1042 Shuffling Machine

1042 Shuffling Machine （20 分）

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

 

/*
解题思路：
1、设置两个数组start[],end[],分别用来存放执行操作前的牌序与执行操作后的牌序
start[i]表示操作前第i个位置的牌的编号。这样在每一次操作中就可以把数组start[]中的每一位置的牌号
存放到数组end[]的对应转换位置中，然后用数组end[]覆盖数组start[]来给下一次操作使用。
这样当执行K轮操作后，数组start[]中即存放了最终的牌序。
2、由于输出需要用花色表示，且每种花色有13张牌，因此不妨设char型数组mp[]={S,H,C,D,J}来建立编号与花色的关系。
假设当前牌号为x,mp[(x-1)/13]就是这张牌对应的花色(1~13号为'S',14`~26号为'H'等)，(x-1)%13+1即为它所属花色下的编号。
当前牌号x减1的原因，假设当前牌号为13，对应花色为S13，那么(13-1)/13=0,对应mp数组mp[0]='S',(13-1)%13+1=13,对应S花色下的编号13
*/

#include <cstdio>

const int N=54;
char mp[5]={'S','H','C','D','J'}; //牌的编号与花色的对应关系
int start[N+1],end[N+1],next[N+1]; //next数组存放每个位置上的牌在操作后的位置,也就是题干中给定的操作序列

int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=N;i++)
    {
        start[i]=i; //初始化牌的编号
    }
    for(int i=1;i<=N;i++)
    {
        scanf("%d",&next[i]); //输入每个位置上的牌在操作后的位置
    }
    for(int step=0;step<n;step++)
    {
        for(int i=1;i<=N;i++)
        {
            end[next[i]]=start[i]; //把第i个位置的牌的编号存放到位置next[i]
        }
        for(int i=1;i<=N;i++)
        {
            start[i]=end[i]; //把end数组赋值给start数组以供下次操作使用
        }
    }
    for(int i=1;i<=N;i++)
    {
        if(i!=1) printf(" ");
        start[i]--; //对应思路中x-1的原因
        printf("%c%d",mp[start[i]/13],start[i]%13+1);
    }
    return 0;
}