Puzzle 8-10-08 (answer)

[BCS]

Reply to Steven,

> "BCS" wrote
> 
>> Reply to wyverex,
>> 
>>> The Abbott's Puzzle
>>> 
>>> "If 100 bushels of corn were distributed among 100 people in such a
>>> manner that each man received three bushels, each woman two, and
>>> each child half a bushel, how many men, women, and children were
>>> there?"
>>> 
>> <M, W, C> = <17,5,78> + i*<-3,5,-2>  for i in [0,5]
>> 
> You missed <20, 0, 80>
> 
> Of course, this assumes that it's possible to have 0 women (unlikely
> :) )
> 
> And what's with the non-D code?  It should be a requirement that you
> have to solve it with D ;)
> 
> -Steve
> 

D can't do the closed form solution (and I can't remember how so I used excel)

how about a D vector solution (untested as I'm still running 1.033)

void main()
{
     short[101] ramp;
     foreach(i,ref j;ramp)j=i;

     short[101][101] grain;
     short[101][101] c;

     foreach(short m, ref short[101] row; grain)
         row[] = m*6 - 200 + ramp[]*4 + (100-m-ramp[])*1;

     foreach(short m, ref short[101] row; c)
         row[] = 100-m-ramp[];

     for(int m = 0; m <= 100; m++)
        for(int w = 0; w <= 100; w++)
            if(c[m][w] >= 0 && grain[m][w] == 0)
                 writef("M=%d  W=%d  C=%d\n",m,w,c[m][w]);
}