find the middle value.c

输入一个数值,找出它的所有因数

// USE DESCRIPTION:
//   - This program will compute the prime divisor of an integer number between 2 and 10,000.
//     The program will first prompt the user to enter an integer number, then show the integer number as the
//     multiplication of its prime divisors. Note that the prime divisors are ordered accendingly.
// DESIGN DESCRIPTION:
//   - 1. ask user to enter a number.
//     2. use two int variables to store the number, one is for estimate if the number just can be divided by itself or
//        else; the other is for estimate which numbers can be divided.
//     3. using "for" repetition to force user to enter a valid number
//     4. "while" repetition combines with "if" selection to divid the number
//        - when the divisor i less than the number then the number goes into the loop
//             - when the number can be divided then number equals to the results, do it again; until the changing
//               number can not be divided by the divisor, the divisor pluses 1 and do the same process untill the
//               divisor equals the number - 1; get out of the loop and do the next step
//        - when the divisor get out of the loop, it equals the number.
//             - if it also equals to the origial number, the number must be the number, which just can be divided by 
//               itself; if it just equals to the final number, it means it can be divided by other numbers.
      

// Required header files
#include <iostream>
using namespace std;

// Start of main function
int main()
{
// Named constants
   const int minNum = 2;                                                         //the minium number
   const int maxNum = 10000;                                                     //the maxium number

// Variable declarations 
   int num;                                                                     //to store the enter number
   int selfDivisor;                                                             //to estimate if the number just can
                                                                                //divided by itself or not
   int i = 2;                                                                   //to limit the loope 
   
   cout << "Please enter an integer number in the range of 2 to 10,000"<<endl;  //require a number
   cin >> num;                                                                  //get a number

   //error checking
   for ( ;num < minNum || num > maxNum; )                                      
   {
    cout << "The integer number must be in the range of 2 to 10,000"<<endl;
    cout << "Please enter an integer number in the range of 2 to 10,000"<<endl;
    cin >> num;
   }
   cout << "You have entered " << num << " and" << endl;
   cout << num << " = ";
   selfDivisor = num;                                                           //for estimating selfdivisor or not

   //dealing with the result
   while ( i < selfDivisor)
   {
    if ( (selfDivisor % i) == 0 )
    { 
     cout << i << " * ";
     selfDivisor /= i;
    }
    else 
      i++;
   }
   if ( i == selfDivisor && i != num )
    cout << selfDivisor << endl;
   else
    cout << "1" << " * " << num <<endl;

  // Successful completion
  return 0;
}