hac*_*cks 41
在一个有点XOR操作:
a b a^b
-----------
0 0 0
0 1 1
1 0 1
1 1 0
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XOR sum指的是对整数的连续XOR运算.
假设你从有一个数字1来N,你必须找到自己的XOR总和然后N = 6,XOR总和会1^2^3^4^5^6 = 7.
1 = 001, 2 = 010, 3 = 011, 4 = 100, 5 = 101, 6 = 110
1^2 = 1^2 = 001^010 = 011 = 3
(1^2)^3 = 3^3 = 011^011 = 000 = 0
(1^2^3)^4 = 0^4 = 000^100 = 100 = 4
(1^2^3^4)^5 = 4^5 = 100^101 = 001 = 1
(1^2^3^4^5)^6 = 1^6 = 001^110 = 111 = 7 --> XOR sum
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希望这会有所帮助.