def isNeedRev(gird,n): if girdRev(gird,n) == gird: return False return Truedef girdRev(gird,n): girdRevT = [] for i in range(n-1,-1,-1): girdRevT.append(gird[i][::-1]) return girdRevTlstT = input().split()ch, n = lstT[0],int(lstT[1])gird = []for i in range(n): gird.append(input())if not isNeedRev(gird,n): print("bu yong dao le")gird = girdRev(gird,n)for i in range(n): for j in range(n): print(ch if gird[i][j] != ' ' else ' ',end="") print()为这段代码加上注释
时间: 2024-03-25 16:39:18 浏览: 16
```
# 定义一个函数,判断翻转后的矩阵是否与原矩阵一致
def isNeedRev(gird, n):
# 判断矩阵是否翻转,如果一致则不需要翻转,返回 False
if girdRev(gird, n) == gird:
return False
# 否则返回 True,需要翻转
return True
# 定义一个函数,将矩阵翻转
def girdRev(gird, n):
girdRevT = []
# 从最后一行开始倒序遍历矩阵,将每一行反转后添加到 girdRevT 中
for i in range(n-1, -1, -1):
girdRevT.append(gird[i][::-1])
# 返回翻转后的矩阵
return girdRevT
# 读取用户输入的字符串列表
lstT = input().split()
# 分别获取字符和矩阵大小
ch, n = lstT[0], int(lstT[1])
# 定义一个空矩阵
gird = []
# 读取用户输入的每一行字符串,并添加到矩阵中
for i in range(n):
gird.append(input())
# 如果不需要翻转,则输出 "bu yong dao le"
if not isNeedRev(gird, n):
print("bu yong dao le")
# 否则翻转矩阵,并输出
gird = girdRev(gird, n)
for i in range(n):
for j in range(n):
# 如果矩阵中的值不为空格,则输出字符 ch,否则输出空格
print(ch if gird[i][j] != ' ' else ' ', end="")
# 每一行输出后换行
print()
```