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)))
