hdu 4919 Exclusive or

Exclusive or

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 327    Accepted Submission(s): 137

Problem Description

Given n, find the value of 


Note: ⊕ denotes bitwise exclusive-or.

 

Input

The input consists of several tests. For each tests:

A single integer n (2≤n<10⁵⁰⁰).

 

Output

For each tests:

A single integer, the value of the sum.

 

Sample Input

3
4

 

Sample Output

6
4

 

Author

Xiaoxu Guo (ftiasch)

 

Source

2014 Multi-University Training Contest 5

 

题解：

以下是官方给的答案：

鉴于比較难以理解官方大神的解说，这里我研究了一下，以下给出求解化简过程：

图片有些不清晰，凑合看吧 T^T

代码：

import java.util.*;
import java.io.*;
import java.math.*;

public class Main {

	public static BigInteger zero=BigInteger.valueOf(0);
	public static BigInteger one=BigInteger.valueOf(1);
	public static BigInteger two=BigInteger.valueOf(2); 
	public static BigInteger four=BigInteger.valueOf(4);
	public static BigInteger six=BigInteger.valueOf(6); 
    
	public static HashMap<BigInteger,BigInteger> map=new HashMap<BigInteger,BigInteger>();
	
	public static BigInteger solve(BigInteger n)
	{
		if(map.containsKey(n))
			return map.get(n);
		BigInteger t=BigInteger.valueOf(0);
		BigInteger k=n.divide(two);
		BigInteger r=n.mod(two);
		
		if(r.compareTo(one)==0)
			t=solve(k).multiply(four).add(k.multiply(six));
		else  {
			t=two.multiply(solve(k));
			t=t.add(two.multiply(solve(k.subtract(one))));
			t=t.add(four.multiply(k));
			t=t.subtract(four);
		}
		map.put(n, t);
		return t;
	}
	
	public static void main(String []args)
	{
        BigInteger n;
        map.put(zero, zero);
        map.put(one,zero);
        Scanner cin = new Scanner(System.in); 
        while(cin.hasNext())
        {
        	n=cin.nextBigInteger();
        	System.out.println(solve(n));
        }
	}
}
/*
借鉴别人的代码，学习了一下java大数和HashMap的使用方法，能够当作模版来使用了。

*转载请注明出处，谢谢。
*/