python的round测试
>>> 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[j:i]
temp.append(do(x[j:i]))
return max(temp)
>>> fun2(x)
[0]
[0, 1 ]
[ 1 ]
[0, 1 , 2 ]
[ 1 , 2 ]
[ 2 ]
[0, 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
def fun4(y):
x = y[:]
temp = []
if len(x) < 1 :
return x
else :
for i in xrange(len(x) - 1 ):
if x[i] < x[i + 1 ]:
temp.append(x[i])
else :
x[i + 1 ] = x[i]
if x[len(x) - 1 ] > x[len(x) - 2 ]:
temp.append(x[len(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[i:]
result.append(do2(x[i:]))
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[i] != ' 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[i] == " 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
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...