A few questions

 

1. Draw n elements from a continuous array of values to find which elements are called

# Include "stdafx. H "# include <iostream> using namespace STD; int _ tmain (INT argc, _ tchar * argv []) {int A [100] = {0 }; // truncate 3, 4, 57 int B [100] = {0}; // original array, [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 [N] = {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 print the value \ n"; for (INT I = 0; I <sizeof (ARR) /sizeof (ARR [0]); ++ I) cout <arr [I] <Endl; return 0 ;}

 

 

 

Running result

 

 

 

 

2. An array a [n + 1] with an element value of [1, N] to find any repeated element.

# Include "stdafx. H "# include <iostream> using namespace STD; /*************************************** * **********************************/* int array arr [n + 1], the value range of the array element is [1, N]. The idea of finding any repeated numbers is: First FAST sort, O (nlogn), then traverse from the beginning, find the first a [I] = A [I + 1], that is, the minimum (large) of the first repeat) value element /************************************* * **********************************/INT partition (int arr [], int low, int high) {int I = low; Int J = high; int priovt = arr [I]; while (I <j) {// find the first element smaller than privot on the right and place it in the low position while (I <J & arr [J]> = priovt) -- J; if (I <j) Arr [I] = arr [J]; // find the first element on the left that is greater than privot, place it in the high position 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 
 
 
 
Running result

 
 
3. The following table lists the users who have performed more than (inclusive) operations since January 1, August 2, 2014.

 
 select * from (  select usr, count(usr) as times from tbl_users where dt>‘2014-08-02‘  group by usr ) t   where times >= 3  
 
 
 
Running result