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【作品展示】

 

【作品介绍】

哈哈

1.可以双人对战

2.可以分辨谁赢

3.缺点：转不了积木模式

请点赞加油，我就继续努力的

谢谢

 

【作品源代码】

   # 1、画棋盘，初始化棋盘
from turtle import *
screensize()
tracer(0)

for i in range(15):  # 画横向
    penup()
    goto(-280, -280+i*40)
    pendown()
    fd(560)
left(90)
for i in range(15):  # 画竖向
    penup()
    goto(-280+i*40, -280)
    pendown()
    fd(560)

penup()  # 画标点
goto(0, 0)
dot()
goto(160, 160)
dot()
goto(-160, 160)
dot()
goto(160, -160)
dot()
goto(-160, -160)
dot()

#如果要使用列表创建一个二维数组，可以使用生成器来辅助实现。
#棋盘上每一个点初始值为0
arr = [[0 for i in range(15)] for j in range(15)]


# 3、判断4个方向是否连成五个子

def fun_null(x, y):
    return


def black_is_end(i, j):
    global arr
    for m in range(11):  # 竖向
        if arr[m][j] == 1 and arr[m+1][j] == 1 and arr[m+2][j] == 1 and \
                arr[m+3][j] == 1 and arr[m+4][j] == 1:
            print("游戏结束，黑子胜利！")
            onscreenclick(fun_null)
    for m in range(11):  # 横向
        if arr[i][m] == 1 and arr[i][m+1] == 1 and arr[i][m+2] == 1 and \
                arr[i][m+3] == 1 and arr[i][m+4] == 1:
            print("游戏结束，黑子胜利！")
            onscreenclick(fun_null)
    if i <= j:  # \向
        m = j-i
        for k in range(m, 11):
            if arr[k-m][k] == 1 and arr[k-m+1][k+1] == 1 and arr[k-m+2][k+2] == 1 and \
               arr[k-m+3][k+3] == 1 and arr[k-m+4][k+4] == 1:
                print("游戏结束，黑子胜利！")
                onscreenclick(fun_null)
    if i > j:  # \向
        m = i-j
        for k in range(m, 11):
            if arr[k][k-m] == 1 and arr[k+1][k-m+1] == 1 and arr[k+2][k-m+2] == 1 and \
               arr[k+3][k-m+3] == 1 and arr[k+4][k-m+4] == 1:
                print("游戏结束，黑子胜利！")
                onscreenclick(fun_null)
    n = i+j  # /向
    if n <= 14:
        for k in range(0, n-4):
            if arr[k][n-k] == 1 and arr[k+1][n-k-1] == 1 and arr[k+2][n-k-2] == 1 and \
               arr[k+3][n-k-3] == 1 and arr[k+4][n-k-4] == 1:
                print("游戏结束，黑子胜利！")
                onscreenclick(fun_null)
    if n > 14:
        for v in range(n-14, 11):
            if arr[v][n-v] == 1 and arr[v+1][n-v-1] == 1 and arr[v+2][n-v-2] == 1 and \
               arr[v+3][n-v-3] == 1 and arr[v+4][n-v-4] == 1:
                print("游戏结束，黑子胜利！")
                onscreenclick(fun_null)


def white_is_end(i, j):
    global arr
    for m in range(11):  # 竖向
        if arr[m][j] == -1 and arr[m+1][j] == -1 and arr[m+2][j] == -1 and \
                arr[m+3][j] == -1 and arr[m+4][j] == -1:
            print("游戏结束，白子胜利！")
            onscreenclick(fun_null)
    for m in range(11):  # 横向
        if arr[i][m] == -1 and arr[i][m+1] == -1 and arr[i][m+2] == -1 and \
                arr[i][m+3] == -1 and arr[i][m+4] == -1:
            print("游戏结束，白子胜利！")
            onscreenclick(fun_null)
    if i <= j:  # \向
        m = j-i
        for k in range(m, 11):
            if arr[k-m][k] == -1 and arr[k-m+1][k+1] == -1 and arr[k-m+2][k+2] == -1 and \
               arr[k-m+3][k+3] == -1 and arr[k-m+4][k+4] == -1:
                print("游戏结束，白子胜利！")
                onscreenclick(fun_null)
    if i > j:  # \向
        m = i-j
        for k in range(m, 11):
            if arr[k][k-m] == -1 and arr[k+1][k-m+1] == -1 and arr[k+2][k-m+2] == -1 and \
               arr[k+3][k-m+3] == -1 and arr[k+4][k-m+4] == -1:
                print("游戏结束，白子胜利！")
                onscreenclick(fun_null)
    n = i+j  # /向
    if n <= 14:
        for k in range(0, n-4):
            if arr[k][n-k] == -1 and arr[k+1][n-k-1] == -1 and arr[k+2][n-k-2] == -1 and \
               arr[k+3][n-k-3] == -1 and arr[k+4][n-k-4] == -1:
                print("游戏结束，白子胜利！")
                onscreenclick(fun_null)
    if n > 14:
        for v in range(n-14, 11):
            if arr[v][n-v] == -1 and arr[v+1][n-v-1] == -1 and arr[v+2][n-v-2] == -1 and \
               arr[v+3][n-v-3] == -1 and arr[v+4][n-v-4] == -1:
                print("游戏结束，白子胜利！")
                onscreenclick(fun_null)


# 2、鼠标事件，点击落子

turn = 0  # 记录该哪方落子


def funclick(x, y):
    global turn
    global arr
    if turn == 0:  # 黑子先下1
        for i in range(15):
            Y = 280-i*40
            for j in range(15):
                X = -280+j*40  # 棋盘上的整点
                if abs(X-x) < 20 and abs(Y-y) < 20 and arr[i][j] == 0:
                    arr[i][j] = 1
                    goto(X, Y)
                    dot(20)
                    turn = 1  # 交替下子
                    black_is_end(i, j)
    if turn == 1:  # 白子-1
        for i in range(15):
            Y = 280-i*40
            for j in range(15):
                X = -280+j*40  # 棋盘上的整点
                if abs(X-x) < 20 and abs(Y-y) < 20 and arr[i][j] == 0:
                    arr[i][j] = -1
                    goto(X+10, Y)
                    pendown()
                    circle(10)
                    penup()
                    turn = 0  # 交替下子
                    white_is_end(i, j)


onscreenclick(funclick)
done()

 

【提示】

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