HDU 4919 Exclusive or （数论 or 打表找规律）

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();
}  

}