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Algorithm

What is the best algorithm to check a number is in which particular range of:
0 to 9
10 to 99
100 to 999
1000 to 9999
10000 to 99999
...rate at exponential of 10

E.g X = 555 is in the range of 100 - 999

I use a if-else-then to do the checking but its looks terribly ugly.

Any advice?
0
engllf
Asked:
engllf
1 Solution
 
chensuCommented:
Which is the best depends on the actual requirements. Here is an alternative. But you still need to check the value of n. You may use n as an index of an array.


#include <math.h>

.

if (X >= 0)
{
    unsigned int n = X == 0 ? 0 : log10((double)X);
    // 0 to 9, n = 0
    // 10 to 99, n = 1
    // 100 to 999, n = 2
    // 1000 to 9999, n = 3
    // ...
}
else
{
   ...
}

0
 
sgantaCommented:


      I have simple and beautiful algorithm and solution for your problem.

      ALGORITHM
      1. Read the number into "n";
      2. Store this number n as a string into test
          i.e., test = string(n);
      3. Get the length of the string into length
          length = string_length(n);
      4. Now
          Range_From = power(10,length-1)
          Range_To     = power(10,length)-1
      5. End of the algorithm

      /* Here the solution in C++ */

      #include <iostream.h>
      #include <string.h>
      #include <math.h>

      void main(void)
      {
         int n,len,ifrom,ito;
         char *test;

         cout << "Enter a number ";
         cin >>n;

         test = itoa(n); /* Converts integer to Ascii */
         len = strlen(test);
         ifrom = pow(10,len-1);
         ito = pow(10,len)-1;
         cout << "The number "<<n<<" is in the range from "<<ifrom<<" to "<<ito<<endl;
      }

      This  algorithm is tested and it works fine. If you are interested with my answer
      Please reject the above answer and accept my answer.
      Thank you.
      JESUS LOVES YOU - sganta





0
 
nietodCommented:
engllf, just so you know, while sganta's solution is pretty short and very slick, it is not very efficient.  If you are concerned about speed, you'll want to use chensu's solution.  But if you aren't concerned about speed, sganta's solution is pretty clear and simple.
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