Card Stacking 队列模拟

题目链接：https://ac.nowcoder.com/acm/contest/993/A
Bessie is playing a card game with her N-1 (2 <= N <= 100) cow friends using a deck with K (N <= K <= 100,000; K is a multiple of N) cards.  The deck contains M = K/N "good" cards and K-M "bad" cards. Bessie is the dealer and, naturally, wants to deal herself all of the "good" cards. She loves winning.
Her friends suspect that she will cheat, though, so they devise a dealing system in an attempt to prevent Bessie from cheating. They tell her to deal as follows:
   1. Start by dealing the card on the top of the deck to the cow to her right
   2. Every time she deals a card, she must place the next P (1 <= P <= 10) cards on the bottom of the deck; and
   3. Continue dealing in this manner to each player sequentially in a counterclockwise manner
Bessie, desperate to win, asks you to help her figure out where she should put the "good" cards so that she gets all of them. Notationally, the top card is card #1, next card is #2, and so on.

输入描述:

* Line 1: Three space-separated integers: N, K, and P

输出描述:

* Lines 1..M: Positions from top in ascending order in which Bessie should place "good" cards, such that when dealt, Bessie will obtain all good cards.

示例1

输入

复制

3 9 2

输出

复制

3
7
8

说明

Bessie should put the "good" cards in positions 3, 7, and 8. The cards will be dealt as follows; the card numbers are "position in original deck":
Card Deck         P1      P2    Bessie
Initial configuration           1 2 3 4 5 6 7 8 9  - - -   - - -   - - -
Deal top card [1] to Player 1   2 3 4 5 6 7 8 9    1 - -   - - -   - - -
Top card to bottom (#1 of 2)    3 4 5 6 7 8 9 2    1 - -   - - -   - - -
Top card to bottom (#2 of 2)    4 5 6 7 8 9 2 3    1 - -   - - -   - - -
Deal top card [4] to Player 2   5 6 7 8 9 2 3      1 - -   4 - -   - - -
Top card to bottom (#1 of 2)    6 7 8 9 2 3 5      1 - -   4 - -   - - -
Top card to bottom (#2 of 2)    7 8 9 2 3 5 6      1 - -   4 - -   - - -
Deal top card [7] to Bessie     8 9 2 3 5 6        1 - -   4 - -   7 - -
Top card to bottom (#1 of 2)    9 2 3 5 6 8        1 - -   4 - -   7 - -
Top card to bottom (#2 of 2)    2 3 5 6 8 9        1 - -   4 - -   7 - -
Deal top card [2] to Player 1   3 5 6 8 9          1 2 -   4 - -   7 - -
Top card to bottom (#1 of 2)    5 6 8 9 3          1 2 -   4 - -   7 - -
Top card to bottom (#2 of 2)    6 8 9 3 5          1 2 -   4 - -   7 - -
Deal top card [6] to Player 2   8 9 3 5            1 2 -   4 6 -   7 - -
Top card to bottom (#1 of 2)    9 3 5 8            1 2 -   4 6 -   7 - -
Top card to bottom (#2 of 2)    3 5 8 9            1 2 -   4 6 -   7 - -
Deal top card [3] to Bessie     5 8 9              1 2 -   4 6 -   7 3 -
Top card to bottom (#1 of 2)    8 9 5              1 2 -   4 6 -   7 3 -
Top card to bottom (#2 of 2)    9 5 8              1 2 -   4 6 -   7 3 -
Deal top card [9] to Player 1   5 8                1 2 9   4 6 -   7 3 -
Top card to bottom (#1 of 2)    8 5                1 2 9   4 6 -   7 3 -
Top card to bottom (#2 of 2)    5 8                1 2 9   4 6 -   7 3 -
Deal top card [5] to Player 2   8                  1 2 9   4 6 5   7 3 -
Top card to bottom (#1 of 2)    8                  1 2 9   4 6 5   7 3 -
Top card to bottom (#1 of 2)    8                  1 2 9   4 6 5   7 3 -
Deal top card [8] to Bessie     -                  1 2 9   4 6 5   7 3 8
Bessie will end up with the "good cards" that have been placed in positions 3, 7, and 8 in the original deck.

题意：n 个人玩 k 张牌，发牌员是 n 号，一共有 k/n 张好牌，发牌员全都要，问需要把好牌放在哪里才能拿到。（发牌规则：从 1 号开始，每次发一张牌，发完之后把牌堆顶部的 p 张牌全部放到牌堆底部。然后继续发牌。）

队列：队列的入队规则，先进先出

#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
vector<int>a[];
queue<int>q;
int main()
{
    int n, k, p, cnt = ;
    cin >> n >> k >> p;
    for (int i = ; i <= k; i++)
        q.push(i);
    while (!q.empty())
    {
        int x;
        x = q.front();
        q.pop();
        a[cnt].push_back(x);
        for (int i = ; i < p&&!q.empty(); i++)//注意要判空，否则会段错误
        {
            int temp;
            temp = q.front();
            q.pop();
            q.push(temp);
        }
        cnt++;
        if (cnt > n)
            cnt=;
    }
    sort(a[n].begin(), a[n].end());

    for (int i = ; i < a[n].size(); i++)
        cout << a[n][i] << endl;

    return ;
}