PAT (Advanced Level) Practice
题目链接.
- 编程题 1042 Shuffling Machine (20分)
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
题意：
有 54 张牌，编号为 1 ～ 54，初始按编号从小到大排序。另外，这些牌按初始排列给定花色，即从左至右分别为 13 张 S、13 张 H、13 张 C、13 张 D、2 张 J。
接下来执行一种操作，这种操作将牌的位置改变为指定位置。需将这种操作执行 K 次，求最后的排列结果。
思路：
步骤 1： 由于题目给出的操作直接明确了每个位置上的牌在操作后的位置，因此不妨设置两个数组 start[] 与 end[]，分别用来存放执行操作前的牌序与执行操作后的排序，（即 start[i] 表示操作前第 i 个位置的牌的编号）。这样在每一次操作中就可以把数组 start[] 中的每一个位置的牌号存放到数组 end[] 的对应转换位置中，然后用数组 end[] 覆盖数组 start[] 来给下一次操作使用。这样 执行 K 轮操作后，数组 start[] 中即存放了最终的排序。
步骤 2： 由于输出需要用花色表示，且每种花色有 13 张牌，因此不妨使用 char 型数组 mp[] = {S, H, C, D, J} 来建立编号与花色的关系。例如，假设当前牌号为 x，那么 mp[(x - 1) / 13] 即为这张牌对应的花色（即 1～13 号为 ‘S’，14～26 号为 ‘H’ 等），而 (x - 1) % 13 + 1 即为它所属花色下的编号。
参考代码
#include <cstdio>
const int N = 54;
char mp[5] = {'S', 'H', 'C', 'D', 'J'}; //牌的编号与花色的对应关系
int start[N+1], end[N+1], next[N+1]; //next 数组存放每个位置上的牌在操作后的位置
int main() {
int k;
scanf ("%d", &k);
for ( int i = 1; i <= N; i++) {
start[i] = i; //初始化牌的序号
}
for ( int i = 1; i <= N; i++) {
scanf ("%d", &next[i]); //输入每个位置上的牌在操作后的位置
}
for (int step = 0; step < k; step++) { //执行 k 次操作
for( int i = 1; i <= N; i++) {
end[next[i]] = start[i]; //把第 i 个位置的牌的编号存于位置 next[i]
}
for ( int i = 1; i <= N; i++) {
start[i] = end[i]; //把 end 数组赋值给 start 数组以供下次操作使用
}
}
for (int i = 1; i <= N; i++) {
if (i != 1) printf(" "); //控制输出格式
start[i]--;
printf ("%c%d", mp[start[i] / 13], start[i] % 13 + 1); //输出结果
}
return 0;
}
tip: 1、注意编号和花色的对应关系，2、注意输出格式的控制，不允许在一行的末尾多出空格，否则会返回“格式错误”。