PAT甲级1042题（刷牌器）

1.题目介绍

2.考察点，难度

模拟现实类，字符串操作，难度易

3.解题代码

#include <iostream>
#include <string>
using namespace std;

int main(){
	int N;
	scanf("%d",&N);
	int order[54];
	int listone[54],listtwo[54]={0};
	for(int i=0;i<54;i++){
		scanf("%d",&order[i]);
		listone[i]=i+1;
	}
	
	for(int i=0;i<N;i++){
		if(i%2==0){
			for(int j=0;j<54;j++){
				listtwo[order[j]-1]=listone[j];
			}
		}
		else{
			for(int j=0;j<54;j++){
				listone[order[j]-1]=listtwo[j];
			}
		}
	}
	
	string out;
	if(N%2==0){
		for(int i=0;i<54;i++){
			int pos=(listone[i]-1)/13,abs=(listone[i]-1)%13;
			switch(pos){
				case 0: out="S";out.append(to_string(abs+1));break;
				case 1: out="H";out.append(to_string(abs+1));break;
				case 2: out="C";out.append(to_string(abs+1));break;
				case 3: out="D";out.append(to_string(abs+1));break;
				default: out="J";out.append(to_string(abs+1));
			}
			if(i==53){
				cout<<out;
				break;
			}
			cout<<out+" ";
		}
	}
	else{
		for(int i=0;i<54;i++){
			int pos=(listtwo[i]-1)/13,abs=(listtwo[i]-1)%13;
			switch(pos){
				case 0: out="S";out.append(to_string(abs+1));break;
				case 1: out="H";out.append(to_string(abs+1));break;
				case 2: out="C";out.append(to_string(abs+1));break;
				case 3: out="D";out.append(to_string(abs+1));break;
				default: out="J";out.append(to_string(abs+1));
			}
			if(i==53){
				cout<<out;
				break;
			}
			cout<<out+" ";
		}
	}
	return 0;
}