POJ 1250 시뮬레이션 문제

드디어 POJ에서 간단한 문제를 만났어요. 쉽지 않았어요.시뮬레이션으로 쉽게 물을 건널 수 있다.제목:
Tanning Salon
Time Limit: 1000MS
 
Memory Limit: 10000K
Total Submissions: 5582
 
Accepted: 3052
Description
Tan Your Hide, Inc., owns several coin-operated tanning salons. Research has shown that if a customer arrives and there are no beds available, the customer will turn around and leave, thus costing the company a sale. Your task is to write a program that tells the company how many customers left without tanning. 
Input
The input consists of data for one or more salons, followed by a line containing the number 0 that signals the end of the input. Data for each salon is a single line containing a positive integer, representing the number of tanning beds in the salon, followed by a space, followed by a sequence of uppercase letters. Letters in the sequence occur in pairs. The first occurrence indicates the arrival of a customer, the second indicates the departure of that same customer. No letter will occur in more than one pair. Customers who leave without tanning always depart before customers who are currently tanning. There are at most 20 beds per salon. 
Output
For each salon, output a sentence telling how many customers, if any, walked away. Use the exact format shown below. 
Sample Input

2 ABBAJJKZKZ
3 GACCBDDBAGEE
3 GACCBGDDBAEE
1 ABCBCA
0

Sample Output

All customers tanned successfully.
1 customer(s) walked away.
All customers tanned successfully.
2 customer(s) walked away.

ac 코드:

#include <iostream>
#include <string>
#include <cstdio>
using namespace std;
int numbed,visted[30];
int main(){
	//freopen("11.txt","r",stdin);
	while(scanf("%d",&numbed),numbed){
	  memset(visted,0,sizeof(visted));
	  string ss;
	  cin>>ss;
	  int num=numbed;
	  int sum=0;
	  int len=ss.size();
	  for(int i=0;i<len;++i){
		  if(visted[ss[i]-'A']==2)
			  continue;
		  else if(num<=0&&!visted[ss[i]-'A']){sum++;visted[ss[i]-'A']=2;}
		  else if(!visted[ss[i]-'A']&&num){
		    visted[ss[i]-'A']=1;
			num--;
		  }
		  else if(visted[ss[i]-'A']){
		    num++;
		  }
	  }
	  if(sum==0)printf("All customers tanned successfully.
");
	  else printf("%d customer(s) walked away.
",sum);
	}
  return 0;
}