SGU 111 Very simple problem

给一个数X (1≤X≤10¹⁰⁰⁰).求出平方后不超过X的最大整数。

 

法一：二分区间[1,10⁵⁰⁰]，寻找符合的数。

 

import java.math.BigInteger;
import java.util.Scanner;

public class Solution {
	public static void main(String[] args) {
		BigInteger l = new BigInteger("1"),r = new BigInteger("10"),x,mid,ans = new BigInteger("0");
		r = r.pow(500);
		Scanner in = new Scanner(System.in);
		x = in.nextBigInteger();
		while(l.compareTo(r)<=0){
			mid = (l.add(r)).shiftRight(1);
			if(mid.pow(2).compareTo(x)<=0){
				ans=l;
				l=mid.add(BigInteger.valueOf(1));
			}
			else r=mid.subtract(BigInteger.valueOf(1));
		}
		System.out.println(ans);
		in.close();
	}
}

 

 

法二：手工开方。。这里有相关证明：http://www.cnblogs.com/Rinyo/archive/2012/12/16/2820450.html

div、ans、temp分别用来存储当前除数、当前所得商（结果）、当前被除数

 

import java.math.BigInteger;
import java.util.Scanner;
public class Solution {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		BigInteger ans=new BigInteger("0"),div,temp=new BigInteger("0"),n;
		int i;
		n=in.nextBigInteger();
		String s="0"+n.toString();
		if(s.length()%2==0) i=1;
		else i=2;
		for(;i<s.length();i+=2)
		{
			div=ans.multiply(BigInteger.valueOf(20));
			temp=temp.multiply(BigInteger.valueOf(100));
			temp=temp.add(new BigInteger(s.substring(i-1,i+1)));
			int k;
			for(k=0;k<10;++k)
				if(div.add(BigInteger.valueOf(k+1)).multiply(BigInteger.valueOf(k+1)).compareTo(temp)==1) break;
			ans=ans.multiply(BigInteger.valueOf(10)).add(BigInteger.valueOf(k));
			temp=temp.subtract(div.add(BigInteger.valueOf(k)).multiply(BigInteger.valueOf(k)));
		}
		System.out.println(ans);
		in.close();
	}
}