UVa 10835,THrowing cards away I,卡牌游戏

原题

Throwing cards away I 
  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. 

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 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

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

 

题目描述:桌上有n(n<=50)张牌,从第一张开始,从上往下依次编号为1-n。当至少还剩下两张牌时进行以下操作:把第一张牌扔掉,然后把新的第一张牌放到整叠牌的最后。输入每行包含一个n,输出每次扔掉的牌以及最后剩下的牌。

使用队列,这道题就很简单了,注意一下输出格式,还有就是等于n=1的情况。

#include<iostream>
#include<queue>

using namespace std;

int main()
{
	int n;
	while(cin>>n && n)
	{
		cout<<"Discarded cards:";
		queue<int> q;
		for(int i=1;i<=n;i++)
		{
			q.push(i);			
		}	
		for(int i=1;i<n;i++)
		{
			cout<<" "<<q.front();
			if(i!=n-1) cout<<',';
			q.pop();
			q.push(q.front());
			q.pop();			
		}
		cout<<endl;
		cout<<"Remaining card: "<<q.front()<<endl;
	}
}

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转载自blog.csdn.net/qq_42761817/article/details/81202053