Euler_Problem-071.b93

v X X C
  X X ????

>020p040p121p141p"}}@"**60pv
v0                         <         >01+*v
>1+ :3*7%!#v_:61p:3*7/:71p7*61g3*-:0`|    >81p61g7*91p41g91g*21g81g*`#v_v
|-g06:     <                        <>01-*^v  p04g16p02g17p12g19p14g18<
>$20g."/ ",,55+,40g.@               ^      <                            <




[20] Result::Denominator
[40] Result::Numerator

[21] Distance::Denominator
[41] Distance::Numerator

[60] LIMIT     1000000

[61] c_n | i
[71] c_d
[81] d_n
[91] d_d


---------------------------------------

For every denominator from `1` to `1 000 000` we generate a possible fraction where the numerator is `(d * 3) / 7`.
(Every other fraction is either greater than `3/7` or has a greater difference than the generated).

Now everything thats left is finding the fraction with the smallest difference to `3/7`.

The difference of two fractions is calculated (without floating points) by:

~~~
n1/d1 - n2/d2  ==  (n1*d2 - n2*d1)/(d1*d2)
~~~

The rest is just iterating through all the possible fractions and in each step remembering the current best.

We don't even need to reduce the result to get a *proper* fraction. 
If it could be reduced we would have got the reduced version first.
(Because it has an smaller denominator).