本文为大家分享了C语言控制台小游戏，打砖块，供大家参考，具体内容如下
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：
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#include "stdafx.h"#include<stdio.h>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：
以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持服务器之家。
原文链接：https://blog.csdn.net/heibai97/article/details/91041646
