Re: Euler problem #48

Marcel Hendrix wrote:

> INCLUDE ../bignum/factor.frt

If you have a 64 bit Forth (and I know you have ;-), you don't need
bignums
for that. All intermediate results have at most 20 digits, and therefore
easily fit in a 128 bit number; the sum with at most 13 digits fits in a
64
bit number. This one should be faster (how much depends on how efficient
your gs^mod actually is). In Gforth with a 2GHz Athlon64, it takes about
1.5ms to run 1000t. I guess iForth64 should do it in less than 1ms, even
when the division for the modulus might be the limiting factor.

\ sum of the last 10 digits of n^n for n=1..1000

Cell 8 < [IF] .( Needs a 64 bit Forth!) cr true abort  [THEN]

&10000000000 Constant mod#

: *mod ( a b -- c )  um* mod# um/mod drop ;

: **mod ( x n -- n )  >r 1 swap
    BEGIN  r@[EMAIL PROTECTED]
 1 and  IF  tuck *mod swap  THEN
        r> 2/ dup  WHILE  >r dup *mod  REPEAT
    2drop ;

: e48 ( n -- g )
    0 swap 1+ 1 DO
        I dup **mod +
    LOOP  mod# mod ;

: Euler48 ( -- )  
         CR ." The last ten digits of the series "
         ." 1^1 + 2^2 + 3^3 + ... + 1000^1000 are " 1000 e48 . cr ;

-- 
Bernd Paysan
"If you want it done right, you have to do it yourself"
http://www.jwdt.com/~paysan/