1042 Shuffling Machine(20)

Shuffling is a procedure used to randomize a deck of playing cards. Because standard shuffling techniques are seen as weak, and in order to avoid “inside jobs” where employees collaborate with gamblers by performing inadequate shuffles, many casinos employ automatic shuffling machines. Your task is to simulate a shuffling machine.

The machine shuffles a deck of 54 cards according to a given random order and repeats for a given number of times. It is assumed that the initial status of a card deck is in the following order:

S1, S2, …, S13,
H1, H2, …, H13,
C1, C2, …, C13,
D1, D2, …, D13,
J1, J2

where “S” stands for “Spade”, “H” for “Heart”, “C” for “Club”, “D” for “Diamond”, and “J” for “Joker”. A given order is a permutation of distinct integers in [1, 54]. If the number at the i-th position is j, it means to move the card from position i to position j. For example, suppose we only have 5 cards: S3, H5, C1, D13 and J2. Given a shuffling order {4, 2, 5, 3, 1}, the result will be: J2, H5, D13, S3, C1. If we are to repeat the shuffling again, the result will be: C1, H5, S3, J2, D13.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer K (≤20) which is the number of repeat times. Then the next line contains the given order. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the shuffling results in one line. All the cards are separated by a space, and there must be no extra space at the end of the line.

Sample Input:

2
36 52 37 38 3 39 40 53 54 41 11 12 13 42 43 44 2 4 23 24 25 26 27 6 7 8 48 49 50 51 9 10 14 15 16 5 17 18 19 1 20 21 22 28 29 30 31 32 33 34 35 45 46 47

Sample Output:

S7 C11 C10 C12 S1 H7 H8 H9 D8 D9 S11 S12 S13 D10 D11 D12 S3 S4 S6 S10 H1 H2 C13 D2 D3 D4 H6 H3 D13 J1 J2 C1 C2 C3 C4 D1 S5 H5 H11 H12 C6 C7 C8 C9 S2 S8 S9 H10 D5 D6 D7 H4 H13 C5

Solution:

shuffling:洗牌
这题题目不怎么理解, 能做出来猜测占了大多数。

初始给出一个打乱次数K, 和一个order序列;

这个order有两个作用,

1, 代表了这张是什么牌, 根据1234..对应S1,S2,S3,S4..;

2, 这是一个始终不变的order, 这个位置的牌下一次就要被放到该位置order表中对应的值的位置中去。

有点抽象, 不过看数据。
只有五张牌 (不是一种特殊情况, 而是为了简化的假设)

1   2   3    4    5
S3, H5, C1, D13, J2
4   2   5    3    1
-------------------  k = 1;
1   2   3    4    5
5   2   4    1    3
J2  H5  D13  S3   C1
4   2   5    3    1
-------------------  k = 2;
1   2   3    4    5
3   2   1    5    4
C1  H5  S3   J2   D13
4   2   5    3    1
------------------- k ..
...

s3从1->4->3, 实际上pos = arr[pos].value;

定一个结点, 存放下标pos和实际元素value, 相当于链表next;这样能求出每个结点最后的pos, 然后对pos从小到大进行统一的排序。

坑点:

注意进位问题, 若13 / 13 = 1; 实际要需要0…13, 于是用整数部分用(i-1)/13, 余数(i-1)%13 +1

排序和遍历都是从1开始到54, 闭区间, 0位置不做处理.

实际代码:

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

struct Node {
  int value;
  int pos;
};

bool cmp(Node n1, Node n2) {
  return n1.pos < n2.pos;
}

int main() {
  int K; scanf("%d", &K);
  Node arr[60];
  memset(arr, 0, sizeof(arr));
  for (int i = 1; i <= 54; i++) {
    Node node;
    node.pos = i;
    scanf("%d", &node.value);
    arr[i] = node;
  }

  for (int i = 1; i <= 54; i++) {
    int pos = i, cnt = 0;
    if (pos == arr[pos].value) continue;
    while (cnt <= K) {
      pos = arr[pos].value;
      cnt++;
    }
    arr[i].pos = pos;
  }

  sort(arr+1, arr+55, cmp);

  char chs[6] = {'S', 'H', 'C', 'D', 'J'};
  for (int i = 1; i <= 54; i++) {
    if (i != 1) printf(" ");
    int a = (arr[i].value-1) / 13,
        b = (arr[i].value-1) % 13 + 1;   // debug02
    printf("%c%d", chs[a], b);           // debug01:b+1 error
  }
  return 0;
}