根据 C++ 题解思路改一下 for 循环就行
by 言琢დ @ 2023-09-12 07:53:15
我之前写过一份 python 的 AC 代码，供你参考：
```python
n = int(input())
for i in range(1, n+1):
for j in range(1, n+1):
print("%.2d" % ((i-1) * n + j), end = "")
# 第 i 行第 j 列的数的值可以用 (i-1)*n+j 表示
print()
print() # 题目描述中：中间有个空行
tot = 0
for i in range(1, n+1):
for j in range(0, n-i):
print(" ", end = " ")
# 第 1 行要打 (n-1)*2 个空格
# 第 2 行要打 (n-2)*2 个空格
# 依次类推，第 n 行要打 0 个空格
for j in range(1, i+1):
tot += 1 # 计数器
print("%.2d" % (tot), end = "")
print()
```
by 言琢დ @ 2023-09-12 07:54:12
```python
n = int(input())
temp = 1
for i in range(n):
for j in range(n):
print("{:02g}".format(temp),end='')
temp += 1
print()
print()
temp = 1
for i in range(n):
for j in range(n-i-1):
print(' '*2,end='')
for j in range(i+1):
print("{:02g}".format(temp),end='')
temp += 1
print()
```
by Winds_Land @ 2024-01-27 10:27:05