uva 227 - Puzzle标准用例的屏幕输出和其他AC代码一模一样，但oj反馈老是WA

Question

代码如下

#include<iostream>
#include<fstream>
using namespace std;
//#define fin cin
int main()
{
    //ifstream fin;
    //fin.open("in.txt");
    char puzzle[10][10];
    char order;
    int x = 0, y = 0;
    int num_of_puzzles = 1;
    cin.getline(puzzle[0], 7);
    while (puzzle[0][0] != 'Z')
    {
        //输入第二至五行
        for (int i = 1; i < 5; i++)
        {
            cin.getline(puzzle[i], 7);
        }

        //for循环找出空格位置，储存为（x，y）
        for (int i = 0; i < 5; i++)
        {
            for (int j = 0; j < 5; j++)
            {
                if ((puzzle[i][j] == ' ')||(puzzle[i][j] == '\0'))
                {
                    x = i;
                    y = j;
                }
            }
        }
        //逐字符输入命令
        cin >> order;
        //声明状态变量
        int status = 1;
        //模拟法执行命令
        while ((order != '0')&&(status))
        {
            status = 1;
            switch (order)
            {
            case 'A':
                if (x == 0)
                {
                    status = 0;
                    break;
                }
                else
                {
                    puzzle[x][y] = puzzle[x - 1][y];
                    puzzle[x - 1][y] = ' ';
                    x--;
                }
                break;
            case 'B':
                if (x == 4)
                {
                    status = 0;
                    break;
                }
                else
                {
                    puzzle[x][y] = puzzle[x + 1][y];
                    puzzle[x + 1][y] = ' ';
                    x++;
                }
                break;
            case 'L':
                if (y == 0)
                {
                    status = 0;
                    break;
                }
                else
                {
                    puzzle[x][y] = puzzle[x][y - 1];
                    puzzle[x][y - 1] = ' ';
                    y--;
                }
                break;
            case 'R':
                if (y == 4)
                {
                    status = 0;
                    break;
                }
                else
                {
                    puzzle[x][y] = puzzle[x][y + 1];
                    puzzle[x][y + 1] = ' ';
                    y++;
                }
                break;
            default:
                cout << "process wrong!\n";
                break;
            }
            cin >> order;
        }
        while (order != '0')
        {
            cin >> order;
        }
        //for循环输出puzzle
        if (num_of_puzzles > 1)cout << endl;
        if (status)
        {
            cout << "Puzzle #" << num_of_puzzles << ":\n";
            for (int i = 0; i < 5; i++)
            {
                for (int g = 0; g < 4; g++)
                    cout << puzzle[i][g] <<" ";
                cout << puzzle[i][4] << endl;
            }
        }
        else
        {
            cout << "Puzzle #" << num_of_puzzles << ":\n";
            cout << "This puzzle has no final configuration.\n";
        }
        //输入第一行
        num_of_puzzles++;
        cin.get();
        cin.getline(puzzle[0], 7);
    }
    //fin.close();
    return 0;
}

Answer 1

https://blog.csdn.net/qq_40738840/article/details/104174748