java中如何对一个数开根号

Answer 1

java实现开根号的运算：\x0d\x0apublic static void main(String[] args) { long start = System.currentTimeMillis(); double target=9876543212345d; double result =sqrt(target);\x0d\x0aSystem.out.println("sqrt耗时："+(System.currentTimeMillis()-start)+",result:"+result);\x0d\x0a\x0d\x0astart=System.currentTimeMillis();\x0d\x0aresult =SqrtByBisection(target, 0);\x0d\x0aSystem.out.println("SqrtByBisection耗时："+(System.currentTimeMillis()-start)+",result:"+result);\x0d\x0a\x0d\x0astart=System.currentTimeMillis();\x0d\x0aresult = SqrtByNewton(target, 0);\x0d\x0aSystem.out.println("SqrtByNewton耗时："+(System.currentTimeMillis()-start)+",result:"+result);\x0d\x0a\x0d\x0a}\x0d\x0a//牛顿迭代法\x0d\x0apublic static double SqrtByNewton(double target,double eps){ double Xa=target,Xb; do {\x0d\x0aXb = Xa;\x0d\x0aXa = (Xa+target/Xa)/2;\x0d\x0a} while (fabsf(Xa,Xb)>eps); return Xa;\x0d\x0a} //二分法 精度是指两次mid值的差值\x0d\x0apublic static double SqrtByBisection(double target,double eps){ double min=1,max=target; double mid =(min+max)/2; double anMid; do { if(mid*mid>target){\x0d\x0amax=mid;\x0d\x0a}else{\x0d\x0amin=mid;\x0d\x0a}\x0d\x0aanMid=mid;\x0d\x0amid=(max+min)/2;\x0d\x0a} while (fabsf(anMid,mid)>eps); return mid;\x0d\x0a}\x0d\x0a\x0d\x0apublic static double sqrt(double d){ double a = 0.1; double x1,x2=0; while (a*a<=d) {\x0d\x0aa+=0.1;\x0d\x0a}\x0d\x0ax1=a; for (int i = 0; i < 10; i++) {\x0d\x0ax2=d;\x0d\x0ax2/=x1;//x2 = x2/x1;\x0d\x0ax2+=x1;\x0d\x0ax2/=2;\x0d\x0ax1=x2;\x0d\x0a} return x2;\x0d\x0a}\x0d\x0apublic static double fabsf(double a,double b){ if(a>b){ return a-b;\x0d\x0a}else\x0d\x0areturn b-a;\x0d\x0a}