三子棋(可变为n子棋)

博客介绍了简单实现三子棋的内容,涉及game.h、game.c、test.c等文件,还展示了运行结果,聚焦于信息技术领域的程序实现。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

简单实现三子棋
game.h

#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#define ROW 3
#define COL 3

game.c

#include"game.h"
void display(char board[ROW][COL],int row,int col){
	int i = 0;
	int j = 0;
	int k = 0;
	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 (k = 0; k < ROW; k++){
			printf("----");
		}
		printf("\n");
	}
	}
}//打印棋盘
void player(char board[ROW][COL], int row, int col){
	int m = 0;
	int n = 0;
	while (1){
		printf("玩家走>\n");
		scanf("%d%d", &m, &n);
		if (board[m-1][n-1] == ' '){
			if (m > 0 && m <= ROW && n > 0 && n <= COL){
				board[m-1][n-1] = '*';
				break;
			}
			else{
				printf("输入非法,请重新输入!\n");
			}
		}
	}
}
//玩家走
void computer(char board[ROW][COL], int row, int col){
	int i = 0;
	int j = 0;
	printf("电脑走>\n");
	while (1){
		i = rand() % ROW;
		j = rand() % COL;
		if (board[i][j] == ' '){
				board[i][j] = '#';
				return;
		}
	}
}

static int is_full(char board[ROW][COL],int row,int col){
	int i = 0;
	int j = 0;
	for (i = 0; i < row; i++)
	{
		for (j = 0; j <col; j++)
		{
			if (board[i][j] == ' ')
			{
				return 0;
			}
		}
	}
	return 1;
//判满
}
char cheak_win(char board[ROW][COL],int row,int col){
	int i = 0;
	int j = 0;
	for (i = 0; i < row; i++)
	{
		for (j = 0; j < col-2; j++){
			if ((board[i][j] == board[i][j+1]) && (board[i][j] == board[i][j+2]) && board[i][j] != ' ')
			{
				return board[i][j];
			}
		}
	}//横向检测
	for (j = 0; j < col; j++)
	{
		for (i = 0; i < row-2; i++){
			if ((board[i][j] == board[i+1][j]) && (board[i+1][j] == board[i+2][j]) && (board[i][j] != ' '))
			{
				return board[i][j];
			}
		}
	}//纵向检测
	j = 0;
		for (i = 0; i < row - 2; i++){
			if ((board[i][j] == board[i + 1][j + 1]) && (board[i + 1][j + 1] == board[i + 2][j + 2]) && (board[i][j] != ' '))
			{
				return board[i][j];
			}
			j--;
			if (j>col - 2){
				break;
			}
	}//左斜向下检测
		j = 0;
		for (i = row-1; i >1; i--){

			if ((board[i][j] == board[i - 1][j + 1]) && (board[i - 1][j + 1] == board[i - 2][j + 2]) && (board[i][j] != ' '))
			{
				return board[i][j];
			}
			j++;
			if (j == col-2){
				break;
			}
	}//右斜向下检测
	if (is_full(board, row, col))
	{
		return ' ';//平局
	}
	return 'q';//游戏继续
}

test.c

#include"game.h"
void menu();
void test();
void game(){
char board[ROW][COL] = { 0 };
int i = 0;
int j = 0;
for (i = 0; i < ROW; i++)
{
	for (j = 0; j < COL; j++)
	{
		board[i][j] = ' ';
	}
}
display(board, ROW, COL);//打印棋盘
char ret;
while (1){
	player(board, ROW, COL);//玩家走
	display(board, ROW, COL);//打印棋盘
	 ret = cheak_win(board, ROW, COL);//检测输赢
	 if (ret != 'q'){
		 break;
	 }
	computer(board, ROW, COL);//电脑走
	display(board, ROW, COL);//打印棋盘
	ret = cheak_win(board, ROW, COL);
	if (ret != 'q'){
		break;
	}
}
	if (ret == '*')
	{
		printf("玩家赢了!\n");
	}
	else if (ret == '#')
	{
		printf("电脑赢了!\n");
	}
	else if (ret == ' ')
	{
		printf("平局!\n");
	}
	printf("*****************************************************************\n");
	printf("*****************************************************************\n");
	printf("*****************************************************************\n");
	test();
}
void menu(){
	printf("*********************\n");
	printf("****   1 .play   ****\n");
	printf("****   0 .exite  ****\n");
	printf("*********************\n");
}
void test(){
	int a = 0;
	menu();
	while (a != 1){
		printf("请选择>");
		scanf("%d", &a);          //选择
		switch (a){
		case 1:game(); break;
		case 0:return; break;
		default:printf("无此选项,请重新输入"); break;
		}
	}
}
int main(){
	srand((unsigned int)time(NULL));
	test();
	//printf("%d", ROW);
	system("pause");
	return 0;
}

