Throwing cards away I--队列

最新推荐文章于 2020-12-09 23:31:13 发布

j597039013

最新推荐文章于 2020-12-09 23:31:13 发布

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本文介绍了一个基于特定规则移除卡片的算法实现，该算法通过不断移除顶部卡片并重新排列剩余卡片来确定最终剩下的卡片及移除顺序。文章提供了一个使用C++编写的程序示例，展示了如何按照题目要求输出移除卡片的序列和最后剩下的卡片。

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

注意输出格式，要求较严格

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

int main()
{
    int n,i,k,t;

    while(scanf("%d",&n),n)
    {
        queue <int> a;
        k=0;
        for(i=1;i<=n;i++)
        {
            a.push(i);
        }
        printf("Discarded cards:");
        int len=a.size();
        while (len>1)
        {
            if(k)
            {
                printf(",");
            }
            printf(" %d",a.front());
            a.pop();
            t=a.front();
            a.pop();
            a.push(t);
            k=1;
            len--;
        }
        printf("\nRemaining card: %d\n",a.front());
    }
    return 0;
}