c多线程面试题同步_高级java多线程面试题_多线程面试题(23)

}*p;

p=0x1000000;

p+0x200=____;

(Ulong)p+0x200=____;

(char*)p+0x200=____;

希望各位达人给出答案和原因，谢谢拉

解答：假设在32位CPU上，

sizeof(long) = 4 bytes

sizeof(char *) = 4 bytes

sizeof(short int) = sizeof(short) = 2 bytes

sizeof(char) = 1 bytes由于是4字节对齐，

sizeof(struct BBB) = sizeof(*p)

= 4 + 4 + 2 + 1 + 1/*补齐*/ + 2*5 + 2/*补齐*/ = 24 bytes (经Dev-C++验证)p=0x1000000;

p+0x200=____;

= 0x1000000 + 0x200*24(Ulong)p+0x200=____;

= 0x1000000 + 0x200(char*)p+0x200=____;

= 0x1000000 + 0x200*4你可以参考一下指针运算的细节

写一段程序，找出数组中第k大小的数，输出数所在的位置。例如{2，4，3，4，7}中，第一大的数是7，位置在4。第二大、第三大的数都是4，位置在1、3随便输出哪一个均可。函数接口为：int find_orderk(const int* narry,const int n,const int k)

要求算法复杂度不能是O(n^2）

谢谢！

可以先用快速排序进行排序，其中用另外一个进行地址查找

代码如下，在VC++6.0运行通过。给分吧^-^//快速排序#include<iostream>usingnamespacestd;intPartition (int*L,intlow,int high)

{

inttemp = L[low];

intpt = L[low];while (low < high)

{

while (low < high && L[high] >= pt)

--high;

L[low] = L[high];

while (low < high && L[low] <= pt)

++low;

L[low] = temp;

}

L[low] = temp;returnlow;

}voidQSort (int*L,intlow,int high)

{

if (low < high)

{

intpl = Partition (L,low,high);QSort (L,low,pl - 1);

QSort (L,pl + 1,high);

}

}intmain ()

{

intnarry[100],addr[100];

intsum = 1,t;cout << "Input number:" << endl;

cin >> t;while (t != -1)

{

narry[sum] = t;

addr[sum - 1] = t;

sum++;cin >> t;

}sum -= 1;

QSort (narry,1,sum);for (int i = 1; i <= sum;i++)

cout << narry[i] << '/t';

cout << endl;intk;

cout << "Please input place you want:" << endl;

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