运行结果
在这里插入图片描述
在这里插入图片描述

### 回答1: C语言代码: #include <stdio.h> int main() { char board[9][9]; //盘数组 int i, j; int x, y; //行数,列数 int player; //当前玩家 //初始化盘 for (i = 0; i < 9; i++) { for (j = 0; j < 9; j++) { board[i][j] = ' '; } } //开始游戏 player = 0; while (1) { //显示盘 for (i = 0; i < 9; i++) { for (j = 0; j < 9; j++) { printf("%c ", board[i][j]); } printf("\n"); } //提示玩家输入 printf("请玩家%d输入行号:", player + 1); scanf("%d", &x); printf("请玩家%d输入列号:", player + 1); scanf("%d", &y); //设置 if (player == 0) board[x][y] = '黑'; else if (player == 1) board[x][y] = '白'; else board[x][y] = '灰'; //判断是否已经获胜 if (checkWin(board, x, y)) { printf("恭喜玩家%d获得胜利!\n", player + 1); break; } //改变玩家 if (player == 2) player = 0; else player++; } return 0; }//判断是否获胜函数 int checkWin(char board[9][9], int x, int y) { //判断横着的 int i; for (i = 0; i < 5; i++) { if (board[x][y - i] != board[x][y] || y - i < 0) break; } if (i == 5) return 1; //判断竖着的 for (i = 0; i < 5; i++) { if (board[x - i][y] != board[x][y] || x - i < 0) break; } if (i == 5) return 1; //判断斜着的 for (i = 0; i < 5; i++) { if (board[x - i][y - i] != board[x][y] || x - i < 0 || y - i < 0) break; } if (i == 5) return 1; //判断反斜着的 for (i = 0; i < 5; i++) { if (board[x - i][y + i] != board[x][y] || x - i < 0 || y + i > 8) break; } if (i == 5) return 1; return 0; }答:我没有具体的代码,但是我可以给你一些建议,你可以创建一个9x9的盘数组,判断每个玩家落的位置,然后判断是否有五连珠,有的话就获胜。 ### 回答2: 下面是一个简单的三人类游戏C代码: ```c #include<stdio.h> #define SIZE 15 // 初始化盘 void initBoard(char board[SIZE][SIZE]) { for(int i=0; i<SIZE; i++) { for(int j=0; j<SIZE; j++) { board[i][j] = ' '; } } } // 打印盘 void printBoard(char board[SIZE][SIZE]) { printf(" "); for(int i=0; i<SIZE; i++) { printf("%d ", i); } printf("\n"); for(int i=0; i<SIZE; i++) { printf("%d ", i); for(int j=0; j<SIZE; j++) { printf("%c ", board[i][j]); } printf("\n"); } } // 判断指定位置是否合法 int isValidMove(int x, int y) { return (x>=0 && x<SIZE && y>=0 && y<SIZE); } // 判断是否胜利 int isWin(char board[SIZE][SIZE], char player) { // 横向判断 for(int i=0; i<SIZE; i++) { for(int j=0; j<SIZE-4; j++) { if(board[i][j] == player && board[i][j+1] == player && board[i][j+2] == player && board[i][j+3] == player && board[i][j+4] == player) { return 1; } } } // 纵向判断 for(int i=0; i<SIZE-4; i++) { for(int j=0; j<SIZE; j++) { if(board[i][j] == player && board[i+1][j] == player && board[i+2][j] == player && board[i+3][j] == player && board[i+4][j] == player) { return 1; } } } // 左上-右下判断 for(int i=0; i<SIZE-4; i++) { for(int j=0; j<SIZE-4; j++) { if(board[i][j] == player && board[i+1][j+1] == player && board[i+2][j+2] == player && board[i+3][j+3] == player && board[i+4][j+4] == player) { return 1; } } } // 右上-左下判断 for(int i=0; i<SIZE-4; i++) { for(int j=SIZE-1; j>=4; j--) { if(board[i][j] == player && board[i+1][j-1] == player && board[i+2][j-2] == player && board[i+3][j-3] == player && board[i+4][j-4] == player) { return 1; } } } return 0; } int main() { // 初始化盘 char