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pr-meltout#45

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me1t0ut wants to merge 16 commits intoluckymark:masterfrom
me1t0ut:master
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pr-meltout#45
me1t0ut wants to merge 16 commits intoluckymark:masterfrom
me1t0ut:master

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@me1t0ut me1t0ut commented Mar 6, 2019

ヽ(○´∀`)ノ♪

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@luckymark luckymark left a comment

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代码清爽,看着真是舒服啊

flag = 0;
for (int k = 1; k <= AimNum; ++k)
{
if (i == AimX[k] && j == AimY[k])
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嵌套层次稍深了一点,可以考虑适当地提取函数

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将判断修改进了原有的CheckStatus函数中

int CheckStatus(int x, int y)
{
    if (x < 0 || x > h + 1 || y < 0 || y > w + 1)return -1;
    if (Maze[x][y] == '#')return 0;
    if (Maze[x][y] == 'o')return 1;
    else
    {
        for (int k = 1; k <= AimNum; ++k)
        {
            if (x == AimX[k] && y == AimY[k])
                return 2;
        }
    }
    return 3;
}

PrintMaze中修改为

if (i == NowX && j == NowY)printf("S");
else if (CheckStatus(i, j) == 2)printf("O");
else printf("%c", Maze[i][j]);

flag = CheckStatus(NowX + tx * 2, NowY + ty * 2);
if (flag == 2)
{
Maze[NowX + tx][NowY + ty] = ' ';
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同上,层次过深

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新增函数MoveCharacter

int MoveCharacter(int tx, int ty)
{
	int flag[2];
	flag[0] = CheckStatus(NowX + tx, NowY + ty);
	flag[1] = CheckStatus(NowX + tx * 2, NowY + ty * 2);
	if (flag[0] == 1 && flag[1] >= 2)
	{
		Maze[NowX + tx][NowY + ty] = ' ';
		Maze[NowX + tx * 2][NowY + ty * 2] = 'o';
		NowX += tx; NowY += ty;
		Score--;
	}
	else if (flag[0] >= 2)
	{
		NowX += tx;
		NowY += ty;
		Score--;
	}
	else return 1;
	return 0;
}

返回1则说明未移动,主函数中不进行刷新

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3 participants