当前位置:网站首页>[C language] Three-pointed chess (classic solution + list diagram)
[C language] Three-pointed chess (classic solution + list diagram)
2022-07-31 03:19:00 【rookieﻬ°】
文章目录
什么是三子棋?
Three-piece chess is in3*3的棋盘中,Both sides carry out containment and ideas,One side reaches the connection first3A pawn is a victory;
Backgammon needs to be masteredCwhere the language is?
According to the author's process of completing three pieces of chess,Backgammon only needs to be mastered二维数组即可.
How to implement three pieces?
1.头文件(函数声明)
#include<stdio.h>
#include<time.h>
#include<stdlib.h>
#define ROW 3
#define COL 3
void menu();
void Init_board(char board[ROW][COL],int row,int col);//初始化数组为空格
void display_board(char board[ROW][COL],int row, int col);//展示数组
void Player(char board[ROW][COL], int row, int col);//玩家下棋
void Computer(char board[ROW][COL],int row,int col);//电脑下棋
char Iswin(char board[ROW][COL],int row,int col);//判断输赢
2.game.c(函数实现)
#define _CRT_SECURE_NO_WARNINGS 1
#include"game.h"
void menu()
{
printf("*****************************\n");
printf("********1.play ************\n");
printf("********0.exit ************\n");
printf("*****************************\n");
}
void Init_board(char board[ROW][COL], int row, int col)//初始化数组为空格
{
int i, j;
for (i = 0; i < row; i++)
{
for (j = 0; j < col; j++)
{
board[i][j] = ' ';
}
}
}
void display_board(char board[ROW][COL], int row, int col)//展示棋盘
{
int i, j;
for (i = 0; i < row; i++)
{
for (j = 0; j < col; j++)
{
printf(" %c ",board[i][j]);
if (j < col - 1)
printf("|");
}
printf("\n");
if (i < row - 1)
{
for (j = 0; j < col; j++)
{
printf("---");
if (j < col - 1)
printf("|");
}
printf("\n");
}
}
}
void Player(char board[ROW][COL], int row, int col)//玩家下棋
{
int x, y;
while (1)
{
printf("玩家下棋:>");
scanf("%d%d", &x, &y);
//判断合法性
if ((x >= 1 && x <= row)&& (y >= 1 && y <= col))
{
//Determine if there is a chess piece
if (board[x-1][y-1] == ' ')
{
board[x-1][y-1] = '*';
break;
}
else
{
printf("This location is already occupied,请重新输入\n");
}
}
else
{
printf("输入非法,请重新输入:>\n");
}
}
}
void Computer(char board[ROW][COL], int row, int col)//电脑下棋
{
printf("电脑下:>\n");
int x, y;
while (1)
{
x = rand() % row;
y = rand() % col;
//Determine if there is a chess piece
if (board[x][y] == ' ')
{
board[x][y] = '#';
break;
}
else
{
;
}
}
}
char Iswin(char board[ROW][COL], int row, int col)//判断输赢
{
//玩家赢
int i;
for (i = 0; i < row; i++)
{
if (board[i][0] == board[i][1] && board[i][1] == board[i][2] && board[i][1] == '*')
{
return '*';
}
if (board[0][i] == board[1][i] && board[1][i] == board[2][i] && board[0][i] == '*')
{
return '*';
}
}
if (board[0][0] == board[1][1] && board[1][1] == board[2][2] && board[2][2] == '*')
{
return '*';
}
if (board[0][2] == board[1][1] && board[1][1] == board[2][0] && board[2][0] == '*')
{
return '*';
}
//电脑赢
for (i = 0; i < row; i++)
{
if (board[i][0] == board[i][1] && board[i][1] == board[i][2] && board[i][1] == '#')
{
return '#';
}
if (board[0][i] == board[1][i] && board[1][i] == board[2][i] && board[0][i] == '#')
{
return '#';
}
}
if (board[0][0] == board[1][1] && board[1][1] == board[2][2] && board[2][2] == '#')
{
return '#';
}
if (board[0][2] == board[1][1] && board[1][1] == board[2][0] && board[2][0] == '#')
{
return '#';
}
//平局
int flag = 0;
for (i = 0; i < row; i++)
{
int j = 0;
for (j = 0; j < col; j++)
{
if (board[i][j] == ' ');
flag++;
}
}
if (flag == 0)
{
return 'Q';
}
else
return 'C';
}
3.text.c(三子棋的测试)
#define _CRT_SECURE_NO_WARNINGS 1
#include"game.h"
void game()
{
char ret;
char board[ROW][COL] = {
"0"};//创建数组
Init_board(board, ROW, COL);//初始化数组为空格
display_board(board, ROW, COL);//展示棋盘
while (1)
{
Player(board, ROW, COL);//玩家下棋
ret = Iswin(board, ROW, COL);//判断输赢
if (ret != 'C')
break;
display_board(board, ROW, COL);//打印棋盘
Computer(board, ROW, COL);//电脑下棋
ret = Iswin(board, ROW, COL);//判断输赢
if (ret != 'C')
break;
display_board(board, ROW, COL);//打印棋盘
}
if (ret == '*')
{
printf("玩家赢\n");
display_board(board, ROW, COL);//打印棋盘
}
else if (ret == '#')
{
printf("电脑赢\n");
display_board(board, ROW, COL);//打印棋盘
}
else
{
printf("平局\n");
display_board(board, ROW, COL);//打印棋盘s
}
}
int main()
{
int input = 0;
srand((unsigned int)time(NULL));//设置时间戳
do
{
menu();
printf("请选择:>");
scanf("%d", &input);
switch (input)
{
case 1:
game();
break;
case 0:
printf("游戏结束\n");
break;
default:
printf("输入错误,请重新选择:>");
break;
}
} while (input);
return 0;
}
An overview idea of chess code
The code block of the main function is the basic routine,Just know how to do it.
如下:
int main()
{
int input = 0;
srand((unsigned int)time(NULL));//设置时间戳
do
{
menu();
printf("请选择:>");
scanf("%d", &input);
switch (input)
{
case 1:
game();
break;
case 0:
printf("游戏结束\n");
break;
default:
printf("输入错误,请重新选择:>");
break;
}
} while (input);
return 0;
}
An overview of the three-piece game is as follows,即game()函数的实现:
创建二维数组
Initialize all spaces
打印棋盘
While(1)
{
玩家下棋
判断输赢
展示棋盘电脑下棋
判断输赢
展示棋盘
}
加油呀!
边栏推荐
- Mycat's master-slave relationship, vertical sub-database, horizontal sub-table, and detailed configuration of mycat fragmented table query (mysql5.7 series)
- Getting Started with CefSharp - winform
- 品牌广告投放平台的中台化应用与实践
- WebSocket Session为null
- 一份高质量的测试用例如何养成?
- C primer plus学习笔记 —— 8、结构体
- MultipartFile文件上传
- 【C语言】表达式求值的一般方法
- MultipartFile file upload
- Detailed explanation of TCP (2)
猜你喜欢

