Q3:工作日每天努力进步1%,周末休息退步1%,一年后,输出结果:4.63

 

# coding=utf-8
dayup = 1.0
dayfactor = 0.01
for i in range(365):
    if i % 7 in [6, 0]:
        dayup = dayup*(1-dayfactor)
    else: dayup = dayup*(1+dayfactor)

print("工作日的力量:{:.2f}".format(dayup))

 

 

Q4:只在工作日努力,休息日退步1%,需要每天努力多少,才能追赶上每天进步1%连续努力365天每的同学

 

# coding=utf-8
def dayUP(df):
    dayup = 1
    for i in range(365):
        if i % 7 in [6, 0]:
            dayup = dayup * (1 - 0.01)
        else:
            dayup = dayup * (1 + df)
    return dayup
df = 0.01
i = 0
while dayUP(df) < 37.78:
    i += 1
    df += 0.001
print("工作日的努力参数是:{:.4f}".format(df))
print("循环的次数是:{}".format(i))

输出结果:

D:\pytest\venv\Scripts\python.exe D:/pytest/SongTian/3.2ddupQ4.py
工作日的努力参数是:0.0190
循环的次数是:9

 

 

Q5:3天打鱼,2天晒网

 

def dayup(df):
    dayup = 1.0
    for i in range(365):
        if i % 5 in [4, 0]:
            dayup = dayup * (1-0.01)
        else:
            dayup = dayup * (1+df)
    return dayup

df = 0.01
while (dayup(df)<37.78):
    df += 0.001

print("{}".format(df))

 

posted on 2021-08-08 01:01  讲道理好嘛  阅读(56)  评论(0编辑  收藏  举报