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)/2
ans=mid.multiply(mid);///ans=mid*mid
sign=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+1
else 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0
String 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();
}
}