查看下面部分代码，指出若是完整正确代码时运行的结果是：......doubl...1:f1}元",a,b);......

在下列4中基本数据类型中选出不属于整数类型的一项（） A: int B: byte C: long D: doubl

C语言允许函数类型缺省定义,此时该函数返回值隐含的类型是()。 A: float型 B: int型 C: long型 D: doubl E: 型

Le prix de l"orge a doublé en un an. Un surcot risque de se ______ sur le prix de la bière. A: répercuter B: repérer C: réflexe D: réformer

有如下函数模板定义： template＜typename T1，typename T2，typename T3) T2 plus(T1 t1，T3 t3){return t1+t3；} 则以下调用正确是( )。 A: plus(3，5； B: plus＜＞(3，5； C: plus＜int＞(3，5； D: plus＜int，doubl(3，5；

for i in range(b.max_row): for j in range(b.max_column): print(b.cell(row=i,column=j).value)上面语句运行的结果是：__________。 A: 1 1 1 1 1 1 1 1 1 1 B: 1111111111 C: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 D: 出现异常

【单选题】CDMA通信的基站,假定基站A的码片序列是+1 +1 +1 -1 -1 +1 -1 -1,基站发射bit为101时,实际发射的信号是 A. +1 +1 +1 -1 -1 +1 -1 -1 +1 +1 +1 -1 -1 +1 -1 -1 –1 –1 –1 +1 +1 –1 +1 +1 B. +1 +1 +1 -1 -1 +1 -1 -1 –1 –1 –1 +1 +1 –1 +1 +1 +1 +1 +1 -1 -1 +1 -1 -1 C. +1 +1 +1 -1 -1 +1 -1 -1 +1 +1 +1 -1 -1 +1 -1 -1 +1 +1 +1 -1 -1 +1 -1 -1 D. –1 –1 –1 +1 +1 –1 +1 +1 –1 –1 –1 +1 +1 –1 +1 +1 –1 –1 –1 +1 +1 –1 +1 +1

for i in range(1,11): for j in range(1,11): b.cell(row=i,column=j).value=1 #b是一个工作表对象for i in range(1,11): for j in range(1,11): print(b.cell(row=i,column=j).value,end=" ") print()上面程序代码运行的结果是（）。 A: 1 B: 1 1 1 1 1 1 1 1 1 1 C: 1111111111 D: 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1

下列哪个码片序列不能用于CDMA通信 A: ( -1 -1 -1 +1 +1 -1 +1 +1 ) B: ( -1 -1 +1 -1 +1 +1 +1 -1 ) C: ( -1 +1 -1 +1 +1 -1 -1 -1 ) D: ( -1 +1 -1 -1 -1 -1 +1 -1 )

请仔细观察下面行列式的计算过程，如果四个等号都理解了，请选择A，否则请选择B。 | a 1 1 1 1 | |a＋4 a＋4 a＋4 a＋4 a＋4| | 1 a 1 1 1 | | 1 a 1 1 1 | | 1 1 a 1 1 |= | 1 1 a 1 1 | | 1 1 1 a 1 | | 1 1 1 a 1 | | 1 1 1 1 a | | 1 1 1 1 a | | 1 1 1 1 1 | | 1 a 1 1 1 | =(a＋4) | 1 1 a 1 1 | | 1 1 1 a 1 | | 1 1 1 1 a | | 1 1 1 1 1 | | 0 a–1 0 0 0 | =(a＋4) | 0 0 a–1 0 0 | | 0 0 0 a–1 0 | | 0 0 0 0 a–1 | =(a＋4)(a–1)^4