hdu-2095 find your present(2)
題目描述如下： 方法一： 位運算 原理：兩個相同的數x，x進行按位與操作，即x^x，結果爲0,某個數y與0按位與則結果爲原數，即y^0=y; so,某個數y，遇到一個數x，y^x,當y^x再次遇到x時就會把x抵消掉，按照這種操作，最後剩下的就是出現次數爲1的數了 1 #include <cstdio>
2 #include <algorithm>
3 using nam
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相關文章
- 1. find your present
- 2. J-find your present(2)
- 3. ACM STEP 1.2.4 find your present (2)
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