Solution -「UNR #5」「UOJ #671」诡异操作

$\mathcal{Desciprtion}$

  Link.

  给定序列 $\{a_n\}$，支持 $q$ 次操作：

1. 给定 $l,r,v$，$\forall i\in[l,r],~a_i\leftarrow\lfloor\frac{a_i}{v}\rfloor$；
2. 给定 $l,r,v$，$\forall i\in[l,r],~a_i\leftarrow a_i\otimes v$，其中 $\otimes$ 表示二进制按位与；
3. 给定 $l,r$，求 $\sum_{i=l}^ra_i$。

  $n\le3\times10^5$​，$q\le2\times10^5$​，值域 $V<2^{128}$​，答案对 __uint128_t 自然溢出。

$\mathcal{Solution}$

  写个“暴力”，证明复杂度，得到正解√

  不考虑 (1) 操作，想想如何支持区间与-区间求和。我们用一棵线段树来维护，对于一个长为 $s$ 的区间，维护一个表格 $S_{0..\log s}$，设某个 bit $x$ 在该区间中出现了 $c$ 次，则在 $S$ 中每个 $c$ 的 bit 处加上 $x$ 的贡献。可以发现 $\sum S_i2^i$ 就是区间和，合并类似高精度加法做到 $\mathcal O(s)$，而区间与直接在令 $S_i\leftarrow S_i\otimes v$ 即可解决，也是 $\mathcal O(s)$。不失为一种奇怪的小 trick√

  取整除，自然而然暴力做。那么至多遍历 $\omega=\log V$ 次线段树，乘上 $\mathcal O(s)$（$s$ 为区间长度）的上传代价，遍历整棵树的上传代价和式 $T(n)=2T(n/2)+\log n=\mathcal O(n)$，所以总复杂度 $\mathcal O(\omega n)$，加上区间与和求答案，最终复杂度 $\mathcal O(q\log^2n+\omega n)$。

$\mathcal{Code}$

  用 template 实现线段树√

/*~Rainybunny~*/

#include <set>
#include <cmath>
#include <cstdio>
#include <cassert>
#include <cstring>

#define rep( i, l, r ) for ( int i = l, rep##i = r; i <= rep##i; ++i )
#define per( i, r, l ) for ( int i = r, per##i = l; i >= per##i; --i )

typedef __uint128_t UI;

inline UI rxint() {
    static char buf[100]; scanf( "%s", buf );
    UI ret = 0;
    for ( int i = 0; buf[i]; ++i ) {
        ret = ret << 4 | ( buf[i] <= '9' ? buf[i] ^ '0' : buf[i] - 'a' + 10 );
    }
    return ret;
}

inline int rdint() {
	int x = 0, f = 1, s = getchar();
	for ( ; s < '0' || '9' < s; s = getchar() ) f = s == '-' ? -f : f;
	for ( ; '0' <= s && s <= '9'; s = getchar() ) x = x * 10 + ( s ^ '0' );
	return x * f;
}

inline void wxint( const UI x ) {
    if ( x >= 16 ) wxint( x >> 4 );
    putchar( ( x & 15 ) >= 10 ? 'a' + ( x & 15 ) - 10 : '0' ^ ( x & 15 ) );
}

inline int imax( const int a, const int b ) { return a < b ? b : a; }

const int MAXN = 3e5, MAXQ = 2e5;
int n, q;

template<const int H>
struct Atom {
	UI val[H + 1];
	
	inline UI& operator [] ( const int k ) { return val[k]; }
	
	friend inline Atom<H + 1> operator + ( Atom<H>& u, Atom<H>& v ) {
		static Atom<H + 1> ret;
		ret[0] = u[0] ^ v[0]; UI up = u[0] & v[0];
		rep ( i, 1, H ) {
			ret[i] = up ^ u[i] ^ v[i];
			up = ( up & u[i] ) | ( up & v[i] ) | ( u[i] & v[i] );
		}
		ret[H + 1] = up;
		return ret;
	}
	
