3.1入门模拟A1042. Shuffling Machine (20)

最新推荐文章于 2022-02-04 19:01:19 发布

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最新推荐文章于 2022-02-04 19:01:19 发布

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本文介绍了一种通过编程模拟洗牌过程的方法，详细解释了如何利用特定的随机顺序对一副包含54张牌的纸牌进行多次洗牌，并提供了完整的代码实现。

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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

#include <iostream>
#include <stdio.h>
#define M 54
using namespace std;
/*
    1、不能直接用char start[55]={'0','S1','S2','S3','S4','S5','S6'...,会报错，因为字符数组中只能存放单个字符，而'S1'算两个字符
        貌似没有字符串数组的说法；得另想它法了：用位置来分开确定字母和数字；
    2、一直没搞明白，怎么控制位置交换的次数，原来其中有重要一步，就是start[i]=end[i];相当于复制了一下；
*/

int main()
{
    char a[5]={'S','H','C','D','J'};
    int start[M+1]={0};
    int end[M+1]={0};
//    int next[M+1]={0,36,52,37,38,3,39,40,53,54,41,11,12,13,42,43,44,2,4,23,24,25,26,27,6,7,8,48,49,50,51,9,10,14,15,16,5,17,18,19,1,20,21,22,28,29,30,31,32,33,34,35,45,46,47};
    int next[55]={0};//存放第一次输入的顺序值
    int N;
    scanf("%d",&N);
    for(int i =1;i<=M;++i)
    {
        start[i]=i;
    }
    for(int i=1;i<=M;++i)
    {
        scanf("%d",&next[i]);
    }
    while(N--)
    {
        for(int i=1;i<=M;++i)
        {
            end[next[i]]=start[i];
        }
        for(int i=1;i<=M;++i)
        {
            start[i]=end[i];
        }
    }
//    int test=0;
    for(int i=1;i<M;++i)
    {
        if(end[i]%13 == 0) //因为当13,26..时，应该显示13，但是else中end[i]%13=0
        {
            printf("%c%d ",a[end[i]/13-1],13);
        }
        else
        {
           printf("%c%d ",a[end[i]/13],end[i]%13);
        }
    }
    printf("%c%d",a[end[M]/13],end[M]%13); //为了保证最后一个数末尾没有空格
    return 0;
}