PAT A1042 Shuffling Machine

1042. Shuffling Machine (20)

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 
#include<cstdio>
//这个S1，H1可以用从1到54的数组来表示，到时候整除取余即可 
//注意我是输出的时候进行调整，减去了1，其实可以一开始从0存储，反正一样的，从0从1，记住调整方法即可 
void print(int num);
int main(){
	int n,a[55],temp[55],loc[55];//1~54可用 
	scanf("%d",&n); 
	for(int i=1;i<55;i++){
		a[i]=i;
		scanf("%d",&loc[i]);//存储下一次的位置 
	}
	for(int i=0;i<n;i++){
		for(int j=0;j<55;j++){
			temp[loc[j]]=a[j];
		}
		//再将temp复制到a[]中？
		for(int j=0;j<55;j++){
			a[j]=temp[j];
			
		} 
	}
	for(int i=1;i<55;i++){
		print(a[i]);
		printf("%c",i==54?'\n':' ');
	}
	return 0;
}

void print(int num){
	int H=(num-1)/13;
	int R=(num-1)%13+1;//不同于num%13 
	switch(H){
		case 0:printf("S");break;
		case 1:printf("H");break;
		case 2:printf("C");break;
		case 3:printf("D");break;
		case 4:printf("J");break;
	}
	printf("%d",R);
}

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