（C++）Shuffling Machine(思路与实现)

题目描述

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.

输入描述:

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.

输出描述:

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.

输入例子:

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

输出例子:

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

思路：

题目解析：这一道题题讲的是有一个顺序的牌，1到45，但是按照花色分类了。从左到右13张是S,  13张是H， 13张是C, 13张是D，13张是J。接下来一种操作，这种操作将牌的位置改变为指定位置，比如现在有54张牌，然后题目执行一个存放顺序，将第一张牌放到34号位置，将第二张牌放到23号位置上，当按照给出的一个放置顺序放完了，那么第一遍就放完了，然后题目中给出一个次数，也就是整个过程要执行多少次这样的顺序。

那么大题的思路则是：利用两个数组start和end数组，start数组存放的是从1到54的数，然后我将每次需要改变位置的数我放到end数组中，然后将end数组复制到这个start，然后进行第二遍的操作，这样一来，只要按照题中给的次数完了之后，最后start这个数组就是我们要的数组，然后我们根据数组中的值计算最后需要输出的(什么要计算：因为这个题目要求的输出带花色和数字），那么我们这样来做，首先创建一个保存花色的数组（字符数组）{‘S’‘H’‘C’‘D’‘J’},然后我们就可以更具存的数字进行得到这个具体的花色，比如我这个是存的14，那么(14-1) / 13 == 1，（i - 1）/13那么我就知道14对应的花色是H，那么对应的数字呢？(i - 1) % 13 +1 == 1,那么综合下来就知道这个14号之前对应的就是H1.通过这种方式就很方便了。

实现：

#include 
       
         const int N = 54; char map[5] = {'S', 'H', 'C', 'D', 'J'}; int start[N + 1], end[N + 1], next[N + 1]; int main(){ int k; scanf("%d", &k); for(int i = 1; i <= N; i ++){ start[i] = i; } for(int i = 1; i <= N; i ++){ scanf("%d", &next[i]); } for(int step = 0; step < k; step ++){ for(int i = 1; i <= N; i ++){ end[next[i]] = start[i]; } for(int j = 1; j <= N; j ++){ start[j] = end[j]; end[j] = 0; } } for(int i = 1; i <= N; i ++){ char ch = map[(start[i] - 1) / 13]; printf("%c%d", ch, (start[i]-1) % 13 + 1); if(i < N){ printf(" "); } } }