判断位数:
i = int(intput('>>>')
if i // 10000:
print(5):
elif i // 1000:
print(4)
elif i // 100:
print(3)
elif i // 10:
print(2)
else:
print(1)
a=int(input(">>>"))
if a<0:
print("Format is wrong")
elif a<100000: ##限定5位
if a<10:
print(1)
elif a<100:
print(2)
elif a<1000:
print(3)
elif a<10000:
print(4)
else:
print(5)
else:
print("请输入一个不超过5位的数")
nnumber=input(">>>>")
length=len(nnumber)
if length>4:
print(5)
elif length>3:
print(4)
elif length>2:
print(3)
elif length>1:
print(2)
else:
print(1)
number=int(input("输入一个不超过5位的正整数:")
if a<=0 or a>=100000:
print('请输入一个不超过5位的正整数')
else:
import math
b=int(math.log10(a)+1)
print(b)
https://runestone.academy/runestone/books/published/pythonds/index.html
http://www.probabilistic-robotics.org/
https://zhuanlan.zhihu.com/p/60356696
这里想说的是,gao's
import math
dayup = math.pow((1.0 + 0.001), 365)
daydown = math.pow((1.0 - 0.001), 365)
print("向上:{:.2f}, 向下:{:.2f}.".format(dayup, daydown))
import math
dayfactor = 0.01
day_up = math.pow((1.0+dayfactor),365)
day_down = math.pow((1.0+dayfactor),365)
print("向上:{:.2f},向下:{:.2f}.".format(day_up,day_down))
pip install tqdm
from tqdm import tqdm
from time import sleep
for i in tqdm(range(1,1000)):
sleep(0.01)
顺手安利大家一个进度条绘制得库
公式:
以上公式接受一个值n,并且通过再每一次迭代中将newguess赋值给oldguess来反复猜测平方根。反复迭代20次左右返回的就是n的平方根。
def func(n):
root = n / 2
for k in range(20):
root = (1 / 2) * (root + n / root)
return root
print(func(9))