UVa---------10935（Throwing cards away I）

题目：

Problem B: 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.

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

7

19

10

6

0

Output for sample input

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

分析：直接用queue进行模拟即可，不过要注意的是当n = 1时，"Discarded cards:"后不能加空格，否则上交时会提示PE错误
代码如下：

#include<cstdio>

#include<queue>

using namespace std;

int main(){

    int n;

    queue<int>cards;

    while(scanf("%d", &n) == 1 && n){

         bool flag = true;

        //将所有纸牌放在队中

        for(int i = 1; i <= n; i++)

            cards.push(i);

        if(n == 1)

            printf("Discarded cards:");

        else

            printf("Discarded cards: ");
　　　　 //模拟过程 while(cards.size() > 1){

            if(flag)

                flag = false;

            else

                printf(", ");

            //第一张卡片出队

            printf("%d",cards.front());

            cards.pop();

            //新的第一张卡片到队尾

            cards.push(cards.front());

            cards.pop();

        }

        printf("\n");

        printf("Remaining card: %d\n", cards.front());

        cards.pop();

    }

    return 0;

}

2015-07-04文