1.一張紙的厚度大約是0.08mm，對折多少次之后能達到珠穆朗瑪峰的高度（8848.13米）？

mountain_h = 8848.13 * 1000 
paper_h = 0.08 
times = mountain_h / paper_h
if times - int(times):
    times = int(times) + 1 
else:
    times = int(times)

print('對折', times, '次之后能達到珠穆朗瑪峰的高度')  # 110601625

2. 古典問題：有一對兔子，從出生后第3個月起每個月都生一對兔子，小兔子長到第三個月后每個月又生一對兔子，假如兔子都不死，問每個月的兔子總數(shù)為多少？

# 方法1， 用for循環(huán)
pre_1 = 1
pre_2 = 1
current = 0
for i in range(1, 13):
    if i < 3:
        current = 1
    else:
        current = pre_1 + pre_2
        pre_1, pre_2 = current, pre_1
    print(i, '月的兔子總數(shù)為', current)

# 方法2: 遞歸算法
def fibonacci(number):
    if number == 1 or number == 2:
        return 1
    else:
        return fibonacci(number - 1) + fibonacci(number - 2)


for month in range(1, 13):
    rabbits = fibonacci(month)
    print(month, '月的兔子總數(shù)為', rabbits)


# 方法3，用列表
rabbits = [1, 1]
for i in range(12):
    if i > 1:
        current = rabbits[i - 1] + rabbits[i - 2]
        rabbits.append(current)
    print(i, '月的兔子總數(shù)為', rabbits[i])

"""
1 月的兔子總數(shù)為 1
2 月的兔子總數(shù)為 1
3 月的兔子總數(shù)為 2
4 月的兔子總數(shù)為 3
5 月的兔子總數(shù)為 5
6 月的兔子總數(shù)為 8
7 月的兔子總數(shù)為 13
8 月的兔子總數(shù)為 21
9 月的兔子總數(shù)為 34
10 月的兔子總數(shù)為 55
11 月的兔子總數(shù)為 89
12 月的兔子總數(shù)為 144
"""

3. 將一個正整數(shù)分解質(zhì)因數(shù)。例如:輸入90,打印出90=2x3x3x5。

# 方法1 循環(huán)
in_number = int(input('求Number的質(zhì)因數(shù)，Number = '))  # 輸入90


# 判斷是不是質(zhì)數(shù)
def is_prime(number_1):
    r = int(number_1 ** 0.5) + 1
    for val in range(2, r):
        if not number_1 % val:
            return False
    else:
        return True


list1 = []  # 保存分解質(zhì)因數(shù)
number_0 = in_number
save_number0_prime = is_prime(number_0)  # 判斷是不是質(zhì)數(shù)
if save_number0_prime:
    print(number_0, '沒有質(zhì)因子')
else:
    while not save_number0_prime:  # 獲取分解質(zhì)因數(shù)的過程
        for i in range(2, number_0):
            if (not number_0 % i) and is_prime(i):
                list1.append(i)
                number_0 = int(number_0 / i)
                save_number0_prime = is_prime(number_0)
                break
    else:
        list1.append(number_0)

print(in_number, '=', '*'.join([str(ls) for ls in list1]))  # 拼接輸出
# 90 = 2*3*3*5

# 方法2 遞歸
# 判斷是不是素數(shù)
def is_prime(n):
    r = int(n**0.5) + 1
    for i in range(2, r):
        if not n % i:
            return False
    else:
        return True


# 遞歸函數(shù)
def func_re_num(num, ls2):
    for i in range(2, num):
        if not num % i and is_prime(i):
            ls2.append(i)
            return func_re_num(int(num / i), ls2)  # 進行遞歸
    else:
        ls2.append(num)
        return


in_num = int(input('求N的分解質(zhì)因數(shù)，N = '))
ls1 = []

if is_prime(in_num):
    print('無法分解因數(shù)')
else:
    func_re_num(in_num, ls1)
    print(in_num, '=', '*'.join([str(ls) for ls in ls1]))

4. 輸入兩個正整數(shù)m和n，求其最大公約數(shù)和最小公倍數(shù)。 程序分析：利用輾除法。

m0 = int(input('m = '))
n0 = int(input('n = '))
m, n = m0, n0
convention = 1
common_multiple = 1

# 判斷最大公約數(shù)(輾除法)
# 方法1，循環(huán)求最大公約數(shù)
if m > n:
    while True:
        if not m % n:
            convention = n
            break
        else:
            m, n = n, m % n
else:
    while True:
        if not m % n:
            convention = n
            break
        else:
            m, n = n, m % n


# 方法2，函數(shù)遞歸
def fun_convention(a, b):
    if a % b:
        return fun_convention(b, a % b)
    else:
        return b


if m > n:
    convention = fun_convention(m, n)
else:
    convention = fun_convention(n, m)


# 判斷最大公倍數(shù)
common_multiple = int(m0 * n0 / convention)

print(convention, common_multiple)

