第1关 判断是否直角三角形
a=eval(input())
b=eval(input())
c=eval(input())
shortest=min(a,b,c)
longest=max(a,b,c)
middle=sum([a,b,c])-shortest-longest
if shortest<=0 or shortest+middle<=longest:
print('NO')
elif shortest**2+middle**2==longest**2:
print('YES')
else:
print('NO')第2关 今年多少天？
year=int(input())
if year%400==0 or year%4==0 and year%100!=0:
print('366')
else:
print('365')第3关 判断三角形并计算面积
a=float(input())
b=float(input())
c=float(input())
if a+b>c and a+c>b and b+c>a:
s=(a+b+c)/2
area=(s*(s-a)*(s-b)*(s-c))**0.50
print('YES')
print(f'{area:.2f}')
else:
print('NO')第4关 身高测算
father_h=int(input())
mother_h=int(input())
child=input()
if child=="男":
child_h=(father_h+mother_h)*1.08/2
print(int(child_h))
elif child=="女":
child_h=((father_h*0.923)+mother_h)/2
print(int(child_h))
else:
print(f'无对应公式')第5关 个税计算器
n=eval(input())
x=0
if n<0:
print("error")
else:
if(0<=n<=5000):
x=0
elif(0<n-5000<=3000):
x=(n-5000)*0.03
elif(3000<=n-5000<=12000):
x=(n-5000)*0.1-210
elif(12000<=n-5000<=25000):
x=(n-5000)*0.2-1410
elif(25000<=n-5000<=35000):
x=(n-5000)*0.25-2660
elif(35000<=n-5000<=55000):
x=(n-5000)*0.30-4410
elif(55000<=n-5000<=80000):
x=(n-5000)*0.35-7160
else:
x=(n-5000)*0.45-15160
y=n-x
print('应缴税款'f'{x:.2f}''元，实发工资'f'{y:.2f}''元。')第6关 判断闰年
year=int(input())
if year%400==0 or year%4==0 and year%100!=0:
print('True')
else:
print('False')第7关 分段函数B
import math
x=int(input())
if x>=-6 and x<0:
y=abs(x)+5
elif 0<=x<3:
y=math.factorial(x)
elif 3<=x<=6:
y=pow(x,x-2)
else:
y=0
print(y)第8关 百分制成绩转换五分制E
grade = eval(input())
if grade < 0 or grade > 100:
result = "data error!"
elif grade >= 90:
result = "A"
elif grade >= 80:
result = "B"
elif grade >= 70:
result = "C"
elif grade >= 60:
result = "D"
else:
result = "E"
print(result)第9关 正负交错数列前n项和
n = int(input())
result = 1
sign = -1
pre,cur = 1,1
for i in range(1,n):
pre,cur = cur,pre + cur
result = result + sign * i / cur
sign = -sign
print('{:.6f}'.format(result))第10关 求数列前n项的平方和
num = int(input())
sum = 0
for i in range(1, num+1):
sum = sum + i ** 2
print(sum)
第11关 百钱买百鸡A
a,b,c = 0,0,0
count = 0
for cock in range(1,20):
for hen in range(1,34):
chick = 100 - cock - hen
if cock * 5 + hen * 3 + chick / 3 == 100:
print('{} {} {}'.format(cock, hen, chick))第12关 用户登录C
for i in range(3):
user = input()
pw = input()
if (user == 'admin' or user == 'administrator') and (pw == '012345'):
print('登录成功')
break
else:
print('登录失败')第13关 鸡兔同笼
h, f = map(int,input().split())
if (h > 0) and (f > 0) and (f >= 2 * h) and ((f - 2 * h) % 2 == 0):
r = (f - 2 * h) // 2
c = h - r
print(f'有{c}只鸡，{r}只兔')
else:
print('Data Error!')第14关 今天是第几天
date = input()
list = date.split('/')
year = int(list[0])
month = int(list[1])
day = int(list[2])
dm = [31,28,31,30,31,30,31,31,30,31,30,31]
if year % 400 == 0 or (year % 4 == 0 and year % 100 !=0):
dm[1] = 29
if month == 1:
n = day
else:
n = sum(dm[0:month - 1 ]) + day
print(f'{year}年{month}月{day}日是{year}年第{n}天')第15关 中国古代数学问题——物不知数
n=int(input())
flag=0
for i in range(n+1):
if (i%3==2) and (i%5==3) and (i%7==2):
flag=1
print(i)
if flag==0:
print('No solution!')第16关 存款买房-B
"""
1.让用户输入半年度加薪的整数百分比，例如输入7表示每半年加薪7%。
2.第6个月后，按该百分比增加薪资。在第12个月、18个月之后，依此类推（只有在第6、12、18……等月份时才加薪）。
写一个程序计算需要多少个月才能攒够钱付首付款。与之前一样，假设所需的首付款百分比为0.30（30%）。
你的程序要给出以下提示并要求用户输入相应的数值：
1. 请输入总房价：total_cost
2. 请输入年薪：annual_salary
3. 请输入月存款比例：portion_saved
4. 每半年加薪比例：semi_annual_raise
测试用例
请输入总房价：1000000
请输入年薪：156800
请输入月存款比例：60
请输入加薪比例：7
输出：需要33 个月可以存够首付
"""
total_cost = float(input()) # total_cost为当前房价
annual_salary = float(input()) # 年薪
portion_saved = float(input()) / 100 # 月存款比例，输入30转为30%
semi_annual_raise = float(input()) /100 # 输入每半年加薪比例，输入7转化为7%
portion_down_payment = 0.3 # 首付比例，浮点数
###################################Begin###################################
# 补充你的代码 # 首付款
###################################Begin###################################
down_payment=total_cost*portion_down_payment
print('首付',down_payment)
current_savings = 0 # 存款金额，从0开始
number_of_months = 0
monthly_salary = annual_salary/12 #月工资
monthly_deposit = monthly_salary * portion_saved # 月存款
# 计算多少个月才能存够首付款，结果为整数，不足1月按1个月计算，即向上取整
###################################Begin###################################
# 补充你的代码
while current_savings<down_payment:
number_of_months=number_of_months + 1
current_savings=current_savings + monthly_deposit
if number_of_months %6 == 0:
monthly_salary=monthly_salary*(1+semi_annual_raise)
monthly_deposit=monthly_salary*portion_saved
###################################Begin###################################
if number_of_months % 12 == 0:
print("第{}个月月末有{:,.0f}元存款".format(number_of_months, current_savings))
print(f'需要{number_of_months}个月可以存够首付')第17关 中国古代数学问题——二鼠打洞
wall=int(input())
b_m,s_m,d,t=1,1,0,1
d_b_m,d_s_m=0,0
while wall>0:
if wall-b_m-s_m<0:
t=wall/(b_m+s_m)
wall=wall-b_m-s_m
d_b_m=d_b_m+t*b_m
d_s_m=d_s_m+t*s_m
b_m=2*b_m
s_m=s_m/2
d=d+1
print(d)
print(round(d_s_m,1),round(d_b_m,1))版权声明：本文为博主作者：不想考试的ID原创文章，版权归属原作者，如果侵权，请联系我们删除！
原文链接：https://blog.csdn.net/LQ370534427/article/details/134807282
