题目内容

为以下程序填空,使得输出结果是{ 40: 'yuwen', 20: ‘yingyu', 30: 'shuxu' }的选项是:tb = { 'yingyu' :20,'shuxue :30, 'yuwen' :40 }stb = { }for it in tb. items( ) :print( it) print( stb) }

A. stb[it[1]] = it[0]
B. stb[it[1]] = stb[i[0]]
C. sb[it[1]] = tb[it[1]]
D. stb[it[1]] = tb[it[0]]

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以下程序的输出结果是:img1 = [12,34,56,78]img2 = [1,2,3,4,5]def modi( ):img1 = img2print( img1)modi( )print( img1 )

A. [12, 34, 56, 78][1,2,3,4,5]
B. [1,2,3,4,5][1,2,3,4,5]
C. [12, 34,56, 78][12, 34, 56,78 ]
D. [1,2, 3,4, 5][12, 34, 56,78 ]

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给以下程序填空,能输出{0: [90,'Aele’], 1: [87, 'Bob'],2: [93,‘lala'l]}结果的选项是:x = [90,87,93]y = ("Aele"," Bob" ," lala" )z = { }for i in range(len(x)):print(z)

A. z[i] = [x[i],y[i]]
B. z[i]= x[i],y[i]
C. z[i] = list(zip(x, y))
D. x[i] = x,y

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ls = [‘1’,’2’,’3’],以下关于循环结构的描述,错误的是:

A. 表达式for i in range( len(ls))的循环次数与for i in ls的循环次数是一样的
B. 表达式for i in range( len(ls))与forr i in ls的循环中i的值是一样的
C. 表达式for i in range(len(ls))的循环次数与for i in range(O,len(ls))的循环次数是一样的
D. 表达式for i in range(len(ls))的循环次数与for i in range(1,len(ls) +1)的循环次数是一样的

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以下程序的输出结果是:a = divmod(5, 2)b = divmod(2, 6)c = set(a + b)print( sum(c) )

A. 3
B. 5
C. 2
D. 0

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为以下程序填空,使得输出结果是{ 40: 'yuwen', 20: ‘yingyu', 30: 'shuxu' }的选项是:tb = { 'yingyu' :20,'shuxue :30, 'yuwen' :40 }stb = { }for it in tb. items( ) :print( it) print( stb) }

A. stb[it[1]] = it[0] B. stb[it[1]] = stb[i[0]] C. sb[it[1]] = tb[it[1]] D. stb[it[1]] = tb[it[0]]

以下程序的输出结果是:img1 = [12,34,56,78]img2 = [1,2,3,4,5]def modi( ):img1 = img2print( img1)modi( )print( img1 )

给以下程序填空,能输出{0: [90,'Aele’], 1: [87, 'Bob'],2: [93,‘lala'l]}结果的选项是:x = [90,87,93]y = ("Aele"," Bob" ," lala" )z = { }for i in range(len(x)):print(z)

ls = [‘1’,’2’,’3’],以下关于循环结构的描述,错误的是:

以下程序的输出结果是:a = divmod(5, 2)b = divmod(2, 6)c = set(a + b)print( sum(c) )

A. stb[it[1]] = it[0]
B. stb[it[1]] = stb[i[0]]
C. sb[it[1]] = tb[it[1]]
D. stb[it[1]] = tb[it[0]]