5. 一個數(shù)如果恰好等于它的因子之和，這個數(shù)就稱為 "完數(shù) "。例如6=1＋2＋3. 編程 找出1000以內(nèi)的所有完數(shù)

sum1 = 1
for i in range(2, 1001):
    sum1 = 1
    for j in range(2, i):
        if not i % j:
            sum1 += j
    if not i - sum1:
        print(i, end=',')

print('\b')
# 6,28,496

6.輸入某年某月某日，判斷這一天是這一年的第幾天？ 程序分析：以3月5日為例，應該先把前兩個月的加起來，然后再加上5天即本年的第幾天，特殊情況，閏年且輸入月份大于3時需考慮多加一天。

while True:
    try:
        now_time = input('請輸入當前時間（格式如 1600年02月21號）：')
        now_time = now_time.replace(' ', '')  # 去掉輸入法里面的空格
        # 字符串變int型
        the_year = int(now_time[:4])
        the_month = int(now_time[5:7])
        the_day = int(now_time[8:10])
        # 判斷月份是否超出
        if not 0 < the_month < 13:
            raise Exception('the_month is out of the range', the_month)
        # 一個月里31天的月份
        months_31 = [1, 3, 5, 7, 8, 10, 12]
        months_30 = [4, 6, 9, 11]
        months_two = [28, 29]
        all_days = 0
        # 判斷是閏年的條件
        year_condition1 = not the_year % 4 and the_day % 100
        year_condition2 = not the_year % 400
        # 閏年時
        if year_condition1 or year_condition2:
            if 0 < the_month < 13:
                for i_month in range(1, the_month):
                    if i_month in months_30:
                        all_days += 30
                    elif i_month in months_31:
                        all_days += 31
                    elif i_month == 2:
                        all_days += months_two[0]
                else:
                    if the_month in months_30:
                        if not 0 < the_day < 31:
                            raise Exception('the_day out of range', the_day)
                    elif the_month in months_31:
                        if not 0 < the_day < 32:
                            raise Exception('the_day out of range', the_day)
                    elif the_month == 2:
                        if not 0 < the_day < 29:
                            raise Exception('the_day out of range', the_day)
                    all_days += the_day
        # 不是閏年
        else:
            if 0 < the_month < 13:
                for i_month in range(1, the_month):
                    if i_month in months_30:
                        all_days += 30
                    elif i_month in months_31:
                        all_days += 31
                    elif i_month == 2:
                        all_days += months_two[1]
                else:
                    if the_month in months_30:
                        if not 0 < the_day < 31:
                            raise Exception('the_day out of range', the_day)
                    elif the_month in months_31:
                        if not 0 < the_day < 32:
                            raise Exception('the_day out of range', the_day)
                    elif i_month == 2:
                        if not 0 < the_day < 90:
                            raise Exception('the_day out of range', the_day)
                    all_days += the_day
        print('今天是這一年的第', all_days, '天')
        break

    except ValueError as e:
        print('輸入格式不正確')
    except Exception as e:
        print(e)
# 》》》1600年03月21號
# 今天是這一年的第 80 天

7. 某個公司采用公用電話傳遞數(shù)據(jù)，數(shù)據(jù)是四位的整數(shù)，在傳遞過程中是加密的，加密規(guī)則如下：每位數(shù)字都加上5,然后用和除以10的余數(shù)代替該數(shù)字，再將第一位和第四位交換，第二位和第三位交換。求輸入的四位整數(shù)加密后的值

in_data = input('請輸入四位數(shù)字：')
str_list = list(in_data)
int_list = list(map(int, str_list))
index = 0
for i in int_list:
    int_list[index] = (i + 5) % 10
    index += 1
index -= 1
int_list[0], int_list[index] = int_list[index], int_list[0]
int_list[1], int_list[index-1] = int_list[index-1], int_list[1]
md_str_list = list(map(str, int_list))
md_str = ''.join(md_str_list)
md_int = int(md_str)
print(md_int)
# 》》》8152
# 7063

8. 獲取第n個丑數(shù)。 什么是丑數(shù): 因子只包含2，3，5的數(shù)

6 =1* 2 * 3 -> 丑數(shù)
2 = 12 -> 丑數(shù)
7 = 17 -> 不是丑數(shù)
1, 2, 3, 4, 5, 6, 8,9,10, 12 ….

n = int(input('獲取第n位丑叔，n = '))
nums = [2, 3, 5]
ugly_numbers = [1]
loc2 = 0
loc3 = 0
loc5 = 0
min_number = 0
len_ugly = 1
while True:
    if n == len_ugly:
        break
    number2 = ugly_numbers[loc2]*2
    number3 = ugly_numbers[loc3]*3
    number5 = ugly_numbers[loc5]*5
    min_number = min(number2, number3, number5)
    if min_number not in ugly_numbers:
        ugly_numbers.append(min_number)
        len_ugly += 1
    if number2 == min_number:
        loc2 += 1
    elif number3 == min_number:
        loc3 += 1
    elif number5 == min_number:
        loc5 += 1

# print(ugly_numbers)   # n >>> 20
# 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36
print('第', n, '位丑數(shù)的值是', ugly_numbers[n-1])
# 第20位丑數(shù)的值是36