几个小题

#include "stdafx.h"
#include <iostream>
using namespace std;

int _tmain(int argc, _TCHAR* argv[])
{
int a[100] = {0};  //抽掉3, 4, 57
int b[100] = {0};  //原始数组, [0, 99]

for (int i = 0; i < sizeof(a) / sizeof(a[0]); ++ i)
{
if (i > 2 && i < 55)
a[i] = i + 2;
else if (i >= 55)
a[i] = i + 3;
else
a[i] = i;

b[i] = i;
}

const int N = 3;

int arr
= {0};
int nIndex = 0;
int k = 0;
int j = 0;

for ( k = 0,   j = 0;
k < (sizeof(a) / sizeof(a[0]) - N) && j < sizeof(b) / sizeof(b[0]); ++ j)
{
if (a[k] != b[j])
arr[nIndex++] = b[j];
else
++ k;

if (sizeof(arr) / sizeof(arr[0]) == nIndex)
break;
}

cout<<"\n打印被抽调的值\n";
for (int i = 0; i < sizeof(arr) / sizeof(arr[0]); ++ i)
cout << arr[i] <<endl;

return 0;
}

#include "stdafx.h"
#include <iostream>
using namespace std;

/************************************************************************/
/* int型数组arr[n + 1]， 数组元素的值域为[1, n]
求任意重复的数字

思路：
先快排， O(nlogn)， 然后从头遍历，找到第一个a[i] == a[i +1]的即为第一个重复
的最小(大)值元素
/************************************************************************/

int Partition(int arr[], int low, int high)
{
int i = low;
int j = high;
int priovt = arr[i];

while(i < j)
{
//从右边找第一个小于privot的元素，将其放在low的位置
while(i < j && arr[j] >= priovt)  --j;
if(i < j)
arr[i] = arr[j];

//从左边找第一个大于privot的元素，将其放在high的位置
while(i < j && arr[i] <= priovt) ++ i;
if(i < j)
arr[j] = arr[i];
}

arr[i] = priovt;

return i;
}

void QuickSort(int arr[], int low, int high)
{
if(NULL == arr)
return;

if (low < high)
{
//分成两部分
int nPos = Partition(arr, low, high);
QuickSort(arr, low, nPos - 1);
QuickSort(arr, nPos + 1, high);
}
}

int _tmain(int argc, _TCHAR* argv[])
{
//arr有6 + 1个元素， 值域为[1, 6]
int arr[] = {3, 1, 2, 4, 5, 6, 3};
int size = sizeof(arr) / sizeof(arr[0]);

QuickSort(arr, 0, size - 1);

cout << "\n打印排序之后的值\n";
for (int i = 0; i < size; ++i)
{
cout << arr[i] << endl;
}

cout << "\n开始从头遍历，直到找到第一个重复的\n";
for (int i = 0; i < size - 1; ++ i)
if(arr[i] == arr[i + 1])
{
cout << "\n第一个重复的元素为: " << arr[i] << endl;
break;
}

cout << endl << endl;

return 0;
}

select * from (
select usr, count(usr) as times from tbl_users where dt>'2014-08-02'  group by usr ) t
where times >= 3

select usr, count(usr) as counts from tbl_users   group by usr  having count(usr)>=3

--在SQL SERVER（作者版本为2008 R2）中不支持下面的写法
select usr, count(usr) as counts from tbl_users   group by usr  having counts>=3

上面这种写法在MYSQL中是支持的