给定层数生成三维金字塔坐标
可以与 算法2 做对比，选择更适合的
当前算法采取理性的思维，做什么事之前，都先做好准备，这样，不管到哪，都能有计划地操作
count是层数，per_length是两个点之间的距离，即单位长度
# -*- coding:utf-8 -*-
# @FileName :金字塔.py
# @DateTime :2022/4/7 21:55
# @Author :wanglb
count = 4 # 层数
per_length = 1 # 单位长度
each_side = [each * 2 + 1 for each in range(count)]
each_count = [each ** 2 for each in each_side]
all_count = sum(each_count)
over_count = []
for each in each_count:
if over_count:
over_count.append(over_count[-1] + each)
else:
over_count.append(each)
# 1, 3, 5, 7, 9
print(f'每层的边长为{each_side}')
print(f'每层的个数 {each_count}')
print(f'总数为 {all_count}')
print(f'截止到当前层数，上面总个数 {over_count}')
for each in range(all_count):
large = [e for e in over_count if e > each]
floor = over_count.index(large[0]) # 第几层
this_count = each - large[0] + each_count[floor] # 当前层第几个
this_side = each_side[floor] # 当前层的边数
print(f'{floor} 第 {this_count}', end='\t')
# 左下角到左上角，向右依次填充
x = this_count // this_side
y = this_count % this_side
if floor > 0:
# 把填充完的平移到中心位置
x -= floor
y -= floor
z = -floor
# 承伤对应的每段长度
x *= per_length
y *= per_length
z *= per_length
print(x, y, z)
解读
- 金字塔的每一层都是正方形，边长第一层是1，第二层是3，第三层是5，一次类推，为连续奇数，即
each_side - 每层的点个数分别为
1^2, 3^2, 5^2 ...，即each_count - 截止当前层，上面点的个数为
1, 1+9, 1+9+25 ...
