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 namedoes 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.
Example
Input
4
abacaba
acaba
abacaba
acab
Output
OK
OK
abacaba1
OK
Input
6
first
first
second
second
third
third
Output
OK
first1
OK
second1
OK
third1
用暴力超出時間限制．所以換了另一種思路，代碼如下
#include<stdio.h>
#include<string.h>
struct node{
char ch[50];
int x;
};//用結構體將字符串和出現的次數存起來
node a[100005],b[100005];
int main()
{
int n;
int t;
int flag;
while(scanf("%d",&n)!=EOF)
{
for(int i=0;i<n;i++)
{
scanf("%s",a[i].ch);
a[i].x=0;
}
b[0]=a[0];
b[0].x=0;//b數組存的是所有字符串中每一種的種類
t=1;
for(int i=1;i<n;i++)
{
flag=0;
for(int j=0;j<t;j++)
{
if(strcmp(a[i].ch,b[j].ch)==0)//如果b中存在則將Ｘ的值改變
{
a[i].x=b[j].x+1;
b[j]=a[i];
flag=1;
break;
}
}//若不存在則將這個字符串１加入到b數組中
if(!flag){
b[t]=a[i];
b[t].x=0;
t++;
}
}
for(int i=0;i<n;i++)
{
if(a[i].x==0)
printf("OK\n");
else
printf("%s%d\n",a[i].ch,a[i].x);
}
}
return 0;
}