题目链接：https://ac.nowcoder.com/acm/contest/993/A/
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld

题目描述

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.

输入

3 9 2

输出

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.

解题思路

题意:有 M=K/N 张“好”卡和 K-M 张“坏”卡。
1.首先将卡片顶部的卡片交给她右边的牛
2.将P张卡片放到下面;
3.以逆时针方式继续以这种方式顺序处理
要求你帮助Bessie应该把“好”牌放在哪里，以便她能得到所有这些牌。
思路:直接模拟。

Accepted Code:

#include <bits/stdc++.h>
using namespace std;
int spt[50005];
int main() {
    int n, k, p;
    queue <int> Q;
    scanf("%d%d%d", &n, &k, &p);
    for (int i = 1; i <= k; i++)
        Q.push(i);
    int cnt = 0, res = 0;
    while (res < k / n) {
        if (!(++cnt % n))
            spt[res++] = Q.front();
        Q.pop();
        for (int i = 0; i < p; i++) {
            Q.push(Q.front());
            Q.pop();
        }
    }
    sort(spt, spt + res);
    for (int i = 0; i < res; i++)
        printf("%d\n", spt[i]);
    return 0;
}