PAT(A）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


注：为了简单和测试，我们设置10张牌并手动验证正确性。
   54只需修改宏定义和应整除和模上的值

#include<stdlib.h>
#include <stdio.h>
#define N 10     //const int N=54;     //牌的编号
int main()
{
    int start[N+1],End[N+1];   //数组下标从0开始，我们初始化大于1，这样就可以存在start[N],从而一一对应
    int Rank[N+1];//牌操作后位置
    char String[5]={'S','H','C','D','J'};//牌的编号与花色的对应关系
    int i;
    //初始化牌的编号
    for(i=1;i<=N;i++)
    {
        start[i]=i;
        scanf("%d",&Rank[i]);
    }
    //对应位置牌操作后位置
    /*for(i=1;i<=N;i++)
    {
        scanf("%d",&Rank[i]);
    }*/

    for(int K=1;K<=2;K++)//两次将第i个位置的数放在End的Rank[i]处
    {
         for(i=1;i<=N;i++)
            {
            End[Rank[i]]=start[i];
            }
         for(i=1;i<=N;i++)
         {
            start[i] =End[i];
         }
    }
   //printf("%d",End[2]);   printf("%d",End[1]);    printf("%d",End[5]);
    for(i=1;i<=N;i++)
    {
        if(i!=1) printf(" ");//控制输出格式
        start[i]--;      //13号牌属于S，13为一组，对13除后取整应为0（牌号与花色的对应关系）
    //printf("%c",String[5]);
        printf("%c%d",String[start[i]/10],End[i]%10+1);
    }
    return 0;
}

运行结果如下：