Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem Statement
	XOR problems became very popular in TopCoder recently. (If you do not know the bitwise operation XOR, see the Notes section for an explanation.) That's why Elly decided to invent one of her own. Fortunately for you, the one she came up with is quite simple. You are given two long longs L and R. She wants you to find the XOR of all numbers between L and R, inclusive.
Definition
	Class: EllysXors Method: getXor Parameters: long long, long long Returns: long long Method signature: long long getXor(long long L, long long R) (be sure your method is public)
Notes
-	XOR (exclusive or) is a binary operation, performed on two numbers in binary notation. First, the shorter number is prepended with leading zeroes until both numbers have the same number of digits (in binary). Then, the result is calculated as follows: for each bit where the numbers differ the result has 1 in its binary representation. It has 0 in all other positions.
-	For example 42 XOR 7 is performed as follows. First, the numbers are converted to binary: 42 is 101010 and 7 is 111. Then the shorter number is prepended with leading zeros until both numbers have the same number of digits. This means 7 becomes 000111. Then 101010 XOR 000111 = 101101 (the result has ones only in the positions where the two numbers differ). Then the result can be converted back to decimal notation. In this case 101101 = 45, so 42 XOR 7 = 45.
-	One of the ways to calculate the XOR of more than two numbers A1, A2, ..., An is "A1 XOR (A2 XOR (... XOR An))..))". Since the function is commutative and associative, you can also express it as "A1 XOR A2 XOR ... XOR An" and group the operands in any way you like.
-	It can be proved that the answer is always less than 2*R for the given constraints.
Constraints
-	L and R will be between 1 and 4,000,000,000, inclusive.
-	L will be less than or equal to R.
Examples
0)
	3 10 Returns: 8 ((((((3 XOR 4) XOR 5) XOR 6) XOR 7) XOR 8) XOR 9) XOR 10 = (((((7 XOR 5) XOR 6) XOR 7) XOR 8) XOR 9) XOR 10 = ((((2 XOR 6) XOR 7) XOR 8) XOR 9) XOR 10 = (((4 XOR 7) XOR 8) XOR 9) XOR 10 = ((3 XOR 8) XOR 9) XOR 10 = (11 XOR 9) XOR 10 = 2 XOR 10 = 8.
1)
	5 5 Returns: 5 The XOR of a single number is the number itself.
2)
	13 42 Returns: 39 A bit larger example.
3)
	666 1337 Returns: 0 The answer can be zero.
4)
	1234567 89101112 Returns: 89998783

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.     

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我的想法跟其他人的寫法應該不太一樣, 我是看每個 bit 的個數的奇偶去決定,
你可以很明顯發現最低位的 bit, 週期是 2, 連續一個 1, 次來 4/2, 8/4, 16/8,
然而除了最低位外都是偶數, 其實就沒有那麼重要, 所以只要去查看 L, R 所碰觸的連續 1 個個數即可

#include <cstdio>
using namespace std;
class EllysXors {
    public:
        long long getXor(long long l, long long r) {
            long long ans = 0, i;
            for(i = 1; i < 40; i++) {
                long long s1 = 0, s2 = 0;
                long long tl = l, tr = r;
                if((l>>i)&1) {
                    tl = ((l>>i)<<i)|((1<<i)-1);
                    if(tl > r)
                        s1 = r-l+1;
                    else
                        s1 = tl-l+1;
                }
                if((r>>i)&1){
                    tr = (r>>i)<<i;
                    if(tr > l)
                        s2 = r-tr+1;

                }
                if((s1+s2)&1)
                    ans |= (long long)1<<i;
            }
            if(l%2 == 0)    l++;
            if(r%2 == 0)    r--;
            if((r-l)/2%2 == 0) ans |= 1;

            return ans;
        }
};