UVa 10935 解题报告

最新推荐文章于 2020-12-09 23:31:13 发布

原创最新推荐文章于 2020-12-09 23:31:13 发布 · 340 阅读

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#uva #uva10593 #队列

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本文通过一道关于队列的应用题目，介绍了如何使用C++实现特定的队列操作，包括丢弃顶部卡片并将新的顶部卡片移至底部的过程。该程序演示了如何输出被丢弃的卡片序列及最终剩余的卡片。

关于队列运用的一道简单题

PE 两次注意·格式

#include <bits/stdc++.h>

using namespace std;

int main()
{
    while(1){
        int n;
        cin >> n;
        if(n==0)break;
        queue<int> q;
        for(int i=1;i<=n;i++){
            q.push(i);
        }
        cout <<"Discarded cards:";
        if(!q.empty()&&q.back()!=q.front()){
            for(int i=1;i<n;i++){
                cout <<" " << q.front();
                if(i!=n-1)cout << ",";
                q.pop();
                int temp = q.front();
                q.push(temp);
                q.pop();
            }
        //cout << endl; 如果在这个地方换行输入数据为 1 的时候格式错误
        }
        cout <<endl; // 在此处换行才AC
        cout << "Remaining card: " << q.front() << endl;
    }

    return 0;

}

原题如下：

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 nd the sequence of discarded
cards and the last, remaining card.
Input
Each line of input (except the last) contains a
number n 50. The last line contains `0' and
this line should not be processed.
Output
For each number from the input produce two
lines of output. The rst line presents the se-
quence of discarded cards, the second line re-
ports 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
Sample Output
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