PAT 1031Shuffling Machine (20)

    xiaoxiao2023-12-02  11

    题目

    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. 输入描述: 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. 输出描述: 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. 输入例子: 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 输出例子: 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

    解题思路

    1.用a[i]来保存原来的序列,其中i和a[i]都是1到54,给出一个orderb[i],题目的意思就是把1到54按照b[i]的序列来重置,我这里就用一个c[i],来保存a[i],因为要循环k次,然后根据b[i]来重置a[i] k次即可。

    代码

    #include<iostream> #include<vector> #include <map> using namespace std; int main() { vector<int> a(55),b(55),c(55); int k; cin >> k; for (int i = 1; i < 55; i++) { //a[i]为初始序列,b[i]为order a[i] = i; cin >> b[i]; } while (k--) { //保存上一个a[i] for (int i = 1; i < 55; i++) { c[i] = a[i]; } //根据order b[i]来重置a[i] for (int i = 1; i < 55; i++) { a[b[i]] = c[i]; } } //输出 for (int i = 1; i < 55; i++) { //cout << a[i] << endl; int k1, k2; k1 = a[i] / 13; k2 = a[i] % 13; if (k2 == 0) { k1--; k2 = 13; } if (i==1) { switch (k1) { case 0: cout << "S" << k2; break; case 1: cout << "H" << k2; break; case 2: cout << "C" << k2; break; case 3: cout << "D" << k2; break; case 4: cout << "J" << k2; } } else { cout << " "; switch (k1) { case 0: cout << "S" << k2; break; case 1: cout << "H" << k2; break; case 2: cout << "C" << k2; break; case 3: cout << "D" << k2; break; case 4: cout << "J" << k2; } } } cout << endl; return 0; }
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