我有一个给定长度的c ++ bitset。我想生成这个bitset的所有可能组合,我想添加1 2 ^ bitset.length次。这该怎么做? Boost库解决方案也是可以接受的
答案 0 :(得分:5)
试试这个:
/*
* This function adds 1 to the bitset.
*
* Since the bitset does not natively support addition we do it manually.
* If XOR a bit with 1 leaves it as one then we did not overflow so we can break out
* otherwise the bit is zero meaning it was previously one which means we have a bit
* overflow which must be added 1 to the next bit etc.
*/
void increment(boost::dynamic_bitset<>& bitset)
{
for(int loop = 0;loop < bitset.count(); ++loop)
{
if ((bitset[loop] ^= 0x1) == 0x1)
{ break;
}
}
}
答案 1 :(得分:2)
所有可能的组合?只需使用64位无符号整数,让您的生活更轻松。
答案 2 :(得分:0)
使用boost库,您可以尝试以下操作:
例如,长度为4的位集
boost::dynamic_bitset<> bitset;
for (int i = 0; i < pow(2.0, 4); i++) {
bitset = boost::dynamic_bitset<>(4, i);
std::cout << bitset << std::endl;
}