1.15、C语言设计简易五子棋

实现一个20*20的简易五子棋，双方使用键盘输入落子的坐标，刷新输出当前棋局状态，并且判断有无五子连心，裁决输赢。

/*****************************************************************
*   Copyright (C) 2018 Sangfor Ltd. All rights reserved.
*   
*   文件名称：wuziqi.c
*   创 建 者：yinfei-hu
*   创建日期：2018年08月07日  星期2 14时39分14秒
*   功能描述：简易五子棋
*****************************************************************/
#include <stdio.h>
#define N 20
#define M 20
int main()
{
    int x,y,e = 0;
    char a[N][M];
    for(int i = 0;i < N;i++)
        for(int j = 0;j < M;j++)
            a[i][j] = ' ';
    for(int i = 0;i <= N;i++) printf("%2d",i);
    printf("\n");
    for(int i = 0;i < N;i++){
        printf("%2d",i+1);
        for(int j = 0;j < M;j++)
            printf("%2c",a[i][j]);
            printf("\n");
    }
    while(e < 400){
        if(e % 2 == 0){
            printf("请选手A输入红棋棋子坐标(1-20)X,Y:\n");
            scanf("%d%d",&x,&y);
            while(x > 20 || y > 20){
                printf("棋子坐标超出范围，请选手A重新输入红棋棋子坐标(1-20)X,Y:\n");
                scanf("%d%d",&x,&y);    
            }
            while(a[x-1][y-1] != ' '){
                printf("该位置已有棋子，请选手A重新输入红棋棋子坐标(1-20)X,Y:\n");
                scanf("%d%d",&x,&y);
            }   
            a[x-1][y-1] = 'O';
        }
        else {
            printf("请选手B输入黑棋棋子坐标(1-20)X,Y:\n");
            scanf("%d%d",&x,&y);
            while(x > 20 || y > 20){
                printf("棋子坐标超出范围，请选手B重新输入黑棋棋子坐标(1-20)X,Y:\n");
                scanf("%d%d",&x,&y);    
            }
            while(a[x-1][y-1] != ' '){
                printf("该位置已有棋子，请选手B重新输入黑棋棋子坐标(1-20)X,Y:\n");
                scanf("%d%d",&x,&y);
            }
            a[x-1][y-1] = 'X';
        }
        for(int i = 0;i <= N;i++) printf("%2d",i);
        printf("\n");
        for(int i = 0;i < N;i++){
            printf("%2d",i+1);
            for(int j = 0;j < M;j++)
                printf("%2c",a[i][j]);
            printf("\n");
        }
        e++;
        int q = 0;
        if(e >= 9){
            for(int i = 0;i < N;i++){
                for(int j = 0;j < M;j++){
                    if(a[i][j] == 'O'&&q == 0){
                        if(a[i][j+1] == 'O'&&a[i][j+2] == 'O'&&a[i][j+3] == 'O'&&a[i][j+4] == 'O')
                            q = 1;
                        else if(a[i+1][j] == 'O'&&a[i+2][j] == 'O'&&a[i+3][j] == 'O'&&a[i+4][j] == 'O')
                                q = 1;
                            else if(a[i+1][j+1] == 'O'&&a[i+2][j+2] == 'O'&&a[i+3][j+3] == 'O'&&a[i+4][j+4] == 'O')
                                    q = 1;
                                else if(a[i+1][j-1] == 'O'&&a[i+2][j-2] == 'O'&&a[i+3][j-3] == 'O'&&a[i+4][j-4] == 'O')
                                        q = 1;
                    }
                }
            }
        }
        if(q == 1){
            printf("选手A红棋赢了！！Game over!!!\n");
            break;
        }
        int w = 0;
        if(e >= 10){
            for(int i = 0;i < N;i++){
                for(int j = 0;j < M;j++){
                    if(a[i][j] == 'X'&&q == 0){
                        if(a[i][j+1] == 'X'&&a[i][j+2] == 'X'&&a[i][j+3] == 'X'&&a[i][j+4] == 'X')
                            w = 1;
                        else if(a[i+1][j] == 'X'&&a[i+2][j] == 'X'&&a[i+3][j] == 'X'&&a[i+4][j] == 'X')
                                w = 1;
                            else if(a[i+1][j+1] == 'X'&&a[i+2][j+2] == 'X'&&a[i+3][j+3] == 'X'&&a[i+4][j+4] == 'X')
                                    w = 1;
                                else if(a[i+1][j-1] == 'X'&&a[i+2][j-2] == 'X'&&a[i+3][j-3] == 'X'&&a[i+4][j-4] == 'X')
                                        w = 1;
                    }
                }
            }
        }
        if(w == 1){
            printf("选手B黑棋赢了！！Game over!!!\n");
            break;
        }    
    }
    return 0;
}