SICP 1.11-1.13

1.11

递归版本

 (define (f n) (cond ((< n ) n)
                     (else (+ (f (- n ))
                              (* (f (- n )) )
                              (* (f (- n )) )))))

 迭代版本

 (define (f n)
   (if (< n )

       (xf n    )))

 (define (xf n a b c d)
   (cond ((= n d) (max a b c))
         (else (cond ((=  (remainder d )) (xf n (+ (*  a)
                                               (*  b)
                                               c) b c (+  d)))
               ((=  (remainder d )) (xf n a (+ (*  b)
                                                 (*  c)
                                                 a) c (+  d)))
               ((=  (remainder d )) (xf n a b (+ (*  c)
                                                   (*  a)
                                                   b) (+  d)))))))

1.12

 (define (pascal h n)
   (cond ((= h ) )
         ((= n ) )
         ((= h n) )
         (else (+ (pascal (- h ) n)
                  (pascal (- h ) (- n ))))))

1.13
证明题看着都烦...
作为补偿我来实现一下1.12的迭代版本 --------> 失败

经过百度发现了一个公式  ------->   (row col)=row!/(col!⋅(row−col)!)   -----> 有这个公式的话那实现起来很简单,就不演示了...