使用Scheme的对数迭代法:
#lang racket ;;N是偶数:b^n = (b^(n/2))^2 (define (square x) (* x x));定义乘积函数 (define (fast-expt b n);筛选 (expt-iter b n 1)) (define (expt-iter b n a) (cond ((= n 0) a);当n= 0,值为1 ((even? n)(expt-iter (square b)(/ n 2) a));判断是否为偶数 ((odd? n)(expt-iter b (- n 1)(* b a)))));判断是否为奇数 (define (even? n) (= (remainder n 2) 0)) (fast-expt 2 15) ;;N是奇数:b^n = b*b^(n-1)
Java实现的递归法和迭代法:
public class Test { public static void main(String args[]){ int ex,ey; ex = expt(122,4); ey = expt_iter(122, 4, 1); System.out.println(ey); System.out.println(ex); } //递归 static int expt(int b,int n){ int sum; if(n == 0) return 1; else { sum = (expt(b,(n-1)))*(b); return sum; } } //迭代 static int expt_iter(int b, int counter,int product){ int sum; if(counter == 0){ return product; } else { sum = expt_iter(b, (counter-1), (b*product)); return sum; } } }
时间: 2024-10-22 11:28:09