1042 Shuffling Machine

一、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.

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

二、题目重述

给定初试序列 S1,S2...S13, H1,H2...H13, C1,C2...C13, D1,D2...D13, J1,J2，共52个字符串，它们分别代表54张牌。显然这个算法是一个特殊的洗牌方法，为了防止有人出老千，是这样的一个背景，所以就有了本题的洗牌题。接下来对这个序列进行这个算法，比如有5张牌S3,H5,C1,D13,J2，然后输入序列{4,2,5,3,1}就会把S3放到4号位置，把H5放到2号位置，C1放到5号位置，D13放到3号位置，J2放到1号位置，于是就得到了{J2,H5,D13,S3,C1}序列，如果再进行一次这样的操作，其实就是在上面变换之后序列的基础上再进行一次这样的变化，{J2,H5,D13,S3,C1}就变成了{C1,H5,S3,J2,D13}。

三、算法思路

3.1、要存储的数据

1. 对输入的数据进行保存：重复操作的次数n，以及54位变换序列（也就是洗牌顺序）
2. 变换前的牌组顺序
3. 变换后的牌组顺序

因此总结来看：

int n;
int before[54],after[54],change[54];

四、参考代码

#include <iostream>
using namespace std;

int main()
{
    int before[54];
    //0~12 - S1,S2,...S13
    //13~25 - H1,H2,...H13
    //26~38 - C1,C2,...C13
    //39~51 - D1,D2,...D13
    //52,53 -J1,J2
    //before[i]=index of array
    for(int i=0;i<54;i++){
        before[i]=i;
    }
    int after[54];
    int change[54];
    //n
    int n;
    cin>>n;

    for(int i=0;i<54;i++){
        cin>>change[i];
        change[i]--;//change-1 to after's index
    }
    //exchange n times
    while(n--){
        for(int i=0;i<54;i++){
            int temp=change[i];
            after[temp]=before[i];
        }
        for(int i=0;i<54;i++){
            before[i]=after[i];
        }
    }
    for(int i=0;i<54;i++){
        if(after[i]>=0 && after[i]<=12){
            //S1--S13
            if(i<53){
                cout<<'S'<<after[i]+1<<' ';
            }
            else{
                cout<<'S'<<after[i]+1;
            }
        }
        if(after[i]>=13 && after[i]<=25){
            //H1--H13
            if(i<53){
                cout<<'H'<<after[i]-12<<' ';
            }
            else{
                cout<<'H'<<after[i]-12;
            }
        }
        if(after[i]>=26 && after[i]<=38){
            //C1--C13
            if(i<53){
                cout<<'C'<<after[i]-25<<' ';
            }
            else{
                cout<<'C'<<after[i]-25;
            }
        }
        if(after[i]>=39 && after[i]<=51){
            //D1--D13
            if(i<53){
                cout<<'D'<<after[i]-38<<' ';
            }
            else{
                cout<<'D'<<after[i]-38;
            }
        }
        if(after[i]==52){
            if(i<53){
                cout<<"J1 ";
            }
            else{
                cout<<"J1";
            }
        }
        if(after[i]==53){
            if(i<53){
                cout<<"J2 ";
            }
            else{
                cout<<"J2";
            }
        }
    }
}

在一些地方可以进行优化，

1. 比如序列与花色的对应，可以用hash的思想进行映射，不必给每种情况一个if，因为他们是有规律的。
2. 再比如每个数组下标从1开始存放数据，在计算对应的时候更加简便。

#include <iostream>
using namespace std;
const int N=54;
int before[N+1],after[N+1],change[N+1];
/*
1~13--S1~S13
14~26--H1~H13
27~39--C1~C13
40~52--D1~D13
53~54--J1~J2
*/
char mp[5]={'S','H','C','D','J'};
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=N;i++){
        before[i]=i;//1-55 stand for S1,S2....
    }
    for(int i=1;i<=N;i++){
        cin>>change[i];
    }
    while(n--){
        for(int i=1;i<=N;i++){
            after[change[i]]=before[i];
        }
        for(int i=1;i<=N;i++){// after to before
            before[i]=after[i];
        }
    }
    //output after[N+1] from 1-54
    for(int i=1;i<=N;i++){
        if(i!=1)cout<<' ';
        cout<<mp[(after[i]-1)/13]<<(after[i]-1)%13+1;
        /*
        after[i]=27,27-1=26,26/13=2,mp[2]=C,26%13+1=1,27--C1,correct
        */
    }
}