@(目录)

就想取个标题

这些操作在状压DP中都很常用，灵活应用可以很好地降低码量，降低常数。

取出x的第i位:

\(y\)= (\(x\)>>(\(i\)-\(1\)))&\(1\);

将x第i位取反

\(x\) ^= \(1\)<<(\(i\)-\(1\));

将x第i位变为1

\(x\) |= \(1\)<<(\(i\)-\(1\));

将x第i位变为0

\(x\) &= ~(\(1\)<<(\(i\)-\(1\)));

将x最靠右的1变成0

\(x\) = \(x\)&(\(x\)-1);

取出x最靠右的1

\(y\) = \(x\)&(-\(x\));

这也是树状数组中我们熟知的\(lowbit\)

把最靠右的0变成1

\(x\)|=\(x\)+\(1\)

判断是否有两个连续的1

\(if\)(\(x\)&(\(x\)<<\(1\))) \(cout\)<<"\(YES\)";

判断是否有\(n\)个连续的\(1\)

\(if\)(\(x\)&(\(x\)<<\(1\))&&\(x\)&(\(x\)<<\(2\))...&&\(x\)&(\(x\)<<\(n\)-\(1\)))

枚举子集

定义\(sta\)为初始状态

\(for\)( \(int\) \(x\) = \(sta\) ; \(x\) ; \(x\) = ( ( \(x\) - \(1\) )&\(sta\)) )
\(cout\)<<\(x\);

位运算-状压DP 常用技巧

原文：https://www.cnblogs.com/hzyhome/p/11658240.html