# 最簡(jiǎn)單的嵌套示例：打印5x5的星號(hào)矩陣
for i in range(5):          # 外層循環(huán)控制行數(shù)
    for j in range(5):      # 內(nèi)層循環(huán)控制每行的列數(shù)
        print("*", end=" ")  # end=" "保持同一行輸出
    print()                 # 每行結(jié)束后換行

# 遍歷二維列表
matrix = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9]
]
 
for row in matrix:          # 遍歷每一行
    for num in row:         # 遍歷行中的每個(gè)元素
        print(num, end="\t")
    print()  # 每行結(jié)束后換行

for i in range(1, 10):      # 1-9的行
    for j in range(1, i+1):  # 每行的列數(shù)等于行號(hào)
        print(f"{j}x{i}={i*j}", end="\t")
    print()

# 模擬用戶(hù)登錄驗(yàn)證（外層while控制整體流程，內(nèi)層for限制嘗試次數(shù)）
correct_pwd = "123456"
attempts = 3
 
while attempts > 0:
    print(f"您還有{attempts}次嘗試機(jī)會(huì)")
    for _ in range(1):  # 實(shí)際只需一次輸入，用for保持結(jié)構(gòu)清晰
        pwd = input("請(qǐng)輸入密碼：")
        if pwd == correct_pwd:
            print("登錄成功！")
            break
    else:  # for循環(huán)正常結(jié)束（沒(méi)有被break中斷）
        attempts -= 1
        continue
    break  # 密碼正確時(shí)跳出while循環(huán)
else:
    print("嘗試次數(shù)過(guò)多，賬戶(hù)已鎖定")

# 找出100-999之間的所有水仙花數(shù)（各位數(shù)字立方和等于本身）
for num in range(100, 1000):
    digits = []
    temp = num
    # 分解各位數(shù)字
    for _ in range(3):      # 固定3位數(shù)分解
        digits.append(temp % 10)
        temp = temp // 10
    # 檢查是否為水仙花數(shù)
    if num == digits[0]**3 + digits[1]**3 + digits[2]**3:
        print(num)

# 優(yōu)化前：內(nèi)層循環(huán)每次迭代都計(jì)算平方
for i in range(100):
    for j in range(100):
        result = i**2 + j**2  # 重復(fù)計(jì)算i的平方
 
# 優(yōu)化后：將不變計(jì)算移到外層
for i in range(100):
    i_square = i**2
    for j in range(100):
        result = i_square + j**2

# 傳統(tǒng)方式：計(jì)算兩個(gè)列表的笛卡爾積和
list1 = [1, 2, 3]
list2 = [4, 5, 6]
result = []
for x in list1:
    for y in list2:
        result.append(x + y)
 
# 優(yōu)化方式：使用生成器表達(dá)式
result = [x + y for x in list1 for y in list2]

# 查找第一個(gè)滿(mǎn)足條件的元素對(duì)
found = False
for i in range(10):
    for j in range(10):
        if i * j > 50:
            print(f"找到第一個(gè)大于50的組合：{i}x{j}={i*j}")
            found = True
            break
    if found:
        break

for i in range(10):
    for j in range(10):
        if i * j > 50:
            print(f"找到第一個(gè)大于50的組合：{i}x{j}={i*j}")
            break
    else:
        continue
    break

# 錯(cuò)誤示例：內(nèi)層循環(huán)修改了外層循環(huán)變量
count = 0
for i in range(3):
    for i in range(2):  # 內(nèi)層i覆蓋了外層i
        count += 1
print(count)  # 輸出6，但邏輯可能不符合預(yù)期

# 錯(cuò)誤示例：while嵌套for時(shí)缺少終止條件
x = 5
while x > 0:
    for i in range(10):
        if i == 5:
            x -= 1  # 只在i=5時(shí)修改x，可能導(dǎo)致意外行為
        print(i)

# 錯(cuò)誤示例：縮進(jìn)錯(cuò)誤導(dǎo)致邏輯完全改變
for i in range(3):
    for j in range(3):
    print(i, j)  # 這行代碼實(shí)際在外層循環(huán)之后執(zhí)行

# 模擬圖像灰度化處理（簡(jiǎn)化版）
image = [
    [255, 100, 50],
    [150, 200, 75],
    [80, 120, 210]
]
 
def grayscale(pixel):
    # 簡(jiǎn)單取RGB平均值作為灰度值
    return sum(pixel) // len(pixel)
 
# 使用嵌套循環(huán)處理每個(gè)像素
for row in image:
    gray_row = []
    for pixel in row:
        # 實(shí)際圖像處理會(huì)更復(fù)雜，這里簡(jiǎn)化演示
        gray_row.append(grayscale([pixel, pixel, pixel]))  # 模擬RGB轉(zhuǎn)灰度
    print(gray_row)

# 模擬銷(xiāo)售數(shù)據(jù)透 視表
sales_data = [
    {"region": "North", "product": "A", "amount": 100},
    {"region": "North", "product": "B", "amount": 150},
    {"region": "South", "product": "A", "amount": 200},
    {"region": "South", "product": "B", "amount": 50},
]
 
# 生成區(qū)域-產(chǎn)品的二維匯總表
regions = list({d["region"] for d in sales_data})
products = list({d["product"] for d in sales_data})
 
pivot_table = {}
for region in regions:
    pivot_table[region] = {}
    for product in products:
        total = 0
        for record in sales_data:
            if record["region"] == region and record["product"] == product:
                total += record["amount"]
        pivot_table[region][product] = total
 
print(pivot_table)
# 輸出：{'North': {'A': 100, 'B': 150}, 'South': {'A': 200, 'B': 50}}

# 用嵌套循環(huán)表示迷宮和尋找路徑
maze = [
    [0, 1, 0, 0],
    [0, 0, 1, 0],
    [1, 0, 0, 0],
    [0, 1, 0, 0]
]  # 0表示通路，1表示障礙
 
def find_path(maze):
    start = (0, 0)
    end = (len(maze)-1, len(maze[0])-1)
    path = [start]
    
    # 簡(jiǎn)化版：只檢查直接向右或向下移動(dòng)
    while path[-1] != end:
        x, y = path[-1]
        # 嘗試向右移動(dòng)
        if y + 1 < len(maze[0]) and maze[x][y+1] == 0:
            path.append((x, y+1))
        # 嘗試向下移動(dòng)
        elif x + 1 < len(maze) and maze[x+1][y] == 0:
            path.append((x+1, y))
        else:
            return "無(wú)路徑可達(dá)"
    return path
 
print(find_path(maze))
# 可能輸出：[(0, 0), (0, 1), (0, 2), (0, 3)]（實(shí)際取決于迷宮結(jié)構(gòu)）

# 使用itertools.product替代雙重循環(huán)
import itertools
for x, y in itertools.product(range(3), repeat=2):
    print(x, y)

# 使用NumPy進(jìn)行矩陣運(yùn)算（比嵌套循環(huán)快100倍以上）
import numpy as np
a = np.array([[1,2],[3,4]])
b = np.array([[5,6],[7,8]])
print(a * b)  # 元素級(jí)乘法

# 樹(shù)形結(jié)構(gòu)遍歷更適合遞歸
tree = {
    "value": 1,
    "children": [
        {"value": 2, "children": []},
        {"value": 3, "children": [
            {"value": 4, "children": []}
        ]}
    ]
}
 
def traverse(node):
    print(node["value"])
    for child in node["children"]:
        traverse(child)
 
traverse(tree)