Exclusive or

题目链接：

http://acm.hust.edu.cn/vjudge/contest/121336#problem/J

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<10^500).

Output

For each tests:

A single integer, the value of the sum.

Sample Input

3

4

Sample Output

6

4

##题意：

求如题所示的和，n的范围是1e500.


##题解：

数据这么大肯定要找规律.
先尝试打出前100个数的表，然后找规律....(弱鸡并不能找出来)
先安利一个网站(http://oeis.org/)这是一个在线整数数列查询网站.
搜一下果然有：(http://oeis.org/A006582)
公式为：a(0)=a(1)=0, a(2n) = 2a(n)+2a(n-1)+4n-4, a(2n+1) = 4a(n)+6n.
由于是个递归公式，可以用dfs来计算，用map去重后应该是O(lgn).

一开始用cpp的大数模版一直出现各种问题(版不太熟悉)，干脆复习一下java语法.
网上找到一份题解有推导过程：


##代码：
``` java
import java.math.BigInteger;
import java.util.HashMap;
import java.util.Scanner;

public class Main {

public static HashMap<BigInteger, BigInteger> myMap = new HashMap<BigInteger,BigInteger>();

public static BigInteger [] num = new BigInteger[10];

public static BigInteger dfs(BigInteger x) {

    if(x == num[0] || x== num[1]) return num[0];

    if(myMap.containsKey(x))return myMap.get(x);

    if(x.mod(num[2]) == num[0]) {

        BigInteger n = x.divide(num[2]);

        BigInteger tmp = num[2].multiply(dfs(n).add(dfs(n.subtract(num[1])))).add(num[4].multiply(n.subtract(num[1])));

        myMap.put(x, tmp);

        return tmp;

    } else {

        BigInteger n = (x.subtract(num[1])).divide(num[2]);

        BigInteger tmp = num[4].multiply(dfs(n)).add(num[6].multiply(n));

        myMap.put(x, tmp);

        return tmp;

    }

}

public static void main(String[] args) {

    Scanner scanner = new Scanner(System.in);  

    for(int i=0; i<10; i++) {

        num[i] = BigInteger.valueOf(i);

    }

    while(scanner.hasNext()){

        myMap.clear();

        BigInteger n = scanner.nextBigInteger();

        BigInteger ans = dfs(n);

        System.out.println(ans);

    }  

    scanner.close();

}

}