[JAVA]大数开平方(模拟手算的方法)

[JAVA]大数开平方(模拟手算的方法)

在 CSDN 看到的问题。　要求算出开平方结果的精确整数部分
想了想，想不到别的好的方法　就用了摸拟手算的方法
public static void main(String[] args) { System.out.print("sqrt(1)=" + sqrt("1")); System.out.print("\tsqrt(25)=" + sqrt("25")); System.out.print("\tsqrt(26)=" + sqrt("26")); // sqrt(26)==5 呵呵,因为只有整数部分 System.out.print("\tsqrt(100)=" + sqrt("100")); System.out.println("\tsqrt(144)=" + sqrt("144")); // 生成一个 5000 位以内的数 StringBuffer sb = new StringBuffer(""); Random rand = new Random(); int length = rand.nextInt(5000)+1; for (int i=0; i<length; i++) sb.append(rand.nextInt(10)); String theNumber = sb.toString(); System.out.println("n:" + sb.toString()); System.out.println("位数:" + length); // 开方 long t = System.currentTimeMillis(); BigInteger result = sqrt(theNumber); t = System.currentTimeMillis()-t; System.out.println("sqrt(n)=" + result.toString()); System.out.println("用时:" + t + "毫秒"); } public static BigInteger sqrt(String theNumber) { int length = theNumber.length(), i; BigInteger res = BigInteger.ZERO; BigInteger twenty = BigInteger.valueOf(20); BigInteger t, x=BigInteger.ZERO, v, few=BigInteger.ZERO; BigInteger hg = BigInteger.valueOf(100); String tmpString = null; int pos = 2-length%2; tmpString = theNumber.substring(0, pos); while (true) { v = few.multiply(hg).add(BigInteger.valueOf(Integer.parseInt(tmpString))); if (res.compareTo(BigInteger.ZERO)==0) i=9; else i = v.divide(res.multiply(twenty)).intValue(); for (; i>=0; i--) { t = res.multiply(twenty).add(BigInteger.valueOf(i)).multiply(BigInteger.valueOf(i)); if (t.compareTo(v)<=0) { x = t; break; } } res = res.multiply(BigInteger.TEN).add(BigInteger.valueOf(i)); few = v.subtract(x); pos++; if (pos>length) break; tmpString = theNumber.substring(pos-1, ++pos); } return res; }