poj 1250(Tanning Salon)

最新推荐文章于 2020-08-27 18:39:06 发布

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最新推荐文章于 2020-08-27 18:39:06 发布

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分类专栏：暴水题北大oj 模拟题文章标签： stdstring integer input each output 存储

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本文介绍了一个使用模拟算法解决旅馆顾客入住与离开问题的方法。通过跟踪每个顾客的行为，该算法可以确定有多少顾客因没有可用床位而离开。文章还提供了一个具体的实现案例。

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题意：一个旅馆有n个位，给出所有旅客到达旅馆和离开旅馆的顺序，问有多少旅客是没有住旅馆就离开的。

由于题目中告诉每个顾客只会出现一次，所以我用一个数组存储每个顾客的情况直接模拟：

Tanning Salon

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 5276		Accepted: 2876

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.

Source

Mid-Central USA 2002

#include<cstdio>
#include<stack>
#include<iostream>
#include<cstring>

using namespace std;

string str[25];
int n,zumu[26];
char ch[70];
int main(){
    while(scanf("%d", &n) && n){
        memset(zumu,0,sizeof(zumu));
        scanf("%s", ch);
        int sum = 0;
        for(int i = 0; ch[i]; i++){
            if(zumu[ch[i] - 'A'] == 0 && n != 0)  {n--;zumu[ch[i] - 'A']++;}
            else if(zumu[ch[i] - 'A'] == 1) {zumu[ch[i] - 'A']--;n++;}
            else if(n == 0){
                if(zumu[ch[i] - 'A'] == 0){
                    zumu[ch[i] - 'A'] = 2;
                    sum++;
                }
            }
        }
        if(sum == 0)  printf("All customers tanned successfully.\n");
        else printf("%d customer(s) walked away.\n", sum);
    }
    return 0;
}