board[SIZE][SIZE]; initBoard(board); // 灰选择支持的一方 char support; printf("请选择支持的一方(输入B表示黑方,W表示白方):"); scanf("%c", &support); char gray = ' '; if(support == 'B') { gray = 'B'; } else if(support == 'W') { gray = 'W'; } // 游戏开始 char currentPlayer = 'B'; // 黑方先行 int x, y; while(1) { printBoard(board); printf("请%c方输入落位置(x, y):", currentPlayer); scanf("%d %d", &x, &y); if(isValidMove(x, y) && board[x][y] == ' ') { board[x][y] = currentPlayer; // 判断胜利条件 if(isWin(board, currentPlayer) || (gray == currentPlayer && isWin(board, gray))) { printf("恭喜%c方获胜!\n", currentPlayer); break; } currentPlayer = (currentPlayer == 'B') ? 'W' : 'B'; // 切换玩家 } else { printf("无效的落位置,请重新输入!\n"); } } return 0; } ``` 这段代码实现了一个基于控制台的三人类游戏。游戏初始化后,玩家可以选择支持黑方还是白方,灰方默认为中立。玩家轮流输入落位置,当任意一方连成五时,即可获胜。灰方支持的一方获胜时,灰方也会获胜。最后输出获胜方,并结束游戏。 ### 回答3: 下面是一个简单的三人类游戏C代码实现,满足规则要求: ```c #include <stdio.h> #define SIZE 10 #define BLACK 'X' #define WHITE 'O' #define GRAY '#' char board[SIZE][SIZE]; char currentPlayer = BLACK; char grayPlayer; void initializeBoard() { for (int i = 0; i < SIZE; i++) { for (int j = 0; j < SIZE; j++) { board[i][j] = ' '; } } } void printBoard() { for (int i = 0; i < SIZE; i++) { for (int j = 0; j < SIZE; j++) { printf("[%c]", board[i][j]); } printf("\n"); } } void switchPlayer() { if (currentPlayer == BLACK) { currentPlayer = WHITE; } else if (currentPlayer == WHITE) { currentPlayer = GRAY; } else { currentPlayer = BLACK; } } int checkWin(int row, int col) { int directions[4][2] = {{1, 0}, {0, 1}, {1, 1}, {1, -1}}; char player = board[row][col]; for (int i = 0; i < 4; i++) { int count = 1; for (int j = 1; j <= 4; j++) { int newRow = row + (directions[i][0] * j); int newCol = col + (directions[i][1] * j); if (newRow >= 0 && newRow < SIZE && newCol >= 0 && newCol < SIZE) { if (board[newRow][newCol] == player) { count++; } else { break; } } else { break; } } if (count == 5) { return 1; } } return 0; } int main() { initializeBoard(); printf("Please choose a player to support (X, O, or #): "); scanf("%c", &grayPlayer); while (1) { printf("\nCurrent player: %c\n", currentPlayer); printBoard(); int row, col; printf("Enter row and column (0-%d, space separated): ", SIZE-1); scanf("%d %d", &row, &col); if (row >= 0 && row < SIZE && col >= 0 && col < SIZE && board[row][col] == ' ') { board[row][col] = currentPlayer; if (checkWin(row, col)) { if (currentPlayer == grayPlayer) { printf("Gray player wins!\n"); } else { printf("Player %c wins!\n", currentPlayer); } break; } switchPlayer(); } else { printf("Invalid move. Try again.\n"); } } return 0; } ``` 这段代码使用一个二维数组 `board` 来表示游戏盘,使用字符 'X','O' 和 '#' 分别代表黑方、白方和灰方的。通过 `initializeBoard()` 函数初始化盘, `printBoard()` 函数打印当前盘状态。 游戏开始时,灰方先选择支持的一方,然后轮流输入行和列来放置。每次放置后,判断是否有玩家连成五,若有则判断是否是灰方支持的一方获胜,输出相应的胜利信息。 代码中的 `checkWin()` 函数用于检查当前位置在四个方向上是否有连续五个相同的, `switchPlayer()` 函数在每一轮结束后切换当前玩家。 请保证编译环境中包含了stdio.h头文件,便于实现代码的编译运行。
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值