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?
engllfAsked:
Who is Participating?
 
yonatConnect With a Mentor Commented:
int n = 100000; // or another max number
while (n > 1)
    if ( x != x % n)
        break;
    else
        n /= 10;

x is in the reange n to n*10.
0
 
sgantaCommented:
Hi !

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
 
alexoCommented:
#include <math.h>
unsigned NumberOfDigits(unsigned n)
{
    return n ? log10(n) + 1 : 1;
}

0
 
Booth882Commented:
here's one thats straight forward and has no limit to the size of the number:

#include <iostream.h>

int Power = 1;

cin >> int TheNumberToBeChecked;

while(1)
{
    if(TheNumberToBeChecked >= 10 ^ Power)
        Power++;
    else
        break;
}

try it and see what you think
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.