python的round测试

86565656 发表于 2017-4-29 10:15:18

>>> def test(x=20):
 a="1.4"+"9"*x
 for i in xrange(3,len(a)):
 print "round(%s)=%s,contains %s '9'" %(a[:i],round(float(a[:i])),(len(a[:i])-3))


>>> test()
round(1.4)=1.0,contains 0 '9'
round(1.49)=1.0,contains 1 '9'
round(1.499)=1.0,contains 2 '9'
round(1.4999)=1.0,contains 3 '9'
round(1.49999)=1.0,contains 4 '9'
round(1.499999)=1.0,contains 5 '9'
round(1.4999999)=1.0,contains 6 '9'
round(1.49999999)=1.0,contains 7 '9'
round(1.499999999)=1.0,contains 8 '9'
round(1.4999999999)=1.0,contains 9 '9'
round(1.49999999999)=1.0,contains 10 '9'
round(1.499999999999)=1.0,contains 11 '9'
round(1.4999999999999)=1.0,contains 12 '9'
round(1.49999999999999)=1.0,contains 13 '9'
round(1.499999999999999)=1.0,contains 14 '9'
round(1.4999999999999999)=2.0,contains 15 '9'
round(1.49999999999999999)=2.0,contains 16 '9'
round(1.499999999999999999)=2.0,contains 17 '9'
round(1.4999999999999999999)=2.0,contains 18 '9'
round(1.49999999999999999999)=2.0,contains 19 '9'
>>>

　　在看到python的round时想到js有三个关于取整的方法Math.round,Math.ceil还有一个没记住,于是做了一些尝试
　　还是有点意思的吧?
　　这个是编程之美里的一个题,我用来熟悉一下python的syntx,不考虑什么算法什么的,just get things done
　　子数组的最大乘积
>>> def do(x):
 return reduce(lambda x,y:x*y,x)
>>> def fun2(x):
 """x is a list"""
 temp=[]
 for i in range(1,len(x)+1):
 for j in range(len(x)+1):
 if(j<i):
 print x
 temp.append(do(x))
 return max(temp)

>>> fun2(x)

1]
[1]
1, 2]
[1, 2]
[2]
1, 2, 3]
[1, 2, 3]
[2, 3]
[3]
6
　　数组中的子数组之和的最大值
def do(x):
 return sum(x);
#改一下这个do函数,继续复用fun2(x)即可
[-10, 2, 3, 1]
>>> fun2(_)
[-10]
[-10, 2]
[2]
[-10, 2, 3]
[2, 3]
[3]
[-10, 2, 3, 1]
[2, 3, 1]
[3, 1]
[1]
6
　　求数组中的递增序列,如1,-1,2,-3,4,-5,6,-7,最长的序列是1,2,4,6
def fun4(y):
 x=y[:]
 temp=[]
 if len(x)<1:
 return x
 else:
 for i in xrange(len(x)-1):
 if x<x+1]:
 temp.append(x)
 else:
 x+1]=x
 if x-1]>x-2]:
 temp.append(x-1])
 return temp

[-1, -2, 9, 6, 10]
>>> fun4(_)
[-1, 9, 10]
>>>
　　这题目走了弯路了,一直在想用reduce,结果进死胡同了,如果是考试肯定答不出了,下面是我错误的代码
def do2(x):
 global temp
 temp=[]
 def nested(a,b):
 #global temp
 print temp
 if(a<b):
 if temp==[]:
 print "temp is []"
 temp.append(a)
 temp.append(b)
 return b
 else:
 temp.append(a)
 return a
 if len(x)>1:
 reduce(nested,x)
 else:
 temp=x[:]
 return temp

def fun3(x):
 result=[]
 for i in xrange(len(x)):
 print "current list=",x
 result.append(do2(x))
 print "haha=",result
 y=result.sort(lambda x,y:cmp(len(x),len(y)))
 print "x=",result
 return result.pop()
　　给一个N!e.g. N!=362800,N!的末尾有2个0.那N=20时，有x个0，求N!的二进制表示中最低位1的位置
>>> def fun5(x):
 temp=str(reduce(lambda a,b:a*b,range(1,x+1)))
 print temp
 for i in xrange(len(temp)-1,-1,-1):
 if temp!='0':
 return len(temp)-i-1


>>> fun5(20)
2432902008176640000
4
>>> def fun6(x):
 yy=reduce(lambda a,b:a*b,xrange(x,0,-1))
 print "yy=",yy
 zz=tobin(yy)
 print "zz=",zz
 for i in xrange(len(zz)-1,-1,-1):
 if zz=="1":
 return len(zz)-i-1


>>>
>>> fun6(5)
yy= 120
zz= 1111000
3
>>> def tobin(x):
 L=[]
 while (x/2)!=0 or (x%2)!=1:
 L.append(str(x%2))
 x=x/2
 else:
 L.append(str(x%2))
 return "".join(L[::-1])

>>>
　　使用递归和非递归的方法计算阶乘
>>> def fun10(x):
 if x>1:
 return x*fun10(x-1)
 else:
 return 1


>>> fun10(3)
6
>>>
>>> def fun9(x):
 return reduce(lambda a,b:a*b,xrange(1,x+1));

>>> fun9(3)
6
　　给一个十进制的整数N,写下从1开始到N的所有整数,然后娄一下其中1出现的次数,例如N=2,时写下1,2,这里只出现1次
　　写一个f(N),返回1到N之间出现"1"的次数,比如f(12)-5
　　满足条件下f(N)=N的最大的N是多少

Code
>>> def fun11(x):
 L=[]
 for i in xrange(1,x+1):
 print i
 L.append(str(i).count("1"))
 return "contains:"+str(sum(L))

>>> fun11(2)
1
2
'contains:1'
　　第二小题我没想出来,难道他是递减的函数吗...

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python的round测试