A君：一年365天，工作一天進(jìn)步1%，不休息一直工作；
B君：一年365天，工作一天進(jìn)步x%，一周休息兩天，休息天下降1%；
問(wèn)：x至少為多少時(shí)，一年后B進(jìn)步的速度超過(guò)A？
Adayupfactors=0.01
Bdaydownfactors=0.01
Adayup=pow(1+Adayupfactors,365)
print("The Adayup is : {:.2f}".format(Adayup))
def Dayup(Bdayupfactors):
Bdayup=1
for i in range(365):
if i%7 in [6,0]:
Bdayup=Bdayup(1-0.01)
else:
Bdayup=Bdayup(1+Bdayupfactors)
return Bdayup
Bdayupfactors=0.01
while Dayup(Bdayupfactors) < eval(str(Adayup)):
Bdayupfactors+=0.0001
print("Bdayupfactors:{:.4f}".format(Bdayupfactors))

代碼為本人親自敲寫，都可以運(yùn)行，轉(zhuǎn)載的朋友請(qǐng)注明出處http://www.lxweimin.com/writer#/notebooks/35294278/notes/43501316，謝謝！