7-43 Shuffling Machine (20 分)（C语言写法，适合初学者）

这题是全英的，相信在一开始可能劝退很多人，但是耐下心看，这题却是不难，可以说是很简单。

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

结尾无空行

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

结尾无空行

思路就是先给出一副牌的数组，然后定义一个存放洗牌的数组，洗牌来回放置。（剩下内容注释说明）

我的写法效率可能会低一点，但是便于理解，很适合初学者理解。

#include <stdio.h>
#include <string.h>
int main()
{
    int k;   //洗牌次数
    int change[54];        //这里是存放洗牌方式
    scanf("%d", &k);
    char temp[54][4] = { 0 };     //和牌组互换的
    char arr[54][4] = { "S1","S2","S3","S4","S5","S6","S7","S8","S9","S10","S11","S12","S13",
        "H1","H2","H3","H4","H5","H6","H7","H8","H9","H10","H11","H12","H13",
        "C1","C2","C3","C4","C5","C6","C7","C8","C9","C10","C11","C12","C13",
        "D1","D2","D3","D4","D5","D6","D7","D8","D9","D10","D11","D12","D13",
        "J1","J2" };
    for (int i = 0; i < 54; i++)
    {
        scanf("%d", &change[i]);
    }
    int flag = 1,x=0;       //这里采用flag变化1和0来实现洗牌的来回替换
    while (k)
    {
        if (flag == 1)
        {
            for (int i = 0; i < 54; i++)
            {
                x = change[i] - 1;
                strcpy(temp[x], arr[i]);
            }
            flag = 0;        //洗好存于temp里面flag==0
            k--;
            continue;
        }
        if (flag == 0)
        {
            for (int i = 0; i < 54; i++)
            {
                x = change[i] - 1;
                strcpy(arr[x], temp[i]);
            }
            flag = 1;         //洗好存于arr里面flag==1
            k--;
            continue;
        }
    }
    if (flag == 0)        //剩下的就只是输出了
    {
        for (int i = 0; i < 54; i++)
        {
            printf("%s", temp[i]);
            if (i != 53)
                putchar(' ');
        }
    }
    if (flag == 1)
    {
        for (int i = 0; i < 54; i++)
        {
            printf("%s", arr[i]);
            if (i != 53)
                putchar(' ');
        }
    }
    return 0;
}

是不是感觉非常简单了，喜欢的给我点个赞吧！