Thread: Finding the closest number to the average and the farthest.

Post the code you have tried first.

The psuedo code for it would be:

- Cycle through the array
- Subtract the current element from the average.
- if the absolute value of the result is less than the current minimum, assign the value to minimum.
- if the absolute value of the result is greater than the current maximum, assign the value to maximum.

That's it.

ok, you will need two variables to start with (along with the array). One to hold the closest, and other for furthest. Then you assign the first number of the array to each. Loop through the array, if the abs(Avg-Closest) > abs(Avg-array[i]) then Closest=array[i]. Flip the greater than sign along with using the Furthest variable and it should also get the furthest number. After the loop, just display Closest and Furthest.

Something like this will find the first closest and farthest values from average:

Code:

#include <iostream>
#include <stdlib.h>

using namespace std;

int main()
{
  int numbers[] = {1,2,3,4,5,6,7,8,9};
  int sum = 0;
  int avg;
  int i;

  for ( i = 0; i < 9; i++ )
    sum += numbers[i];
  avg = sum / 9;
  cout<<"Average: "<< avg <<'\n';
  int closest = numbers[0];
  int farthest = numbers[0];
  for ( i = 1; i < 9; i++ ) {
    if ( abs ( avg - numbers[i] ) < abs ( avg - closest ) )
      closest = numbers[i];
    if ( abs ( avg - numbers[i] ) > abs ( avg - farthest ) )
      farthest = numbers[i];
  }
  cout<<"First closest to average: "<< closest <<'\n';
  cout<<"First farthest from average: "<< farthest <<endl;
}

[edit]
I'm too slow I suppose.
[/edit]

GAR! Sebastiani, you beat me

Not sure, make your own absolute function

Code:

float absolute(float number)
{
if (number < 0)
   number *= -1;
return number;
}

just use absolute() instead of abs() and it should work fine

>Any way to fix this?
fabs is the equivalent absolute value function for floating-point. It resides in math.h.