B09 Answer

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>

/* Puzzle B09 -- print the number out of of N random doubles 0 <= d < 100
  that is closest to 50.0*/

double randDoubleRange( double min, double max )
{
  return (max-min)*(rand()/(RAND_MAX+1.0)) + min;    
}

const int limit = 10000 ;
const double MIN=0.0, MAX= 100.0;
const double target=50.0;

int main(int argc, char *argv[])
{
  int j=0 ;
  double r, f;
  double minDifference = MAX;
  double closest;
  
  srand( time(NULL) );

  for ( j=0; j<limit; j++ )
  {
    r = randDoubleRange(MIN, MAX);
    
    if ( (f = fabs(target-r) ) < minDifference )
    {
      minDifference = f ;
      closest = r;
    }
  }
    
  printf("Closest: %lf; difference: %lf\n", closest, minDifference );
  system("pause");	
  return 0;
}

Comments: fabs(double) evaluates to the double precision absolute value of its argument. You may need to inclue the header funcion math.h

The expression in the if statement might look funny, but this is a common way of doing things like this. Assignment expressions evaluate to the value that is assigned to the variable, so the expression

(f = fabs(target-r))