# another fine solution by misof

# variables x and y have to be set
# (x / y) is the exact fraction we want to approximate, y is a power of 10
# e.g. for the input 0.33456 set
#   x =  33456
#   y = 100000
#
# we seek the first fraction that falls into the interval < (10x-5)/10y , (10x+5)/10y )

# fraction comparison 
define less(a,b,c,d) { return (a*d < b*c); }

# check whether c/d approximates a/b well 
define matches(a,b,c,d) { 
  if (less(c,d,10*a-5,10*b)) return 0;
  if (less(c,d,10*a+5,10*b)) return 1;
  return 0;
}

# max() will come in handy
define max(a,b) { if (a>b) return a; return b; }

# set initial bounds for the search 
a = 0
b = 1
c = 1
d = 1

while (1) {
  # compute the number of successive steps to one side, do them all at once
  if (b<d) { 
    u = (c*y - d*x); v = (b*x - a*y); if (u<0) { u=-u; v=-v; }
    k = (u-1)/v;
    if (k==0) k=1;
    l = 1;
    e = k*a + l*c; 
    f = k*b + l*d; 
  } 
  if (b>d) { 
    u = (a*y - b*x); v = (d*x - c*y); if (u<0) { u=-u; v=-v; }
    k = 1;
    l = (u-1)/v; 
    if (l==0) l=1;
    e = k*a + l*c; 
    f = k*b + l*d;
  }
  if (b==d) { e=a+c; f=b+d; k=1; l=1; }
 
  # if we hit the bull's eye, get the result
  if (matches(x,y,e,f)) { 
   
    # we could've skipped the best solution
    # now we find it using binary search on the number of steps
    m=0; n=max(k,l);
    while (m+1 < n) {
      o = (m+n)/2
      p = k-o; q = l-o;
      if (p<1) p=1; if (q<1) q=1;
      e = p*a + q*c; 
      f = p*b + q*d;
      if (matches(x,y,e,f)) { m=o; } else { n=o; } 
    }
    # compute the best result
    o = m;
    p = k-o; q = l-o;
    if (p<1) p=1; if (q<1) q=1;
    e = p*a + q*c; 
    f = p*b + q*d;
    
    print e," ",f,"\n"; 
    break; 
  }
  if (less(x,y,e,f)) { c=e; d=f; } else { a=e; b=f; }
}