PAT A1042. Shuffling Machine (20)

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

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

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.
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.

样例输入

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

样例输出

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

题意重述

假设有5张牌,A B C D E,变换序列每一个元素代表变换前相应位置的元素,在变换后的位置。例如给定一个变换序列[2 3 1 5 4],

第一轮变换后 C A B E D,第二轮变换后 B C A D E

解题思路

  1. 设置三个数组记录始态,变化量和终态,每一轮变换根据始态和变化量求出终态,并为下一轮变换更新始态
  2. 花色是求除数,牌号是求余,注意思考为什么要加上+1和-1的偏移

AC代码

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#include<iostream>
using namespace std;
const int N = 54;
char str[]={'S', 'H', 'C', 'D', 'J'};

int main()
{
int K = 0;
while(cin>>K)
{
// 始态数组
int a1[N+1]={0};
// 变化量数组
int a2[N+1]={0};
// 终态数组
int a3[N+1]={0};
for(int i=1; i<=N; i++)
{
a1[i]=i;
cin>>a2[i];
}
// K轮变换
for(int i=0; i<K; i++)
{
for(int j=1; j<=N; j++)
{
// 始态 + 变化量 -> 终态
a3[a2[j]]=a1[j];
}
for(int j=1; j<=N; j++)
{
// 更新始态
a1[j]=a3[j];
}
}
for(int i=1; i<=N; i++)
{
// 求除得花色 求余得
cout<<str[(a1[i]-1)/13]<<(a1[i]-1)%13+1;
if(i<N)
{
cout<<" ";
}
}
cout<<endl;
}
return 0;
}
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