Big Root

国内做题，从来没有用过JAVA。介于北美题对时间卡得没那么紧，而java对大数的操作比较方便，所以练习了一下。

题目的意思是，给定exp和bigNum，问是否存在answer，answer的exp次方是bigNum。其中bigNum的范围是1到10^100。

有了java的BigInteger，用二分搜索做这题很舒服。

import java.io.File;
import java.io.FileNotFoundException;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.Scanner;


public class Pro3 {

	/**
	 * @param args
	 */
	public static Scanner in=null;
	public static PrintWriter out=null;
	public static void main(String[] args) throws FileNotFoundException {
		// TODO Auto-generated method stub
		in = new Scanner (new File("D:\\ACM_UNL\\region2009\\pkg\\InputData\\in3.txt"));
		out = new PrintWriter (new File("D:\\ACM_UNL\\region2009\\pkg\\MyOutput\\out3.txt"));
		for(int i=1;;i++)
		{
			int exp=in.nextInt();
			if(exp==0) break;
			BigInteger bigNum=in.nextBigInteger();
			out.println("Case "+i+": "+Process(exp,bigNum));
			out.println();
		}
		in.close();
		out.close();
	}
	public static String Process(int exp, BigInteger bigNum)
	{
		BigInteger low=BigInteger.ONE;
		BigInteger high=bigNum;
		BigInteger answer=null;
		while(low.compareTo(high)<=0)
		{
			answer=low.add(high).divide(new BigInteger("2"));
			BigInteger test=answer.pow(exp);
			if(bigNum.compareTo(test)==0) return answer.toString();
			else if(bigNum.compareTo(test)<0) high=answer.subtract(BigInteger.ONE);
			else low=answer.add(BigInteger.ONE);
		}
		return "No solution";
	}
}