codeforces 828A. Restaurant Tables(水题)

本文介绍了一个简单的算法,用于解决小餐厅中如何为顾客群体分配单人桌和双人桌的问题。该算法考虑了不同顾客群体的规模,并描述了在各种情况下如何分配桌子,以及无法服务的顾客数量。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

题目链接:http://codeforces.com/problemset/problem/828/A点击打开链接

A. Restaurant Tables
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

In a small restaurant there are a tables for one person and b tables for two persons.

It it known that n groups of people come today, each consisting of one or two people.

If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table. If there are none of them, it is seated at a two-seater table occupied by single person. If there are still none of them, the restaurant denies service to this group.

If a group consist of two people, it is seated at a vacant two-seater table. If there are none of them, the restaurant denies service to this group.

You are given a chronological order of groups coming. You are to determine the total number of people the restaurant denies service to.

Input

The first line contains three integers na and b (1 ≤ n ≤ 2·1051 ≤ a, b ≤ 2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables.

The second line contains a sequence of integers t1, t2, ..., tn (1 ≤ ti ≤ 2) — the description of clients in chronological order. If ti is equal to one, then the i-th group consists of one person, otherwise the i-th group consists of two people.

Output

Print the total number of people the restaurant denies service to.

Examples
input
4 1 2
1 2 1 1
output
0
input
4 1 1
1 1 2 1
output
2
Note

In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, all clients are served.

In the second example the first group consists of one person, it is seated at the vacant one-seater table. The next group consists of one person, it occupies one place at the two-seater table. It's impossible to seat the next group of two people, so the restaurant denies service to them. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, the restaurant denies service to 2 clients.



水题 注意当接待一个人时 如果有剩余一个位子的两位桌和没有人的两位桌 要选择没有人的 没看清就这样wa了。。

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include<algorithm>
#include <math.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <vector>
using namespace std;
int main()
{
    int a[3];
    a[1]=0;a[2]=0;a[0]=0;
    int n=0;int sum=0;
    scanf("%d%d%d",&n,&a[1],&a[2]);
    for(int i=0;i<n;i++)
    {
        int nn=0;
        scanf("%d",&nn);
        if(nn==2)
        {
            if(a[2]>0)
                a[2]--;
            else
            {
                sum+=2;
            }
        }
        else if(nn==1)
        {
            if(a[1]>0)
                a[1]--;
            else
            {
                if(a[2]>0)
                {
                    a[2]--;
                    a[0]++;
                }
                else if(a[0]>0)
                    {
                        a[0]--;
                    }
                else
                    sum++;
            }
        }
    }
    printf("%d",sum);

}




### Codeforces Div.2 比赛难度介绍 Codeforces Div.2 比赛主要面向的是具有基础编程技能到中级平的选手。这类比赛通常吸引了大量来自全球不同背景的参赛者,包括大学生、高中生以及一些专业人士。 #### 参加资格 为了参加 Div.2 比赛,选手的评级应不超过 2099 分[^1]。这意味着该级别的竞赛适合那些已经掌握了一定算法知识并能熟练运用至少一种编程语言的人群参与挑战。 #### 目设置 每场 Div.2 比赛一般会提供五至七道目,在某些特殊情况下可能会更多或更少。这些目按照预计解决难度递增排列: - **简单(A, B 类型)**: 主要测试基本的数据结构操作和常见算法的应用能力;例如数组处理、字符串匹配等。 - **中等偏难(C, D 类型)**: 开始涉及较为复杂的逻辑推理能力和特定领域内的高级技巧;比如图论中的最短路径计算或是动态规划入门应用实例。 - **高难度(E及以上类型)**: 对于这些问,则更加侧重考察深入理解复杂概念的能力,并能够灵活组合多种方法来解决问;这往往需要较强的创造力与丰富的实践经验支持。 对于新手来说,建议先专注于理解和练习前几类较容易的问,随着经验积累和技术提升再逐步尝试更高层次的任务。 ```cpp // 示例代码展示如何判断一个数是否为偶数 #include <iostream> using namespace std; bool is_even(int num){ return num % 2 == 0; } int main(){ int number = 4; // 测试数据 if(is_even(number)){ cout << "The given number is even."; }else{ cout << "The given number is odd."; } } ```
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值