共计 1292 个字符，预计需要花费 4 分钟才能阅读完成。
题目粗心：
有 54 张牌，编号为 1~ 54，初始按编号从小到大排列。另外，这些牌按初始排列给定花色，即从左至右别离为 13 张 S、13 张 H、13 张 C、13 张 D、2 张 J，如下所示:
S1, S2.., S13, H1, H2,.., H13, C1, C2,..，CI3, D1, D2,., D13,J1,J2接下来执行一种操作，这种操作将牌的地位扭转为指定地位。例如有 5 张牌 S3, H5, C1,D13, J2, 而后给定操作序列 {4,2, 5,3,1}，因而把 S3 放到 4 号位、把 H5 放到 2 号位、C1 放到 5 号位、D13 放到 3 号位、J2 放到 1 号位，于是就变成了 J2, H5, D13, S3, C1。当初须要将这种操作执行 K 次，求最初的排列后果。例如下面的例子中，如果执行第二次操作，那么序列 J2, H5, D13, S3,C1 就会变成 C1, HS, S3, J2, D13.
算法思路：
题目要求模仿洗牌过程，给定初始牌组为 cards, 起始下标地位从 1 开始，给定洗牌程序 orders，和洗牌次数 K, 要求得出最终洗牌后果。
题目比较简单，就是应用 hash 的思维，咱们每次洗牌就创立一个新的数组 new_cards 暂存最终洗牌后果, 该数组的值取决于 orders 与 cards,
其对应关系为 new_cards[orders[j]] = cards[j]; 在 new_cards 数组求解结束后，将 new_cards 数组赋值给 cards 即可，每次循环执行该过程，直到完结
提交后果：
AC 代码：
#include<cstdio>
#include<string>
using namespace std;
string cards[55]={"","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"};
int main(){
int K;// 反复次数
scanf("%d",&K);
int orders[55];
for(int j=1;j<55;++j){scanf("%d",&orders[j]);
}
for(int i=0;i<K;++i){string new_cards[55];
for(int j=1;j<55;++j){new_cards[orders[j]] = cards[j];
}
// 将 new_cards 的值赋值给 cards
for(int j=1;j<55;++j){cards[j] = new_cards[j];
}
}
for(int i=1;i<55;++i){printf("%s",cards[i].c_str());
if(i<54) printf(" ");
}
return 0;
}