hdu-2095 find your present(2)
题目描述如下： 方法一： 位运算 原理：两个相同的数x，x进行按位与操作，即x^x，结果为0,某个数y与0按位与则结果为原数，即y^0=y; so,某个数y，遇到一个数x，y^x,当y^x再次遇到x时就会把x抵消掉，按照这种操作，最后剩下的就是出现次数为1的数了 1 #include <cstdio>
2 #include <algorithm>
3 using nam
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相关文章
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