Java 递归解决 "仅仅能两数相乘的计算器计算x^y" 问题
2017-07-18 21:57
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/** * 求一个数的乘方 * 求x^y,y是一个正整数。 设计算器仅仅能计算两数相乘,不能一次计算n个数相乘。 * 知:2^5=(2^2)^2*2; 2^6=(2^2)^3=((4)^2)*4; 2^8=(2^2)^4= (4^2)^2= 16^2 * 得到规律:x^y= (x^2)^(y/2),定义a=x^2,b=y/2, 则得到形如: x^y= a^b; * y假设是奇数,则分解的最后还要再乘以a(如上面2^6分解成4^3时):x^y=a^b*x. * * 用递归来解:那么每次x都传入一个新的值。即是a。y值倍减,即是b * * @author stone * @date 2015-7-2 上午11:31:53 */ public class Power { private static long pow(long x, long y) { if (x == 0 || x == 1) { return x; } long tx = 0; long ty = 0; if (y > 1) { ty = y / 2; tx = x * x; System.out.println(tx); if (y % 2 == 0) { return pow(tx, ty); } else { return pow(tx, ty) * x; } } else { return x; } } public static void main(String[] args) { long result = pow(2, 10); System.out.println(result); } }
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