10 Permission introduction

STM32问题合集

What is a distributed lock?Three ways of implementing distributed lock

YOLOV5 study notes (2) - environment installation + operation + training

LeetCode简单题之两个数组间的距离值

Database implements distributed locks

12 Disk related commands

The distance value between two arrays of LeetCode simple questions

Getting Started with CefSharp - winform
![[Compilation principle] Lexical analysis program design principle and implementation](/img/eb/035234a402edf18beb7e2f82ebec17.png)
[Compilation principle] Lexical analysis program design principle and implementation
随机推荐
Thesis framework of the opening report
C primer plus学习笔记 —— 8、结构体
Chapter 9 SVM Practice
TCP详解(二)
Detailed explanation of TCP and UDP
11. Redis implements follow, unfollow, and follow and follower lists
【C语言】求两个整数m和n的最大公因数和最小公倍数之和一般方法,经典解法
IIR滤波器和FIR滤波器
Difference between unallocated blocks and unused blocks in database files
C#远程调试
【HCIP】ISIS
Use of QML
C# remote debugging
2022牛客多校联赛第四场 题解
C primer plus study notes - 8, structure
return in try-catch
测试中的误报和漏报同样的值得反复修正
Database implements distributed locks
[Compilation principle] Design principle and implementation of recursive descent parsing
IDEA comment report red solution