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>>>
?
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 '
>>> ?
????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時想到j(luò)s有三個關(guān)于取整的方法Math.round,Math.ceil還有一個沒記住,于是做了一些嘗試
還是有點意思的吧?
?
這個是編程之美里的一個題,我用來熟悉一下python的syntx,不考慮什么算法什么的,just get things done
子數(shù)組的最大乘積
?
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>>>
?
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
???? 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
?
數(shù)組中的子數(shù)組之和的最大值
?
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def
?do(x):
???? return ?sum(x);
# 改一下這個do函數(shù),繼續(xù)復(fù)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
???? return ?sum(x);
# 改一下這個do函數(shù),繼續(xù)復(fù)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
?
求數(shù)組中的遞增序列,如1,-1,2,-3,4,-5,6,-7,最長的序列是1,2,4,6
?
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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 ]
>>> ?
????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 ]
>>> ?
?
這題目走了彎路了,一直在想用reduce,結(jié)果進死胡同了,如果是考試肯定答不出了,下面是我錯誤的代碼
?
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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()
???? 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()
給一個N!e.g. N!=362800,N!的末尾有2個0.那N=20時，有x個0，求N!的二進制表示中最低位1的位置
?
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>>>
?
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
????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
?
?
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>>>
?
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 ])
>>> ?
????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 ])
>>> ?
?
使用遞歸和非遞歸的方法計算階乘
?
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>>>
?
def
?fun10(x):
???? if ?x > 1 :
???????? return ?x * fun10(x - 1 )
???? else :
???????? return ? 1
????
>>> ?fun10( 3 )
6
>>> ?
???? if ?x > 1 :
???????? return ?x * fun10(x - 1 )
???? else :
???????? return ? 1
????
>>> ?fun10( 3 )
6
>>> ?
<!--<br /> <br /> Code highlighting produced by Actipro CodeHighlighter (freeware)<br /> http://www.CodeHighlighter.com/<br /> <br /> -->
>>>
?
def
?fun9(x):
???? return ?reduce( lambda ?a,b:a * b,xrange( 1 ,x + 1 ));
>>> ?fun9( 3 )
6
???? return ?reduce( lambda ?a,b:a * b,xrange( 1 ,x + 1 ));
>>> ?fun9( 3 )
6
?
?給一個十進制的整數(shù)N,寫下從1開始到N的所有整數(shù),然后婁一下其中1出現(xiàn)的次數(shù),例如N=2,時寫下1,2,這里只出現(xiàn)1次
寫一個f(N),返回1到N之間出現(xiàn)"1"的次數(shù),比如f(12)-5
滿足條件下f(N)=N的最大的N是多少
?
Code
第二小題我沒想出來,難道他是遞減的函數(shù)嗎...
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