在c中简单连接四个游戏

Question

在此游戏中，您有9列6行的木板。首先，第一列和最后一列填充有交替的玩家标记。计算机将随机选择首先移动哪个玩家（人或计算机）。该玩家将其代币放入棋盘的中间7列之一。 （最左边和最右边的列已满）。然后下一位玩家做同样的事情。在任何时候，如果玩家丢下一个令牌，该令牌连续（水平，垂直或对角线）完成一组5个令牌，则该玩家将赢得比赛。

我无法使用printf()，scanf()，putc()和getc()以外的库函数。对角线检查遇到麻烦，请帮帮我这是我当前的代码

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define BOARD_ROWS 6
#define BOARD_COLS 9

void printBoard(char *board);
int takeTurn(char *board, int player, const char*);
int checkWin(char *board);
int checkfour(char* board , int,int,int,int);
int horizontalCheck(char *board);
int verticalCheck(char *board);
int diagonalCheck (char *board);

int main(int argc, char *argv[]){
    const char *PIECES = "XO";
    char board[BOARD_ROWS*BOARD_COLS];
    for(int i =0; i< BOARD_ROWS*BOARD_COLS ; i ++){
         board[i] = '.';
    }
    int turn, done=0;

    for (turn=0; turn < BOARD_ROWS*BOARD_COLS && !done; turn++){
        printBoard(board);
        while(!takeTurn(board,turn %2, PIECES))
        {
            printBoard(board);
            puts("**Column full!**\n");
        }
    done = checkWin(board);
    }
    printBoard(board);
    if (turn == BOARD_ROWS*BOARD_COLS && !done)
    {
        puts("It's a tie!");
    }

    else{
        turn--;
        printf("player %d (%c) wins!\n", turn %2 + 1, PIECES[turn%2]);
    }
    return 0;
}

void printBoard(char *board){
    int row,col;
    puts("\n *****Connect tour*****\n");
    for(row=0; row < BOARD_ROWS; row++){
        for(col =0; col <BOARD_COLS;col++){
            printf(" %c",board[BOARD_COLS * row+col]);
        }
        puts(" ");
        //puts(".......");
    }
puts(" ");
puts(" 1 2 3 4 5 6 7 8 9 \n");
}

int takeTurn(char*board, int player, const char *PIECES)
{
    int row,col = 0;
    printf("Player %d (%c): \nEnter number Coordinate:", player+1, PIECES[player]);
    while(1){
        if (1 != scanf("%d", &col) || col < 1 || col >9){
            while(getchar() != '\n');
            puts("number out of bounds! Try again.");
        }
        else{
            break;
        }
    }
    col--;
    for(row = BOARD_ROWS-1; row>= 0; row--){
        printf("here");
         if(board[BOARD_COLS * row+col] == '.'){
            board[BOARD_COLS * row+col] = PIECES[player];
            return 1;
         }
    }
    return 0;
    }
int checkWin(char *board){
    return(horizontalCheck(board)|| verticalCheck(board) || diagonalCheck(board));
}

int checkfour(char *board, int a, int b, int c , int d){
    if (board[a] == board[b] && board[b]==board[c] && board[c]==board[d]&&board[a] != '.'){
        return 1;
    }
    return 0;
}
int horizontalCheck(char *board){
    int row,col,idx;
    const int WIDTH = 1;
    for (row=0; row<BOARD_ROWS; row++){
        for(col = 0; col < BOARD_COLS-3; col++){
            idx = BOARD_COLS * row+col; 
            if(checkfour(board, idx, idx + WIDTH, idx + WIDTH *2, idx + WIDTH*3)){
                return 1;
            }
        }
    }
    return 0;
}

int verticalCheck(char*board){
    int row,col,idx;
    const int HEIGHT = 7;
    for(row = 0; row< BOARD_ROWS-3; row++){
        for(col = 0; col < BOARD_COLS;col++){
            idx = BOARD_COLS*row+col;
            if (checkfour(board, idx,idx+HEIGHT, idx+ HEIGHT*2, idx+ HEIGHT*3)){
                return 1;
            }
        }
    }
    return 0;
}

int diagonalCheck(char *board){
    int row,col,idx,Count = 0;
    const int DIAG_RGT = 6, DIAG_LFT = 8;
    for (row = 0; row < BOARD_ROWS-3; row++){
        for(col = 0; col<BOARD_COLS;col++){
            idx = BOARD_COLS*row+col;
            if (Count <=3 && checkfour(board, idx, idx+DIAG_RGT *2, idx+DIAG_RGT*2, idx + DIAG_RGT*3)){
                return 1;
            }
            else if(Count >=3 && checkfour(board,idx,idx+DIAG_RGT,idx+DIAG_RGT*2, idx+DIAG_RGT*3)){
                return 1;
            }
            Count++;
            }
            Count = 0;
        }
    return 0;
}

Answer 1

您的内部for for(col = 0; col<BOARD_COLS;col++){会一直显示到BOARD_COLS，而不是BOARD_COLS-3。在内部if中，您需要检查一个方向（DIAG_RGT）或另一方向（DIAG_LFT）是否存在这样的对角线。在第一个if中，您将使用DIAG_RGT来计算索引，在另一个if中，您将使用DIAG_LFT。