\(\mathcal{Desciprtion}\)

给定序列 \(\{a_n\}\)，支持 \(q\) 次操作：

给定 \(l,r,v\)，\(\forall i\in[l,r],~a_i\leftarrow\lfloor\frac{a_i}{v}\rfloor\)；
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；
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;

}

巴特西

Solution -「UNR #5」「UOJ #671」诡异操作

\(\mathcal{Desciprtion}\)

\(\mathcal{Solution}\)

\(\mathcal{Code}\)

最新文章

热门文章