We are enclosing a full set of copy documents ______this first consignment
We are enclosing a full set of copy documents ______this first consignment
set2={1,2,3}set2.add(5)=set2.remove(1)=set2.discard(1)=len(set2)=2 in set2=2 not in set2=set6=set2.copy(); set6=set2.clear();set2=
set2={1,2,3}set2.add(5)=set2.remove(1)=set2.discard(1)=len(set2)=2 in set2=2 not in set2=set6=set2.copy(); set6=set2.clear();set2=
If these trousers are too big, buy a smaller ____. A: set B: one C: copy D: pair
If these trousers are too big, buy a smaller ____. A: set B: one C: copy D: pair
复制命令COPY的快捷键是( )(1)C (2)CO (3)COP (4)COPY A: (1)C B: (2)CO C: (3)COP D: (4)COPY
复制命令COPY的快捷键是( )(1)C (2)CO (3)COP (4)COPY A: (1)C B: (2)CO C: (3)COP D: (4)COPY
She took good care of public property and ______ a good example to all of us. A: copy B: set C: take D: make
She took good care of public property and ______ a good example to all of us. A: copy B: set C: take D: make
接上题,(2)应该填写( )。 A: (rst_n==1'b0) B: (set) C: (set==1'b0) D: (!set)
接上题,(2)应该填写( )。 A: (rst_n==1'b0) B: (set) C: (set==1'b0) D: (!set)
declare @x int,@y int,@t int set @x=1 set @y=2 begin set @t=@x set @x=@y set @y=@t end print @x
declare @x int,@y int,@t int set @x=1 set @y=2 begin set @t=@x set @x=@y set @y=@t end print @x
下列语句,哪个不能创建一个集合? A: set( ) B: (1, 2, 3) C: set((1, 2, 3)) D: {1, 2, 3}
下列语句,哪个不能创建一个集合? A: set( ) B: (1, 2, 3) C: set((1, 2, 3)) D: {1, 2, 3}
If the CIF price of a product is USD100/set, freight charge USD10/set, insurance premium USD 10/set, commission rate 2%, the commission payment based on CIF price should be ( ) A: USD1.60/set B: USD1.63/set C: USD2.00/set D: USD2.04/set
If the CIF price of a product is USD100/set, freight charge USD10/set, insurance premium USD 10/set, commission rate 2%, the commission payment based on CIF price should be ( ) A: USD1.60/set B: USD1.63/set C: USD2.00/set D: USD2.04/set
Map对于map和filter方法使用正确的是() A: Map本身没有map和filter方法 B: constmap0=newMap().set(1,’a’).set(2,’b’).set(3,’c’);constmap1=newMap([...map0].filter(([k,v])=>k<3));//产生Map结构{1=>’a’,2=>’b’} C: newMap().set(1,’a’).set(2,’b’).set(3,’c’)..filter(([k,v])=>k<3)//产生Map结构{1=>’a’,2=>’b’} D: constmap2=newMap([...map0].map(([k,v])=>[k*2,’_’+v]));//产生Map结构{2=>’_a’,4=>’_b’,6=>’_c’}
Map对于map和filter方法使用正确的是() A: Map本身没有map和filter方法 B: constmap0=newMap().set(1,’a’).set(2,’b’).set(3,’c’);constmap1=newMap([...map0].filter(([k,v])=>k<3));//产生Map结构{1=>’a’,2=>’b’} C: newMap().set(1,’a’).set(2,’b’).set(3,’c’)..filter(([k,v])=>k<3)//产生Map结构{1=>’a’,2=>’b’} D: constmap2=newMap([...map0].map(([k,v])=>[k*2,’_’+v]));//产生Map结构{2=>’_a’,4=>’_b’,6=>’_c’}