Uva 10935 Throwing cards away I

本文介绍了一个关于有序卡片堆的编程问题,通过队列操作来模拟特定规则下卡片的丢弃与移动过程,并展示了如何使用C++实现这一过程,包括读取输入、处理逻辑以及输出结果。

Given is an ordered deck of n cards numbered 1 to n with card 1 at the top and card n at the bottom. The following operation is performed as long as there are at least two cards in the deck:

Throw away the top card and move the card that is now on the top of the deck to the bottom of the deck.

Your task is to find the sequence of discarded cards and the last, remaining card.

Each line of input (except the last) contains a number n ≤ 50. The last line contains 0 and this line should not be processed. For each number from the input produce two lines of output. The first line presents the sequence of discarded cards, the second line reports the last remaining card. No line will have leading or trailing spaces. See the sample for the expected format.

Sample input

7
19
10
6
0

Output for sample input

Discarded cards: 1, 3, 5, 7, 4, 2
Remaining card: 6
Discarded cards: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 4, 8, 12, 16, 2, 10, 18, 14
Remaining card: 6
Discarded cards: 1, 3, 5, 7, 9, 2, 6, 10, 8
Remaining card: 4
Discarded cards: 1, 3, 5, 2, 6
Remaining card: 4

这道题就是一个队列的问题,之所以把这道题拉上来,是想说这个题PE的事。 

我之前把回车换行放在了discarded cards的数字后面,结果一直PE。因为这样n=1的时候,输出会变成一排不会输出一个回车换行。不过这道题居然也没有提n=1的特殊情况应该怎么输出,只能去尝试它到底是什么格式了.....

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        queue<int> que;
        int a[55];
        for(int i=1;i<=n;i++)
            que.push(i);
        for(int i=1;i<=n-1;i++)
        {
            int x=que.front();
            que.pop();
            a[i]=x;
            x=que.front();
            que.pop();
            que.push(x);
        }
        int y=que.front();
        printf("Discarded cards:");
        for(int i=1;i<=n-1;i++)
        {
            if(i!=n-1) printf(" %d,",a[i]);
            else printf(" %d",a[i]);
        }
        printf("\nRemaining card: %d\n",y);
    }
    return 0;
}


在 Java 中,`readLine` 方法通常来自 `BufferedReader` 类,用于从输入流中读取一行文本。以下是几种使用或存储 `readLine` 方法返回值而非丢弃该值的常见方法: #### 存储到字符串变量中 可以将 `readLine` 方法的返回值存储在一个字符串变量中,以便后续使用。 ```java import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; public class StoreReadLineValue { public static void main(String[] args) { try (BufferedReader reader = new BufferedReader(new InputStreamReader(System.in))) { String line = reader.readLine(); System.out.println("读取的内容是: " + line); } catch (IOException e) { e.printStackTrace(); } } } ``` #### 存储到列表中 如果需要读取多行文本,可以将每行内容存储到一个列表中。 ```java import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.ArrayList; import java.util.List; public class StoreReadLinesInList { public static void main(String[] args) { List<String> lines = new ArrayList<>(); try (BufferedReader reader = new BufferedReader(new InputStreamReader(System.in))) { String line; while ((line = reader.readLine()) != null) { lines.add(line); } } catch (IOException e) { e.printStackTrace(); } for (String line : lines) { System.out.println(line); } } } ``` #### 在循环中直接使用 在读取每行内容时,也可以直接在循环中对每行内容进行处理,而不是先存储起来。 ```java import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; public class UseReadLineInLoop { public static void main(String[] args) { try (BufferedReader reader = new BufferedReader(new InputStreamReader(System.in))) { String line; while ((line = reader.readLine()) != null) { System.out.println("处理每行内容: " + line.toUpperCase()); } } catch (IOException e) { e.printStackTrace(); } } } ```
评论
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
红包 添加红包
表情包 插入表情
 条评论被折叠 查看
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