前言
正文
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.
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
思路:
- 由于输出需要用花色+数字表示，且每种花色有13张牌，想到使用char数组huase[]={‘S’,‘H’,‘C’,‘D’,‘J’}来表示花色，故对54张牌从1~54编号，假设当前牌号为x，则huase[(x-1)/13]即为这张牌的花色，而(x-1)%13+1即为它所属花色下的编号
- 由于初始序列已给出，而每次的操作也已经确定，故可得到每个位置上的牌被操作后的位置。这里使用start[]，end[]数组，分别来存放操作前的牌序和操作后的牌序。即start[i]表示操作前第i个位置上的牌号是start[i]；end[i]表示操作后第i个位置上的牌号是end[i]；next[i]表示第i个位置的牌在操作后所处的位置是next[i]。然后再把数组end[]赋给start[]，给循环的下一次的操作初始化start[]
参考题解:
/*
注意找到花色和花色下的编号与牌号 之间的联系
*/
#include<cstdio>
const int N=54;
char huase[]={'S','H','C','D','J'};
int main(){
int start[N+1],next[N+1],end[N+1],times;
//初始化牌序
for(int i=1;i<=N;i++){
start[i]=i;
}
scanf("%d",×);
//读入下一次第i张牌所在的位置
for(int i=1;i<=N;i++){
scanf("%d",&next[i]);
}
for(int step=0;step<times;step++){
for(int i=1;i<=N;i++){
end[next[i]]=start[i];
}
for(int i=1;i<=N;i++){
start[i]=end[i];
}
}
for(int i=1;i<=N;i++){
if(i!=1)printf(" ");
printf("%c%d",huase[(start[i]-1)/13],(start[i]-1)%13+1);
}
return 0;
}
后记
两年前的今天是和你的第一次遇见，你的眼中，明暗交杂，一笑生花
京公网安备 11010502036488号