Write the game Tic Tac Toe. This game involves two players, where each player takes a turn to place either a X or an O (depending on the player) in a 2-dimensional grid. The x and y coordinates lie within the range [0-2] (inclusive). A player (X or O) can win if they have 3 in a row, column or diagonal.
程序几乎是正确的，当“X”或“O”获胜时，程序给出了正确的输出，即在“X 获胜!”之前不打印一个空行。或'O wins'，但是，程序在绘制时在打印'Draw'之前打印一个空行，预期的输出不应该在打印'Draw'之前打印一个空行。我不知道如何修复它。
def print_matric(eri):
for j in range(0,3):
for i in range(0,3):
if eri[i][j]==-1:
print(" ",end="")
if eri[i][j]==0:
print("O",end="");
if eri[i][j]==1:
print("X",end="");
if not i==2:
print("|",end="");
print()
# prints ("_ _ _") two times only not third time
if not j==2:
print("-----",end="")
print()
def compute(eri):
if eri[0][0]== eri[0][1] and eri[0][1] == eri[0][2] and eri[0][0] != -1:
return eri[0][0]
if eri[1][0]== eri[1][1] and eri[1][1] == eri[1][2] and eri[1][0] != -1:
return eri[1][0]
if eri[2][0]== eri[2][1] and eri[2][1] == eri[2][2] and eri[2][0] != -1:
return eri[2][0]
if eri[0][0]== eri[1][0] and eri[1][0] == eri[2][0] and eri[0][0] != -1:
return eri[0][0]
if eri[0][1]== eri[1][1] and eri[1][1] == eri[2][1] and eri[0][1] != -1:
return eri[0][1]
if eri[0][2]== eri[1][2] and eri[1][2] == eri[2][2] and eri[0][2] != -1:
return eri[0][2]
if eri[0][0]== eri[1][1] and eri[1][1] == eri[2][2] and eri[0][0] != -1:
return eri[0][0]
if eri[0][2]== eri[1][1] and eri[1][1] == eri[2][0] and eri[0][2] != -1:
return eri[0][2]
return -1
def find_whether_empty(eri):
for j in range(0,3):
for i in range(0,3):
if eri[j][i]==-1:
return 1
return 0
eri = [[-1,-1,-1],[-1,-1,-1],[-1,-1,-1]]
count=0
while True:
a,b = input().split()
a=int(a)
b=int(b)
if a<0 or a>2 or b<0 or b>2:
print("\ncheck your coordinates and enter again")
continue
if not eri[a][b]==-1:
print("\nposition already occupied, please enter again")
continue
if count%2==0:
eri[a][b]=1
else:
eri[a][b]=0
count=count+1
if compute(eri)==-1:
print()
x=compute(eri)
if not x==-1:
if x==0:
print("O wins!\n")
else:
print("X wins!\n")
if not find_whether_empty(eri):
print("Draw\n")
print_matric(eri)
break
print_matric(eri)
if not x==-1:
break
预期输出:
0 0
X| |
-----
| |
-----
| |
1 1
X| |
-----
|O|
-----
| |
1 0
X|X|
-----
|O|
-----
| |
2 0
X|X|O
-----
|O|
-----
| |
0 2
X|X|O
-----
|O|
-----
X| |
0 1
X|X|O
-----
O|O|
-----
X| |
2 1
X|X|O
-----
O|O|X
-----
X| |
2 2
X|X|O
-----
O|O|X
-----
X| |O
1 2
Draw
X|X|O
-----
O|O|X
-----
X|X|O
实际输出:
0 0
X| |
-----
| |
-----
| |
1 1
X| |
-----
|O|
-----
| |
1 0
X|X|
-----
|O|
-----
| |
2 0
X|X|O
-----
|O|
-----
| |
0 2
X|X|O
-----
|O|
-----
X| |
0 1
X|X|O
-----
O|O|
-----
X| |
2 1
X|X|O
-----
O|O|X
-----
X| |
2 2
X|X|O
-----
O|O|X
-----
X| |O
1 2
Draw
X|X|O
-----
O|O|X
-----
X|X|O
请您参考如下方法:
只需删除这两行注释，并添加仅在游戏未结束时运行的print():
if count%2==0:
eri[a][b]=1
else:
eri[a][b]=0
count=count+1
# if compute(eri)==-1: # REMOVED THESE
# print() # REMOVED THESE
x=compute(eri)
if not x==-1:
if x==0:
print("O wins!\n")
else:
print("X wins!\n")
elif not find_whether_empty(eri): # CHANGED if -> elif
print("Draw\n")
print_matric(eri)
break
else: # ADDED THESE
print() # ADDED THESE
Then it will give the correct output when drawn:
...... ...... O|X| ----- X|O|O ----- X| |X 1 2 O|X| ----- X|O|O ----- X|O|X 2 0 Draw O|X|X ----- X|O|O ----- X|O|X
