Java大数开方模板及判断一个大数是否是平方数模板

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判断是否是平方数

	public static int judge (BigInteger x)///通过二分在区间(0,x)寻找是否有平方等于n的数{BigInteger left=BigInteger.ZERO,right=x,mid,ans;int sign;while(left.compareTo(right)<=0)///l<=r 注意有带等号{mid=left.add(right).shiftRight(1);//求出中间值mid=(right+left)/2ans=mid.multiply(mid);///ans=mid*midsign=ans.compareTo(x);///比较区间中间值的平方与x的大小if(sign==0) return 1;///相等else if(sign<0) left=mid.add(BigInteger.ONE);///ans小，左端点位mid+1else left=mid.subtract(BigInteger.ONE);///ans大，右端点为mid-1}return 0;}

算出开方数（参考博客）

import java.math.BigInteger;
import java.util.Scanner;public class Main {final static BigInteger NUM20 = BigInteger.valueOf(20);// 将后面使用的参数定义为final常量public static void main(String[] args) {Scanner input = new Scanner(System.in);System.out.println(kaifang(input.next));}// 手算法开方public static String kaifang(String s) {String result = "0";// 结果初始化为字符串0String remainder = "";// 余数if (s.length() % 2 != 0)// 将字符串长度变为偶数s = "0" + s;for (int i = 0; i < s.length() / 2; i++) {// 两两分组remainder += s.substring(i * 2, (i + 1) * 2);// 余数为之前的结果加上后面2个数字int shang = f1(new BigInteger(remainder), new BigInteger(result));remainder = f2(new BigInteger(remainder), new BigInteger(result), BigInteger.valueOf(shang)) + "";result += shang;}return result.substring(1);// 去掉结果之前的0}//  public static int f1(int remainder, int result) {
//      int i;
//      for (i = 9; i >= 0; i--)
//          if ((result * 20 + i) * i <= remainder)
//              break;
//      return i;
//  }
//
//  public static int f2(int remainder, int result, int shang) {
//      return remainder - (result * 20 + shang) * shang;
//  }public static int f1(BigInteger remainder, BigInteger result) {int i;for (i = 9; i >= 0; i--) {BigInteger NUMI = BigInteger.valueOf(i);if (result.multiply(NUM20).add(NUMI).multiply(NUMI).compareTo(remainder) <= 0)break;}return i;}public static String f2(BigInteger remainder, BigInteger result, BigInteger shang) {return remainder.subtract(result.multiply(NUM20).add(shang).multiply(shang)).toString();}
}