《计算思维导论——一种跨学科的方法》第二章习题解答

v_out=v_in*pow(r_in/r_out,2)

from math import pi
r_cube = (d*p)/(pi*S)
r = r_cube**(1/3)

r = r/12
money = (P*r*((1+r)**N))/(((1+r)**N)-1)

y_0 = y-(14-m)/12
x = y_0+y_0/4-y_0/100+y_0/400
m_0 = m+12*((14-m)/12)-2
d_0 = (d+x+(31*m_0)/12) % 7

from math import acos, sin, cos
R = 6367000
x_1, y_1 = 49.87, -2.33
x_2, y_2 = 37.8, -122.4
d = R*acos(sin(x_1)*sin(x_2)+cos(x_1)*cos(x_2)*cos(y_1-y_2))
print(f'巴黎和旧金山两点间的大圆弧距离是：{d}m.')
# 巴黎和旧金山两点间的大圆弧距离是：5330001.193839m.

v = float(input())
def get_level(v):
    if 74<=v<=95:
        return 1
    elif 96<=v<=100:
        return 2
    elif 101<=v<=130:
        return 3
    elif 131<=v<=154:
        return 4
    elif v>=155:
        return 5
    else:
        return 0

print(f'飓风的级别是：{get_level(v)}')

card_num = list(map(int, list(input())))

def is_valid(card_num):
    nums_1 = card_num[::-2]
    sums_1 = sum(nums_1)
    nums_2 = card_num[::2]
    nums_2 = [str(2*x) for x in nums_2[::-1]]
    nums_2 = "".join(nums_2)
    sums_2 = sum(map(int, list(nums_2)))
    sums = sums_1+sums_2
    if str(sums)[-1] == 0:
        return True
    else:
        return False

if is_valid(card_num):
    print(f'信用卡号有效')
else:
    print(f'信用卡号无效')

isbn = list(map(int, list(input())))

sums = 0
for i, num in enumerate(isbn[::-1], 2):
    sums += i*num

for j in range(0, 11):
    if (j+sums) % 11 == 0:
        if j == 10:
            check_sum = 'X'
        else:
            check_sum = j
isbn.append(check_sum)
ans = ''.join(list(map(str, isbn)))
print(f'ISBN号为{ans}.')

def majority(x,y,z):
    lt=[x,y,z]
    if lt.count(True)>=2:
        return True
    else:
        return False

参考链接：LeetCode-Python-273. 整数转换英文表示

num = input()

def get_english(num):
    n = int(num)
    if n < 100:  # 两位数的情况
        return subhelper_1(num)
    else:  # 三位数的情况
        return subhelper_2(num)

def subhelper_1(num):
    n = int(num)
    l1 = ["Zero", "One", "Two", "Three", "Four",
          "Five", "Six", "Seven", "Eight", "Nine"]
    l2 = ["Ten", "Eleven", "Twelve", "Thirteen", "Fourteen",
          "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"]
    l3 = ["Twenty", "Thirty", "Forty", "Fifty",
          "Sixty", "Seventy", "Eighty", "Ninety"]
    if n < 10:
        return l1[n]
    if 10 <= n < 20:
        return l2[n-10]
    if 20 <= n < 100:
    	# 判断最后一位是否为0
        if num[1] != 0:  # 不是整十
            return l3[int(num[0])-2]+' '+l1[int(num[1])]
        else:  # 整十
            return l3[int(num[0])-2]
def subhelper_2(num):
    l1 = ["Zero", "One", "Two", "Three", "Four",
          "Five", "Six", "Seven", "Eight", "Nine"]
    if num[1:] != '00':  # 不是整百
        return l1[int(num[0])] + ' hundred '+subhelper_1(num[1:])
    else:  # 整百
        return l1[int(num[0])] + ' hundred '

print(f'{get_english(num)}')

清华大学-文泉学堂-《计算思维导论——一种跨学科的方法》

来源：CSDN

作者：huxuedan01

链接：https://blog.csdn.net/m0_37586991/article/details/104144896