK
OK
abacaba1

OK

题意：给定N个字符串，对于每个字符串如果之前出现过，输出出现的次数，否则输出OK。

题解：BKDRhash，map存储次数即可。

AC代码：

/*
* @Author: 王文宇
* @Date:   2018-06-25 23:52:54
* @Last Modified by:   王文宇
* @Last Modified time: 2018-06-26 00:08:51
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 100006;
const ll mod = 1998585857;
const int k = 131;
#define _for(i,a,b) for(int i=a;i<=b;i++)
int n;
char s[maxn];
map<ll,int> Q;
ll getnum(char *s)
{
	ll ans = 0;
	while(*s)
	{
		ans=(ans*k+(*s++)+mod)%mod;
	}
	return ans%mod;
}
int main(int argc, char const *argv[])
{
	cin>>n;
	_for(i,1,n)
	{
		cin>>s;
		ll now = getnum(s);
		if(!Q.count(now))
		{
			cout<<"OK"<<endl;
			Q[now]=1;
		}
		else
		{
			cout<<s<<Q[now]<<endl;
			Q[now]++;
		}
	}
	return 0;
}