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今天小编给大家分享一下C语言如何实现控制台打砖块小游戏的相关知识点，内容详细，逻辑清晰，相信大部分人都还太了解这方面的知识，所以分享这篇文章给大家参考一下，希望大家阅读完这篇文章后有所收获，下面我们一起来了解一下吧。

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"#includeint 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])//Print brick function { 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)//Arkanoid function { 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;}

I use the compiler is VS2013, C language to write console procedures, freshman C language students can take a look at this programming idea.

The final effect is this:

The above is "C language how to achieve console Arkanoid Mini games" all the content of this article, thank you for reading! I believe everyone has a great harvest after reading this article. Xiaobian will update different knowledge for everyone every day. If you want to learn more knowledge, please pay attention to the industry information channel.

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