[开发测试]PAT A1042

方法一：借鉴别人的，满分通过版本

#include<stdio.h>
const int N = 54;
char mp[5] = { 'S', 'H', 'C', 'D', 'J'};
int start[N+1], end[N+1], next[N+1];
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]);//操作后牌的位置
    }
    for(int step = 0; step < K; step++)
    {
        for(int i = 1; i <=N; i++)
        {
            end[next[i]] = start[i];//i位置的牌存在next[i]
        }
        for(int i = 1; i <= N; i++)
        {
            start[i] = end[i];//非常有用的一步
        }
    }
    for(int i = 1; i<= N; i++)
    {
        if(i != 1) printf(" ");//控制格式
        start[i]--;
        printf("%c%d",mp[start[i] / 13] , start[i] % 13 + 1);
    }
    return 0;
}

方法二：输出结果与答案完全一致，但是所有测试用例均显示答案错误，可能是每次循环都输出了’\0’的缘故。

#include<cstdio>

const int CARDS = 54;
int a[60] = {0};//存放洗牌顺序 
int b[2][60] = {0};//存放每次洗牌后的位置 

char * getCard(int a){
	char c[4];
	char * result = c; 
	switch((a - 1) / 13){
		case 0:c[0] = 'S';break;
		case 1:c[0] = 'H';break;
		case 2:c[0] = 'C';break;
		case 3:c[0] = 'D';break;
		case 4:c[0] = 'J';break;
	}
	int num = (a - 1) % 13 + 1;
	if(num >= 10){
		c[1] = '1';
		c[2] = num - 10 + '0';
		c[3] = '\0';
	}else{
		c[1] =  num + '0';
		c[2] = '\0';
	}
	return result;
}
int main(){
	
	int k;
	scanf("%d",&k);
	for(int i = 1; i <= CARDS; i++){
		scanf("%d",&a[i]);
	}
	for(int i = 0;i < k; i++){
		if(i == 0){
			for(int i = 1; i <= CARDS; i++){
				b[0][a[i]] = i;
			}
		} else{
			if(i % 2 == 0){
				for(int i = 1; i <= CARDS; i++){
					b[0][a[i]] = b[1][i];
				}
			}else{
				for(int i = 1; i <= CARDS; i++){
					b[1][a[i]] = b[0][i];
				}
			}
		}
	}
	for(int i = 1; i <= CARDS; i++){
		if(i != 1){
			printf(" ");
		}
		printf("%s",getCard(b[1 - k % 2][i]));
	} 
	return 0;
}

方法三：在方法二的基础上结合方法一最后的处理，满分通过

#include<cstdio>

const int CARDS = 54;
int a[60] = {0};//存放洗牌顺序 
int b[2][60] = {0};//存放每次洗牌后的位置 
char mp[5] = { 'S', 'H', 'C', 'D', 'J'};
char * getCard(int a){
	char c[4];
	char * result = c; 
	switch((a - 1) / 13){
		case 0:c[0] = 'S';break;
		case 1:c[0] = 'H';break;
		case 2:c[0] = 'C';break;
		case 3:c[0] = 'D';break;
		case 4:c[0] = 'J';break;
	}
	int num = (a - 1) % 13 + 1;
	if(num >= 10){
		c[1] = '1';
		c[2] = num - 10 + '0';
		c[3] = '\0';
	}else{
		c[1] =  num + '0';
		c[2] = '\0';
	}
	return result;
}
int main(){
	
	int k;
	scanf("%d",&k);
	for(int i = 1; i <= CARDS; i++){
		scanf("%d",&a[i]);
	}
	for(int i = 0;i < k; i++){
		if(i == 0){
			for(int i = 1; i <= CARDS; i++){
				b[0][a[i]] = i;
			}
		} else{
			if(i % 2 == 0){
				for(int i = 1; i <= CARDS; i++){
					b[0][a[i]] = b[1][i];
				}
			}else{
				for(int i = 1; i <= CARDS; i++){
					b[1][a[i]] = b[0][i];
				}
			}
		}
	}
	for(int i = 1; i <= CARDS; i++){
		if(i != 1){
			printf(" ");
		}
		int index = b[1 - k % 2][i];
		printf("%c%d",mp[(index - 1)/13],(index - 1)%13+1);
	} 
	return 0;
}