215 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
What is the sum of the digits of the number 21000?

2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
What is the sum of the digits of the number 2^1000?
唉又一道无聊的题。
加点难度，编写大数乘法运算，然后算出2^1000并求出各位数字和。

此程序效率低下仅仅就是试验下fft-mul算法的正确与否~~~
;;;每日一题016
;;;fft乘法，见之前关于介绍fft乘法的帖子
;;(fft-muln num/numstr ...) -> numstr
(require "../mod/fftmulmod.rkt")
(require "../mod/constrmod.rkt")
;-----------------------
(define (fft-expt n m)
(let loop ((n n) (m m) (r 1))
(cond ((= m 0) r)
(else
(loop n
(sub1 m)
(fft-muln n r))))))
(define (prog016 n m)
(foldr + 0
(string->listnum
(number->string
(inexact->exact
(string->number
(fft-expt n m)))))))
;;;test
(prog016 2 15)
(prog016 2 1000)
;26
;1366

(define (prog016 n)
(foldr + 0
(map (lambda (x) (- (char->integer x) 48))
(string->list
(number->string
(expt 2 n))))))
(prog016 15);26
(prog016 1000);1366