The source code for solving the problem of the maximum number of questions under multithreading

#include <iostream> #include <windows.h> #include <time.h>using namespace std;// The array of the Declaration prime table is the global variable int a[10000]={2,0};//prime table initializes the number of int flagnumber=0;//in the list of prime numbers int minnumber=0;int  maxnumber=0;//generally need to declare the function that will be called int simple_count (int a,int b);//Tired Division Int numlist (int a[])/ /build Prime Table {  int flag=0;//determines the position of the current number of stored primes in the array  bool repeat=true;//loop flag bit setting  int j=0;  for (int i=2;i<10000;i++)//will expand the scope of the search by setting the upper limit of I  {    while (repeat)    {   if (J==flag && (i%a[flag]!=0))//To determine whether the last element of the current prime number table is not divisible, the current numbers are pushed into the prime table     {    repeat=false;    flag++;    a[ Flag]=i;   }   else   {    if (i%a[j]==0)// If you can divide out the while statement     {     repeat=false;    }     else//otherwise continue to judge     {     j++;    }   }   }  repeat=true;//i before each test, you must set the value to True   j=0;//the next number of divisor should start from a[0], so J should be 0  } return  flag;//returns the number of calculations  //in fact there may be errors, if this contains the largest number of numbers there is a large prime factor is not detected  //or there is a number because its prime factor is too many, so less calculation, there may be errors  //but on the other hand, if this number really contains the most approximate, then its prime factor must be very small  //therefore, the probability of error must be related to the number of the existence of A and B}dword winapi test ( Lpvoid pparam) { int jump=0; * ((int*) pparam) =jump;     bool  repeat=true; int testnumber=minnumber;//storage contains the maximum number of approximations for inspection (with the use of a tired division check)  int numberflag=0;//set comparison bit  int statist=1;//should set the statistical value to how much need to parse   int b[10000]={0};//initialize the current flag bit to record how many of these same primes, and initialize  int j=0;//saves the current cursor to the array of flag bits  int temp=0;//the vector  for of the cyclic division (int i=jump;i<=maxnumber;i= I+2)  {  //j can not exceed the presence of flag   for ( j=0;i>=a[j];j++)//cannot exceed the bounds of the prime table, but must exclude this number less than 10000, When this number is very small, there is no need to compare it down  &nbsP; {   if (J>flagnumber)//When the range of the out-of-prime list indicates that all elements of the prime table have been traversed    {     break;   }   else   {    temp=i;     while (temp%a[j]==0)//If divisible, flag bit plus 1, remember to remove the number of times that have been added     {      b[j]++;     //Remember that the subscripts of A and B are corresponding      temp=temp/a [j];    }   }  }  //now counts the number of public factors in this count   for (int  k=0;k<=j;k++)   {   if (b[k]!=0)    {     statist=statist* (b[k]+1);       }     }    if (numberflag<statist)//Take one of the largest numbers   {   numberflag=statist;    testnumber=i;     }  statist=1;//Reset   for ( k=0;k<=j; k++)   {   b[k]=0;    } }  cout<< "The maximum of approximately several numbers obtained by the method of using the prime table is:" <<numberflag <<endl; jump=numberflag; return 0;} Int simple_count (INT&NBSP;A,INT&NBSP;B)//When the number between A and B is around 50, you can use class division directly (by calculating the test time to indicate that {//when the value of a is greater than 100000000, each count is calculated, It takes nearly 1s of time to be ignorant: it takes about half an hour to calculate the number of  int number=0;//statistics for 100,000 numbers  int flag=0;//set the comparison number, when the next number is about a number greater than the previous number, Reset   for (int i=a;i<=b;i++)//outer loop from A to B {  for (int j=1;j<=i/2;j=j+1)// Inner loop detection from 2 to self   {      if (i%j==0)    {     number++;   }  }  number++;//itself is also an approximate, need to add in   if (number> Flag)//reset flag bit   {   flag=number;  }  number=0;//just forgot to restore the count value to 0 }  return flag;} Int main () { clock_t start,finish; start=clock ();    handle  Hhandle[2]; int handleresult1,handleresult2;&nbSp;flagnumber=numlist (a);//Get the current Prime number table, and the current prime list of flag+1, indicating that the current primes table starts from 0  cout<<flagnumber<<endl;  cout<<a[1000]<<endl;     int result;//getting the current results  cout< < "Please enter two numbers in order to get the current range";  cout<<endl; cin>>minnumber>>maxnumber; handleresult1= Minnumber; handleresult2=minnumber+1; if (MAXNUMBER-MINNUMBER&LT;50)  {  result=Simple_ Count (Minnumber,maxnumber);  } else {  hhandle[0]=createthread (NULL,0,    test, (void*) (&AMP;HANDLERESULT1), 0,null);   hhandle[1]=createthread (null,0,    Test, (void*) (&AMP;HANDLERESULT2), 0,null);  }    waitformultipleobjects (2,hHandle,TRUE, INFINITE);  finish=clock (); cout<< "THE&NBSP;TIME&NBSP;HAS&NBSP;BEEN&NBSP;USED&NBSP;:" << finish-start<<endl; return 0;}

The source code for solving the problem of the maximum number of questions under multithreading