noi.openjudge 2406:Card Stacking

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http://noi.openjudge.cn/ch0304/2406/

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 right2. Every time she deals a card, she must place the next P (1 <= P <= 10) cards on the bottom of the deck; and3. 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

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

输出

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.

样例输入

3 9 2

样例输出

3
7
8

提示

INPUT DETAILS:

Bessie is playing cards with two cow friends and a deck of 9 cards. She must place two cards on the bottom of the deck each time she deals one.

OUTPUT DETAILS:

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    BessieInitial 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.

代码

/*
题意 贝斯和奶牛们玩牌共n个队员玩k张牌，分牌时候，每轮先分给奶牛，再分给贝斯，每次分给一个队员牌后，将牌前面的q张牌按顺序放在后面，给出最后贝斯的排,按照从小到大排好序。
*/
#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;int n,k,p,i,j;
int q[6000010]={0},head,tail,cnt=0,a[100011];int main()
{scanf("%d%d%d",&n,&k,&p);memset(q,0,sizeof(q));for(i=1; i<=k; i++)q[i]=i;head=1;tail=k+1;cnt=0;for(i=1; i<=k; i++){if(i%n==0)//到贝斯的牌 a[++cnt]=q[head];head++;//发牌，队首出队 for(j=1;j<=p;j++) //移动p张牌到后面去 q[tail++]=q[head++];}sort(a+1,a+1+cnt);for(i=1; i<=cnt; i++)printf("%d\n",a[i]);return 0;
}