At the end of an interview, when the candidates are going to ask the interviewers some questions, they must be very careful to avoid some commonly asked __________

A．REATE PROCEDURE sp5（IN num INT）BEGIN END

B．REATE PROCEDURE sp5（out num INT）BEGIN END

C．REATE PROCEDURE sp5（INOut num INT）BEGIN END

D．REATE PROCEDURE sp5（OUTIN num INT）BEGIN END

A.INT

B.END

C.EXT

D.CEN

A.INT

B.END

C.EXT

D.CEN

A.130402

B.1342

C.13-04-02

D.13-4-2

A.and=””

B.end=””

C.int=””

D.true=””

namespace std; int main() { int a ;int b; try { cin＞＞a＞＞b; if(b==0) throw b; cout＜＜a＜＜" cin＞＞a＞＞b; if(b==0) throw b; cout＜＜a＜＜" cin＞＞a＞＞b; if(b==0) throw b; cout＜＜a＜＜" } catch(int) { return 0;＞ A、8/5:1 6/3:2 end

B、8/5:1 end

C、8/5 6/3:2 end

D、8/5 5/0 end

A.* * * *

B.*

C.****

D.***

A.INT(交点)

B.MID(中间点)

C.CEN(中心点)

D.END(终点)

What input can cause this loop to end？ int Number, ComputerNumber; do{ printf("Please enter a number: "); scanf("%d",&Number); ComputerNumber = Number+1; printf("%d", Number); }while(Number＜=computernumber); printf("the end");＞ A、0

B、1

C、2147483648(即)

D、Nothing would cause this loop to end; it is an infinite loop.

下面给出了一个C函数，用于计算某个范围内的整数之和。可以假定begin总是小于或等于end(begin＜=end)： int summation(int begin，int end)(int result=begin； begin=begin+1； while(begin＜=end){ result=result+begin； begin=begin+1； } return result； } 重写这个函数，使之使用递归而不是迭代。