PAT A1042. Shuffling Machine

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本文介绍了一种模拟洗牌机工作的算法实现，该算法通过指定的随机顺序对一副54张的扑克牌进行洗牌，并可重复指定次数。文章提供了完整的代码示例，包括输入输出规格说明。

1042. Shuffling Machine (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

抬头看时间，又很晚了，昨日周日休息一天（其实是因为打扫卫生弄太晚了，没来得及==），这题写的复杂了，用mp【5】记录花色，%13、/13可以简单点

#include <cstdio>
#include <cstring>

int main(){
	int start[55],order[54],end[54];
	//把数字扩大一倍，在后面加一位标记，尾数1是s，2是h，3是c，4是d 
	for(int i=0;i<13;i++) start[i]=10*(i+1)+1;
	for(int i=13;i<26;i++)	start[i]=10*(i+1-13)+2;
	for(int i=26;i<39;i++)	start[i]=10*(i+1-26)+3;
	for(int i=39;i<52;i++)	start[i]=10*(i+1-39)+4;
	start[52] = 1;
	start[53]= 2;
	
	int k;
	
	scanf("%d",&k);
	for(int j=0;j<54;j++){
		scanf("%d",&order[j]);
	}
	
	//用end[i]记录按输入顺序读入的数。 
	while(k--){
		int temp;
		for(int i=0;i<54;i++){
			temp = order[i]-1;
			end[temp] = start[i];
		}
		for(int i=0;i<54;i++){
			start[i] = end[i];
		}
	}
	
	// 求余数，根据尾数判断shcd，除10恢复原来数字 
	for(int i=0;i<53;i++){
		if(start[i]<10)	printf("J%d ",start[i]);
		else if(start[i]%10==1)	printf("S%d ",start[i]/10);
		else if(start[i]%10==2)	printf("H%d ",start[i]/10);
		else if(start[i]%10==3)	printf("C%d ",start[i]/10);
		else if(start[i]%10==4)	printf("D%d ",start[i]/10);
	}
	
	//最后一个数不带空格 
			if(start[53]<10)	printf("J%d",start[53]);
		else if(start[53]%10==1)	printf("S%d",start[53]/10);
		else if(start[53]%10==2)	printf("H%d",start[53]/10);
		else if(start[53]%10==3)	printf("C%d",start[53]/10);
		else if(start[53]%10==4)	printf("D%d",start[53]/10);
	return 0;
}