#天天向上e3.1
import math
dayup = math.pow((1.0+0.001),365) #每天提高0.001
daydown = math.pow((1.0-0.001),365) #放任0.001
print("向上:{:.2f},向下:{:.2f}.".format(dayup,daydown))
#天天向上0.005
import math
dayup = math.pow((1.0+0.005),365) #每天提高0.005
daydown = math.pow((1.0-0.005),365) #放任0.005
print("向上:{:.2f},向下:{:.2f}.".format(dayup,daydown))
#一年365天,如果好好学习时能力值相比前一天提高1%,当放任时相比前一天下降1%。
from math import pow
dayfactor = 0.01
dayup = pow((1 + dayfactor),365)
daydown = pow((1 - dayfactor),365)
print("向上:{:.2f},向下:{:.2f}.".format(dayup,daydown))
#一年365天,一周5个工作日努力,可提高1%,周末放任能力值下降1%,效果如何 ?
dayup,dayfactor=1.0, 0.01
for i in range(365):
if i%7 in [6,0]: #表示i%7=6或0
dayup = dayup *(1 -dayfactor)
else:
dayup = dayup *(1 +dayfactor)
print("向上5天向下2天的力量:{:.2f}.".format(dayup))
#对于上个实例,若要工作日努力到什么程度,一年后的水平才与每天努力1%取得效率一样
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
dayfactor = 0.01
while(dayUP(dayfactor)<37.78):
dayfactor += 0.001
print("每天的努力参数是:{:.3f}.".format(dayfactor))
#一年365天,初始水平值1,每工作一天水平增加N,不工作时水平不下降,一周连续工作4天
a = eval(input("请输入N的值:"))
dayup = 1
for i in range(365):
if i%7 in [1,2,3,4]:
dayup = dayup*(1 + a)
else:
dayup = dayup
print(round(dayup,2)) #round函数功能是用来四舍五入,保留两位小数
#一年365天,初始水平值1,每工作一天水平增加N,不工作时水平不下降,一周连续工作5天
a = eval(input("请输入N的值:"))
dayup = 1
for i in range(365):
if i%7 in [6,0]:
dayup = dayup
else:
dayup = dayup*(1 + a)
print(round(dayup,2)) #round函数功能是用来四舍五入,保留两位小数
#一年360天,初始为1,以每个月30天计算,在每个月初连续工作10天,每天水平增加\
#N,该月其他时间工作与否都不增加水平值。
a = eval(input("请输入N的值:"))
dayup = 1
for i in range(360):
if i%30 in range(10):
dayup = dayup*(1+a)
else:
dayup = dayup
print(round(dayup,2))
Examples of the upward power in python
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Origin blog.csdn.net/langezuibang/article/details/106157856
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