习题1.25

 

 对吗？实践出真知，运行看看。代码如下。

(defn square [x] (* x x))

(defn fast-expt[b n]
    (println "call iter" n)
    (cond (= 1 n) b
          (= 2 n) (square b)
          (even? n) (square (fast-expt b (/ n 2)))
          :else (* b (fast-expt b (- n 1)))
    )
  )

(defn expmod [base exp m]
    (mod (fast-expt base exp) m)
)

再看看运行结果

 这执行结果有点奇怪，需要再次检查代码逻辑及输出.

(defn divisor? [ a b](= (mod b a) 0 ))
(defn square [x](* x x ))
(defn find-divisor [n test-divisor]
  (cond (> (square test-divisor) n) n
        (divisor? test-divisor n) test-divisor
        :else (find-divisor n (+ test-divisor 1))
        )