1042. Shuffling Machine (20) - sstream实现数字转字符串

本文介绍了一种模拟洗牌机工作的算法实现,通过输入特定的洗牌顺序和重复次数,实现了对一副54张牌的扑克进行洗牌。文章详细展示了如何利用C++中的vector容器来生成原始牌序,并按照指定顺序重新排列。

题目例如以下:

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

题目看起来较为复杂。事实上本质就是实现一个1-54依次排列的序列的按需求又一次排列。
使用三个vector来。一个用来生成原序列、记录变更后序列,还有一个用来记录变更的中间过程,第三个用来记录变更要求就可以,注意索引从0開始,因此要求的变更序列的全部值应当-1再被记录。

#include <iostream>
#include <sstream>
#include <vector>
#include <stdio.h>

using namespace std;

string getCard(int pos){
    // pos 1 - 13 S
    // pos 14 - 26 H
    // pos 27 - 39 C
    // pos 40 - 52 D
    // pos 53 - 54 J
    string result = "";
    int value;
    stringstream ss;
    if(pos < 53){
        int stand = (pos - 1)/ 13;
        value = pos % 13;
        if(value == 0) value = 13;
        switch(stand){
        case 0:
            ss << 'S';
            break;
        case 1:
            ss << 'H';
            break;
        case 2:
            ss << 'C';
            break;
        case 3:
            ss << 'D';
            break;
        }
        ss << value;
        ss >> result;
        return result;
    }
    value = pos - 52;
    ss << 'J' << value;
    ss >> result;
    return result;
}

int main()
{
    vector<int> origin,handle;
    vector<int> ask;
    handle.resize(54);
    for(int i = 1; i <= 54; i++)
        origin.push_back(i);
    int N;
    cin >> N;
    int num;
    for(int i = 0; i < 54; i++){
        scanf("%d",&num);
        ask.push_back(num-1);
    }
    for(int i = 0; i < N; i++){
        for(int j = 0; j < 54; j++){
            handle[ask[j]] = origin[j];
        }
        origin = handle;
    }
    cout << getCard(origin[0]);
    for(int i = 1; i < origin.size(); i++){
        cout << " " << getCard(origin[i]);
    }
    cout << endl;
    return 0;
}


转载于:https://www.cnblogs.com/bhlsheji/p/5322037.html

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