POJ1250：Tanning Salon

Description

Input

Output

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.

题意：有n个位置，每个字母第一次出现代表客人的进来，第二次出现代表离开
统计流失了几个客户

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

char s[100005];

int main()
{
    int n;
    while(~scanf("%d",&n),n)
    {
        int l=0,i,j,flag,len;
        scanf("%s",s);
        len = strlen(s);
        char z[25];//Z数组存床是否有人
        memset(z,‘#‘,sizeof(z));
        for(i=0; i<len; i++)
        {
            flag=0;//flag=0表示进入
            for(j=1; j<=n; j++) //离开后位置清空
                if(z[j]==s[i])
                {
                    z[j]=‘#‘;
                    flag=1;
                    break;
                }
            if(flag==0)
                for(j=1; j<=n; j++) //有空位置就让顾客进去
                    if(z[j]==‘#‘)
                    {
                        z[j]=s[i];
                        break;
                    }
            if(j>n)//没有空位置流失人数加一
                l++;
        }
        if(l==0)
            printf("All customers tanned successfully.\n");
        else
            printf("%d customer(s) walked away.\n",l/2);
    }
    return 0;
}

POJ1250：Tanning Salon,布布扣,bubuko.com

POJ1250：Tanning Salon

原文：http://blog.csdn.net/libin56842/article/details/20238601