手写bitset

手写bitset

Update: stl 的 bitset 并不慢（大概比手写还快一些？？），除以 32 还是 64 看机子的位数，需要注意的是 stl 的构造函数在不开 O2 的情况下还是比较慢的。

rt, 有一道玄学题好像可以用bitset, 但时间复杂度为\(\Theta(\frac{1e5*1e6}{32})\), 于是手写bitset, 发现用__int128反而比ull慢, 所以一下代码用了ull, 本地\(i78750h\)三秒跑完

const int N = 1000000;

ull mi[200];
struct Bit {
	ull bit[15626]; int len = 15626;
	#define I inline 
	I void reset() {memset(bit, 0, sizeof(bit));}
	Bit() {memset(bit, 0, sizeof(bit));}
	I void set1(int x) { bit[x>>6] |= mi[x&63];}
	I void set0(int x) { bit[x>>6] &= ~mi[x&63];}
	I void flip(int x) { bit[x>>6] ^= mi[x&63];}
	bool operator [](int x) { 
		return (bit[x>>6] >> (x&63)) & 1;
	}
	
	#define re register
	Bit operator ~ (void) const {
		Bit res;
		for (re int i = 0;i < len; i++) res.bit[i] = ~bit[i];
		return res;
	}
	
	Bit operator & (const Bit &b) const {
		Bit res;
		for (re int i = 0;i < len; i++) res.bit[i] = bit[i] & b.bit[i];
		return res;
	}
	
	Bit operator | (const Bit &b) const {
		Bit res;
		for (re int i = 0;i < len; i++) res.bit[i] = bit[i] | b.bit[i];
		return res;
	}	
	
	Bit operator ^ (const Bit &b) const {
		Bit res;
		for (re int i = 0;i < len; i++) res.bit[i] = bit[i] ^ b.bit[i];
		return res;
	}
	
	void operator &= (const Bit &b)  {
		for (re int i = 0;i < len; i++) bit[i] &= b.bit[i];
	}
	
	void operator |= (const Bit &b)  {
		for (re int i = 0;i < len; i++) bit[i] |= b.bit[i];
	}
	
	void operator ^= (const Bit &b)  {
		for (re int i = 0;i < len; i++) bit[i] ^= b.bit[i];
	}
	
	Bit operator << (const int t) const {
		Bit res; int high = t >> 6, low = t & 63;
		ull last = 0;
		for (register int i = 0;i + high < len; i++) {
			res.bit[i + high] = (last | (bit[i] << low));
			if (low) last = (bit[i] >> (64 - low));
		}
		return res;
	}
	
	Bit operator >> (const int t) const {
		Bit res; int high = t >> 6, low = t & 63;
		ull last = 0;
		for (register int i = len - 1;i >= high; i--) {
			res.bit[i - high] = last | (bit[i] >> low);
			if (low) last = bit[i] << (64 - low);
		}
		return res;
	}
	
	void operator <<= (const int t) {
		int high = t >> 6, low = t & 63;
		for (register int i = len - high - 1; ~i; i--) {
			bit[i + high] = (bit[i] << low);
			if (low && i) bit[i + high] |= bit[i - 1] >> (64 - low);
		}
		for (register int i = 0;i < high; i++) bit[i] = 0;
	}
	
};