UVA - 10935 Throwing cards away I

Given is an ordered deck of n cards numbered 1 to n with card 1 at the top and cardn 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

7191060

Output for sample input

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

题意：桌子上面有一叠牌，从上到下依次编号为 1 - n;做如下的操作：把第一张牌扔掉，把新的第一张牌放到最后一直操作：
#include <iostream>#include <algorithm>#include <cstdio>#include <queue>using namespace std;    const int maxn = 100; queue<int> q;    int main() {    int n;     int ans[maxn];     while(scanf("%d", &n) != EOF && n)     {         for(int i = 1; i <= n; i++)   q.push(i);         int k = 0;         while(!q.empty())         {             ans[k++] = q.front();             q.pop();            int t = q.front();             q.pop();             q.push(t);         }         printf("Discarded cards:");         for(int i = 0; i < n-1; i++)         {             if(i)   printf(",");             printf(" %d", ans[i]);        }         printf("\nRemaining card: %d\n", ans[n-1]);     }     return 0; }