年前面趋逝科技的一套比试题求解

比试时间是2个小时 前面的智力题和后面的网络题都没记住 只记住了几道而已 大家做做看吧
题目1：（这题我没看懂英文）
There is binary search tree which is used to store characters 'A', 'B',
'C','D','E','F','G','H',which of the following is post-order tree walk
result? (有一个二叉搜索树用来存储字符'A', 'B', 'C','D','E','F','G','H'下
面哪个(些)结果是后序树遍历结果)
A. ADBCEGFH
B. BCAGEHFD
C. BCAEFDHG
D. BDACEFHG
E. All of above

题目2：
At time 0, process A has arrived in the system, in that order; at time
30, both progress B and C have just arrived; at time 90, both progress D
and E have also arrived.The quantum or timeslice is 10 units.
（在0时刻，进程A进入系统，按照这个顺序，在30时刻，进程B和进程C也抵达；在90时刻，进程D和进程E也抵达。一个时间片是10个单元）
Process A requires 50 units of time in the CPU;
Process B requires 40 units of time in the CPU;
Process C requires 30 units of time in the CPU;
Process D requires 20 units of time in the CPU;
Process E requires 10 units of time in the CPU;
进程A需要占用CPU50个单元;
进程B需要占用CPU40个单元;
进程C需要占用CPU30个单元;
进程D需要占用CPU20个单元;
进程E需要占用CPU10个单元;
Which of the process will be the LAST to complete, if scheduling policy
is preemptive SJF (Short Job First).please describe the principle.
(如果按照短作业优先级的方法，哪个进程最后结束。请描述原理)


题目3：Function club is used to simulate guest in a club. With 0 guests
initially and 50 as max occupancy, when guests beyond limitation, they need to wait outside;when some guests leave the waiting list will decrease. The function will print out number of guests in the club and waiting outside. The function declaration as follows:
void club(int x); positive x stands for guests arrived, nagative x stands for guests left from within the club
（club函数用来模拟一个俱乐部的顾客。初始化情况下是0个顾客，俱乐部最大规模只能有50个顾客，当用户超过了最大规模，他们必须等在外面。当一些顾客离开了等待队列将减少。这个club函数将打印在俱乐部里面的顾客人数，和外面的等待人数。函数声明如下：
void club(int x);正数x代表客人来了，负数x代表客人离开了俱乐部）
For example, club (40) prints 40,0; and then club (20) prints 50,10; and
then club (-5) prints 50,5; and then club (-30) prints 25,0; and then
club (-30) prints N/A; since it is impossible input.
（举例而言，club (40)打印40，0；接着club (20)打印50，10；接着club (-5)打印50，5；接着club (-30)打印25，0；接着club (-30)打印N/A；因为这是不可能实现的。）
Please write the void club(int x) with c++;
To make sure this function works as defined, we have following set of
data to pass into the function and check the result are correct.
(请用c++编程实现club函数。为了确保函数工作正常，我们使用下列数据来测试函数是否正常，你认为该选哪个选项)

a 60
b 20 50 -10
c 40 -30
d 60 -5 -10 -10 10
e 10 -20
f 30 10 10 10 -60
g 10 10 10
h 10 -10 10

A a d e g
B c d f g
C a c d h
D b d g h
E c d e f


题目4：The following C++ code tries to count occurence of each ASCII charcater
in given string and finally print out the occurrency numbers:（下面C++代码用来统计每个ASCII字符的出现次数，最后给出出现数值）
void histogram(char* src)
{
int i;
char hist[256];
for (i=0;i<256;i++)
{
hist[i]=0;
}
while(*src!='\0')
{
hist[*src]++;
}
for(i=0;i<=256;i++)
{
printf("%d\n", hist[i]);
}
}
If there may be some issue in the code above, which line(s) would be
（如果上面代码有错，将在哪行出现）
A 1 and 3
B 3 and 7
C 9
D 5 and 7
E 4 and 8


题目5：（说实话这道题我没看懂）
Consider the grammar （如下语法）
S--> aSbS|bSaS|V (S是变量)
where V means a terminator for grammar.（V是一个语法的终止符）
which statement(s) below is(are) corrects?（下列哪种说法是正确的）
1 This grammar is not an ambiguous grammar （这个语法不是一个含糊的语法）
2 Sentence 'abba' can be constructed by only one parse tree（语句abba能被构造成一个分析树）
3 This grammar construct sentence contain any 'a' (1 and more) and any 'b' (1 and more)（这个语法构造句包含一个或多个a和1个或多个b）
4 The grammar equals to regular expression "(a|b)?(a|b)+"（这个语法等于正则表达式(a|b)?(a|b)+）
A 1 B 2 3 C 3 D 1 3 E 2 4

题目6：
struct data
{
unsigned char x;
unsigned short y;
unsigned int z;
}
已知变量var类型为struct data;程序在32位上编，不考虑对齐，请补齐下面汇编语
言，以完成相应C语言功能
var.x=0;
LEA EBX, var
XOR EAX, ______
MOV [EBX], ______
var.y = (unsigned short) (var.z+1)
MOV ECX, ______
INC ECX
MOV ______, ______
var.z = var.y-1;
MOV DX, ______
AND EDX, ______
DEC, EDX
MOV ______, EDX


题目7：下列程序中能编译通过么？如果不能该如何修改使其通过。
是定义#define VOID_P 1 还是 #defineVOID_P 0，mem才能得到正确地址。
#include <iostream>
#define VOID_P 0
#if VOID_P
void allocm(void *pout, unsigned long size)
{
pout = malloc(size);
}
#else
void allocm(void **pout, unsigned long size)
{
*pout = malloc(size);
}
#elseif

int main(int argc, char *argv[]){
void *mem = 0;
#if VOID_P
allocm(mem, 100);
#else
allocm(&mem, 100);
#endif
return 0;

}

题目8：请编C++实现：函数可变长有序数组的插入（无重复数据节点）
int *head = NULL;
int Insert = (int **pHead, int n);

//参数：*pHead 数组首地址 n 插入数值
//返回值： 0成功， 1失败