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a^b^b = a^0 = a 即利用异或可以成对消去重复元素
相关题目: 389
318 Maximum Product of Word Lengths
1009 Complement of Base 10 Integer &取数字二进制最后一位 << 乘2 >> 除2
868 >>除2 &获取二进制最后一位
693 Binary Number with Alternating Bits &获取二进制最后一位
461 Hamming Distance
476 Number Complement
477
338 Counting Bits 右移 &
371 位运算实现加法
201
260 (^ a^b^b=a)
136 xor
190 &取数字二进制最后一位 左移 右移
191
397
762(求连续数二进制1的个数)
1239 利用32位表示用过的字符 有点意思
The text was updated successfully, but these errors were encountered:
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异或
a^b^b = a^0 = a 即利用异或可以成对消去重复元素
相关题目: 389
&
318 Maximum Product of Word Lengths
1009 Complement of Base 10 Integer &取数字二进制最后一位 << 乘2 >> 除2
868 >>除2 &获取二进制最后一位
693 Binary Number with Alternating Bits &获取二进制最后一位
461 Hamming Distance
476 Number Complement
477
338 Counting Bits 右移 &
371 位运算实现加法
201
260 (^ a^b^b=a)
136 xor
190 &取数字二进制最后一位 左移 右移
191
397
762(求连续数二进制1的个数)
1239 利用32位表示用过的字符 有点意思
The text was updated successfully, but these errors were encountered: