UVa 10935

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

题解：很显然这是约瑟夫环问题，用队列来做就简单多了，每次抽两张卡片，第一张踢出队列另一张加入队列尾部，一直循环直至只有一个元素为止。

#include <bits/stdc++.h>using namespace std;int main (){    int n,a,b;    while(cin>>n&&n)    {        queue<int>q;        for (int i=1; i<=n; i++)  //将所有数字入队            q.push(i);        if(n==1) //控制格式        printf("Discarded cards:\n");        else        printf("Discarded cards: ");        while(!q.empty())        {            a=q.front();            if(q.size()==1) //只有一个元素就跳出                break;            if(q.size()!=2)//控制格式                printf("%d, ",a);            else                printf("%d\n",a);            q.pop();            b=q.front();            q.pop();            q.push(b); //将第二数加入队尾        }        printf("Remaining card: %d\n",a);    }    return 0;}


题目链接：https://vjudge.net/contest/169852#problem/C