C语言控制台小游戏，打砖块

C语言控制台小游戏，打砖块
这个问题是我在领扣上面看到的一道困难问题，原题是这样的：

#include "stdafx.h"
#include
int a[10][10] = { { 0, 0, 1, 0, 0, 0, 0, 0, 1, 0 }, 
				  { 0, 0, 1, 1, 1, 1, 0, 1, 1, 0 },
				  { 0, 0, 0, 0, 1, 1, 0, 1, 1, 0 },
				  { 0, 1, 1, 1, 1, 1, 0, 0, 1, 0 },
				  { 0, 0, 0, 0, 0, 0, 0, 1, 1, 0 },
				  { 0, 0, 0, 0, 0, 0, 1, 1, 1, 1 },
				  { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },
				  { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },
				  { 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 },
				  { 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 } };//初始化二维数组，写成这个形状便于一目了然

void down(int a[10][10])//负责控制砖块下落的函数，使被赋值为3的砖块下落，下落到下界或值为1的方块之上
{
	int i, j;
	int m, n;
	for (i = 9; i >=0; i--)
	for (j = 0; j < 10; j++)
	if (a[i][j] == 3)
	{
		m = i;
		n = j;
		while (a[m + 1][n] != 1&&m!=9)
		{
			a[m + 1][n] = 1;
			a[m][n] = 0;
			m++;
		}
	}
}

void freshen(int a[10][10])//刷新函数，用于每次打过砖块之后，检查所有砖块的松动情况，过程大概是这样的，先将全部为1的砖块赋值为3，之后将四周与墙壁相连并且值为3的砖块赋值为·1，然后再进行一次全体砖块的循环遍历，这一次将所有与1相连接（1上下左右连接的砖块并且值为3的）的砖块赋值为1，这样的操作要做四遍，为什么要做这么多遍，这个问题留给读者体会。
{
	int i, j;
	for ( i = 0; i < 10; i++)
	for ( j = 0; j < 10; j++)
		if (a[i][j]==1)
			a[i][j] = 3;
		for (i = 0; i < 10; i++)
		{
			j = 0;
			while (a[i][j] != 0)
			{
				a[i][j] = 1;
				j++;
			}
		}
		for (i = 0; i < 10; i++)
		{
			j = 9;
			while (a[i][j] != 0)
			{
				a[i][j] = 1;
				j--;
			}
		}
		for (j = 0; j < 10; j++)
		{
			i = 0;
			while (a[i][j] != 0)
			{
				a[i][j] = 1;
				i++;
			}
		}
		for (j = 0; j < 10; j++)
		{
			i = 9;
			while (a[i][j] != 0)
			{
				a[i][j] = 1;
				i--;
			}
		}
		for (i = 0; i < 10; i++)
		for (j = 0; j < 10; j++)
		if (a[i][j] == 1)
		{
			if (a[i - 1][j] == 3)
				a[i - 1][j] = 1;
			else
			if (a[i + 1][j] == 3)
				a[i + 1][j] = 1;
			else
			if (a[i ][j-1] == 3)
				a[i ][j-1] = 1;
			else
			if (a[i ][j+1] == 3)
				a[i ][j+1] = 1;
		}
		for (i = 9; i >=0; i--)
		for (j = 9; j >=0; j--)
		if (a[i][j] == 1)
		{
			if (a[i - 1][j] == 3)
				a[i - 1][j] = 1;
			else
			if (a[i + 1][j] == 3)
				a[i + 1][j] = 1;
			else
			if (a[i][j - 1] == 3)
				a[i][j - 1] = 1;
			else
			if (a[i][j + 1] == 3)
				a[i][j + 1] = 1;
		}
		for (i = 9; i >= 0; i--)
		for (j = 9; j >= 0; j--)
		if (a[i][j] == 1)
		{
			if (a[i - 1][j] == 3)
				a[i - 1][j] = 1;
			else
			if (a[i + 1][j] == 3)
				a[i + 1][j] = 1;
			else
			if (a[i][j - 1] == 3)
				a[i][j - 1] = 1;
			else
			if (a[i][j + 1] == 3)
				a[i][j + 1] = 1;
		}
		for (i = 9; i >= 0; i--)
		for (j = 9; j >= 0; j--)
		if (a[i][j] == 1)
		{
			if (a[i - 1][j] == 3)
				a[i - 1][j] = 1;
			else
			if (a[i + 1][j] == 3)
				a[i + 1][j] = 1;
			else
			if (a[i][j - 1] == 3)
				a[i][j - 1] = 1;
			else
			if (a[i][j + 1] == 3)
				a[i][j + 1] = 1;
		}
}

void view(int a[10][10])//打印砖块函数
{
	for (int i = -1; i < 10; i++)
	{
		printf("0%d ", i);
	}
	printf("\n");
	for (int i = 0; i < 10; i++)
	{
		printf("%d:  ", i);
		for (int j = 0; j < 10; j++)
		{
			if (a[i][j] == 1)
				printf("*  ");
			else
				printf("   ");
		}
		printf("\n");
	}
}

void beat(int a[10][10],int i,int j)//打砖块函数
{
	a[i][j] = 0;
}

void main()
{
	int p,q;
	view(a);
	for (int w = 0; w < 18; w++)
	{
		printf("beat whichp?\n");
		scanf("%d", &p);
		printf("beat whichq?\n");
		scanf("%d", &q);
		beat(a, p, q);
		freshen(a);
		down(a);
		view(a);
	}
	getchar();
	return;
}

我用到的编译器是VS2013，C语言写控制台程序，大一初学C语言的同学们可以看一下这个编程思想。
最后的效果是这样的：