1042 Shuffling Machine

零碎时间竟然花了一天，自己的代码也没运行出来555~

先贴一下我的代码，依旧是C哈哈哈

#include
char pat[55][3],pat1[55][3];
int main(){
	int n,i,j,k;
	scanf("%d",&n);
	int seq[55];
	for(i=1;i<55;i++){
        scanf("%d",&seq[i]);		
	}
	for(i=1;i<14;i++){
		pat[i][1]='S';
		pat[i][2]=(char)i;		
	}
	for(i=14;i<27;i++){
		pat[i][1]='H';
		pat[i][2]=(char)i;		
	}
	for(i=27;i<40;i++){
		pat[i][1]='C';
		pat[i][2]=(char)i;		
	}
	for(i=40;i<53;i++){
		pat[i][1]='D';
		pat[i][2]=(char)i;		
	}
	for(i=53;i<55;i++){
		pat[i][1]='J';
		pat[i][2]=(char)i;		
	}
	for(k=0;k 
没运行出来的原因在于int型数据到char类型的转换，丢失了数据，导致结果不对，怎么解决呢？我不知道hh，如果有大佬知道还望指教哦。后来我还用了指针，依旧没有运行出来，还显得麻烦，就作罢了。
 
解决输入输出数据涉及多种类型的问题，用C++就好了呀，我来贴个别人的代码
 
#include
#include
using namespace std;
int main() {
	string arr1[] = {"","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"};
	string arr2[55];

	int repeat;
	cin >> repeat;

	int a[55];
	for (int i = 1; i <= 54; i++) {
		cin >> a[i];
	}

	for (int i = 0; i < repeat; i++) {
		//arr1重排放入arr2
		for (int i = 1; i <= 54; i++) {
			arr2[a[i]] = arr1[i];
		}
		//arr2复制到arr1
		for (int i = 1; i <= 54; i++) {
			arr1[i] = arr2[i];
		}
	}
	//输出
	for (int i = 1; i <= 54; i++) {
		cout << arr2[i];
		if (i != 54)cout << " ";
	}

	system("pause");
}
 
除了输入输出，其他的基本一样，嗯，思路一样嘛！
 
所以C语言怎么解题？晚点更~
 
C语言大神版来了
 
#include 
const int N=54;
char mp[5]={'S','H','C','D','J'}; //牌的编号与花色的对应关系
int start[N+1],end[N+1],next[N+1];//next数组存放每个位置上的牌在操作后的位置
int main(){
int K;
scanf("%d",&K);
for(int i=1;i<=N;i++){
start[i]=i;//初始化牌的编号
}
for(int i=1;i<=N;i++){
scanf("%d",&next[i]);//输入每个位置上的牌在操作后的位置
}
for(int step=0;step 
这个的话是要数学好hhh，去分析牌号与花色以及花色下的编号之间的关系啦！