【待复习】PAT甲级1042. Shuffling Machine (20)

第一遍做的时候为了测试，把重复覆盖次数设置成了2，结果一直AC不过。。。

花了很长时间在字符串创建上，如何把“S”和"1"一个二维字符串数组后拼接一个字符串，找了很多方法都没有搞定

只好一个一个手写

后来看到《晴神宝典》里的除法才发现这种方法确实简单高效，于是按照晴神的方法试了一遍

我的代码：

#include <bits/stdc++.h>
using namespace std;

void show() {

}

int main () {

	char *cacheArray[55];
	int commandArray[55];
	int RepeatTimes = 0;
	cin>>RepeatTimes;


	char *Array[55]
	    = {"no",
	       "S1","S2","S3","S4","S5","S6","S7","S8","S9","S10","S11","S12","S13",
	       "H1","H2","H3","H4","H5","H6","H7","H8","H9","H10","H11","H12","H13",
	       "C1","C2","C3","C4","C5","C6","C7","C8","C9","C10","C11","C12","C13",
	       "D1","D2","D3","D4","D5","D6","D7","D8","D9","D10","D11","D12","D13",
	       "J1","J2"
	      };
	//Array是原顺序，cacheArray是变化后的顺序，command是指令


//	printf("\n原:\n");
//	printf("%s",Array[1]);
//	for(int i = 2 ; i < 55 ; ++i) {
//		printf("  %d:%s",i,Array[i]);
//	}
//	printf("\n------------\n");

	for(int i = 1 ; i < 55 ; ++i) {
		cin>>commandArray[i];
	}

//  cacheArray = Array[0][0];
	memcpy(cacheArray,Array,sizeof(cacheArray));

	for(int i = 0 ; i < RepeatTimes; ++i) {

		for(int i = 1 ; i < 55 ; ++i) {
			Array[commandArray[i]]= cacheArray[i];
		}
		memcpy(cacheArray,Array,sizeof(cacheArray));

//		printf("\n第%d次翻转:\n",i+1);
//		printf("%s",Array[1]);
//		for(int i = 2 ; i < 55 ; ++i) {
//			printf("  %d:%s",i,Array[i]);
//		}
//		printf("\n------------\n");
	}

	printf("%s",Array[1]);
	for(int i = 2 ; i < 55 ; ++i) {
		//  printf("  %d:%s",i,Array[i]);
		printf(" %s",Array[i]);
	}

//  printf("\n--------\n");
//  printf("%s",cacheArray[1]);
//  for(int i = 2 ; i < 55 ; ++i) {
//    printf(" %s",cacheArray[i]);
//  }

	return 0;
}

晴神思路（修改成mencpy）：

#include <bits/stdc++.h>
using namespace std;

int main () {
	char Color[5] = {'S','H','C','D','J'}; //花色
	int Array[55],cacheArray[55],command[55];
	//Array是原顺序，cacheArray是变化后的顺序，command是指令

	for(int i = 1 ; i < 55 ; ++i) {
		Array[i] = i;
	}


//复制，下同
//	for(int i = 1 ; i < 55 ; ++i) {
//		cacheArray[i] = Array[i];
//	}
	memcpy(cacheArray,Array,sizeof(cacheArray));


	int repeatTimes = 0;
	cin>>repeatTimes;
	for(int i = 1 ; i < 55 ; ++i) {
		cin>>command[i];
	}

//	printf("\n原:\n");
//	printf("1:%c%d",Color[(Array[1]-1)/13],(Array[1]-1)%13+1);
//	for(int i = 2 ; i < 55 ; ++i) {
//		printf(" %d:%c%d",i,Color[(Array[i]-1)/13],(Array[i]-1)%13+1);
//	}
//	printf("\n------------\n");

	for(int i = 0 ; i < repeatTimes ; ++i) {
		for(int j = 1 ; j < 55; ++j ) {
			Array[command[j]] = cacheArray[j];
			//变换
		}
//		for(int i = 1 ; i < 55 ; ++i) {
//			cacheArray[i] = Array[i];
//		}
		memcpy(cacheArray,Array,sizeof(cacheArray));

//		printf("\n第%d次翻转:\n",i+1);
//		printf("1:%c%d",Color[(Array[1]-1)/13],(Array[1]-1)%13+1);
//		for(int i = 2 ; i < 55 ; ++i) {
//			printf(" %d:%c%d",i,Color[(Array[i]-1)/13],(Array[i]-1)%13+1);
//		}
//		printf("\n------------\n");
	}


	printf("%c%d",Color[(Array[1]-1)/13],(Array[1]-1)%13+1);
	for(int i = 2 ; i < 55 ; ++i) {
		printf(" %c%d",Color[(Array[i]-1)/13],(Array[i]-1)%13+1);
	}
	//很巧妙的利用了(Array[1]-1)/13，判断花色，注意i的顺序从1~54，并不是1~13
	//(Array[i]-1)%13+1)保证了不会出现0;

	return 0;
}

晴神的思路主要是

很巧妙的利用了(Array[1]-1)/13，判断花色

原题地址：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


  

   


  

 

 


  
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