	inline Atom& operator &= ( const UI& x ) {
		rep ( i, 0, H ) val[i] &= x;
		return *this;
	}
	
	inline UI get() const {
		UI ret = 0;
		rep ( i, 0, H ) ret += val[i] << i;
		return ret;
	}
};

template<const int H>
struct SegmentTree {
	Atom<H> sum; bool zero; UI tag;
	SegmentTree<H - 1> lch, rch;
	
	inline void pushup() {
		sum = lch.sum + rch.sum, zero = lch.zero && rch.zero;
	}
	
	inline void pushan( const UI& v ) {
		tag &= v, sum &= v;
	}
	
	inline void pushdn() {
		if ( ~tag ) {
			lch.pushan( tag ), rch.pushan( tag );
			tag = ~UI( 0 );
		}
	}
	
	inline void build( const int l, const int r ) {
		int mid = l + r >> 1; tag = ~UI( 0 );
		lch.build( l, mid ), rch.build( mid + 1, r );
		pushup();
	}
	
	inline void secAnd( const int l, const int r,
	  const int ql, const int qr, const UI& v ) {
		if ( zero ) return ;
		if ( ql <= l && r <= qr ) return void( pushan( v ) );
		int mid = l + r >> 1; pushdn();
		if ( ql <= mid ) lch.secAnd( l, mid, ql, qr, v );
		if ( mid < qr ) rch.secAnd( mid + 1, r, ql, qr, v );
		pushup();
	}
	
	inline void secDiv( const int l, const int r,
	  const int ql, const int qr, const UI& v ) {
		if ( zero || v == 1 ) return ;
		if ( l == r ) return void( zero = !( sum[0] /= v ) );
		int mid = l + r >> 1; pushdn();
		if ( ql <= mid ) lch.secDiv( l, mid, ql, qr, v );
		if ( mid < qr ) rch.secDiv( mid + 1, r, ql, qr, v );
		pushup();
	}
	
	inline UI query( const int l, const int r, const int ql, const int qr ) {
	  	if ( zero ) return 0;
		if ( ql <= l && r <= qr ) return sum.get();
		int mid = l + r >> 1; UI ret = 0; pushdn();
		if ( ql <= mid ) ret += lch.query( l, mid, ql, qr );
		if ( mid < qr ) ret += rch.query( mid + 1, r, ql, qr );
		return ret;
	}
};

template<>
struct SegmentTree<0> {
	Atom<0> sum; bool zero; UI tag;
	
	inline void pushan( const UI& v ) { tag &= v, sum &= v; }
	
	inline void build( const int l, const int r ) {
		assert( l == r );
		tag = ~UI( 0 ), zero = !( sum[0] = r < n ? rxint() : 0 );
	}
	
	inline void secAnd( const int l, const int r,
	  const int ql, const int qr, const UI& v ) {
		if ( zero ) return ;
		assert( ql <= l && r <= qr ), pushan( v );
	}
	
	inline void secDiv( const int l, const int r,
	  const int ql, const int qr, const UI& v ) {
		if ( zero || v == 1 ) return ;
		zero = !( sum[0] /= v );
	}
	
	inline UI query( const int l, const int r, const int ql, const int qr ) {
	  	if ( zero ) return 0;
		return assert( ql <= l && r <= qr ), sum.get();
	}
};

SegmentTree<19> sgt;

int main() {
//	freopen( "machine.in", "r", stdin );
//	freopen( "machine.out", "w", stdout );
	
	n = rdint(), q = rdint();
	int R = 1 << 19;
	sgt.build( 0, R - 1 );
	
	for ( int op, l, r; q--; ) {
		op = rdint(), l = rdint() - 1, r = rdint() - 1;
		if ( op == 1 ) sgt.secDiv( 0, R - 1, l, r, rxint() );
		else if ( op == 2 ) sgt.secAnd( 0, R - 1, l, r, rxint() );
		else wxint( sgt.query( 0, R - 1, l, r ) ), putchar( '\n' );
	}
	return 0;
}