\(\mathcal{Description}\)

维护一列二元组 \((a,b)\)，给定初始 \(n\) 个元素，接下来 \(m\) 次操作：

在某个位置插入一个二元组；
翻转一个区间；
区间 \(a\) 值加上一个数；
区间 \(a\) 值乘上一个数；
区间 \(a\) 值赋为一个数；
询问 \(\sum_{i=l}^r\sum_{j=i}^ra_j^3\bmod10086001\)。

特别地，若区间操作指名类型为 \(1\)，则需要将输入的左端点替换为输入区间内次大二元组 \((b_i,i)\) 的 \(i\)（二元组传统偏序关系比较；保证存在）。

\(n,m\le2\times10^5\)。

\(\mathcal{Solution}\)

出题人都写不对的码农题系列。（

很明显 \(a\) 的维护和 \(b\) 的维护完全不相关，先考虑处理 \(a\) 上的操作。区间加法和一个次数不算高的幂和，提示我们暴力维护 \(0\sim3\) 次幂和来处理修改。所以维护一个阶梯幂和（值 \(\times\)下标，应对“后缀和之和”的询问），和一个普通幂和（乘若干倍后与前者相减就能做到翻转区间）。

对于 \(b\)，直接维护次大值相关信息非常麻烦。可以考虑仅维护最大值，然后利用非旋 Treap 的操作处理。首先，找到区间最大值，把树裂乘最大值左侧树和最大值右侧树，分别在两棵树里找最大值，显然次大值比为其中之一。

于是 \(\mathcal O((n+m)\log(n+m))\) 就口胡完了，至于 \(8K+\) 的码量嘛……/xyx

\(\mathcal{Code}\)

/* Clearink */

#include <cstdio>

#include <cassert>

#include <cstdlib>

#include <iostream>

#include <algorithm>

#define rep( i, l, r ) for ( int i = l, repEnd##i = r; i <= repEnd##i; ++i )

#define per( i, r, l ) for ( int i = r, repEnd##i = l; i >= repEnd##i; --i )

inline int rint () {

	int x = 0, f = 1; char 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;

}

template<typename Tp>

inline void wint ( Tp x ) {

	if ( x < 0 ) putchar ( '-' ), x = -x;

	if ( 9 < x ) wint ( x / 10 );

	putchar ( x % 10 ^ '0' );

}

const int MAXN = 4e5, MOD = 10086001, INF = ( 1ll << 31 ) - 1;

int n, q, a[MAXN + 5];

inline void iswp ( int& a, int& b ) { a ^= b ^= a ^= b; }

inline void muleq ( int& a, const int b ) { a = 1ll * a * b % MOD; }

inline int mul ( const long long a, const int b ) { return a * b % MOD; }

inline int sub ( int a, const int b ) { return ( a -= b ) < 0 ? a + MOD : a; }

inline int add ( int a, const int b ) { return ( a += b ) < MOD ? a : a - MOD; }

inline void addeq ( int& a, const int b ) { ( a += b ) >= MOD && ( a -= MOD, 0 ); }

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

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

inline int sqr ( const int n ) { return ( n * ( n + 1ll ) >> 1 ) % MOD; }

struct Cube {

	int pwr[3];

	Cube (): pwr {} {}

	Cube ( const int v ): pwr { v, mul ( v, v ), mul ( v, mul ( v, v ) ) } {}

	Cube ( const int v1, const int v2, const int v3 ): pwr { v1, v2, v3 } {}

	inline int& operator [] ( const int k ) { return pwr[k]; }

	inline Cube operator + ( Cube c ) const {

		return {

			add ( pwr[0], c[0] ), add ( pwr[1], c[1] ), add ( pwr[2], c[2] )

		};

	}

	inline Cube operator - ( Cube& c ) const {

		return {

			sub ( pwr[0], c[0] ), sub ( pwr[1], c[1] ), sub ( pwr[2], c[2] )

		};

	}

	inline Cube operator * ( const int c ) const {

		return {

			mul ( pwr[0], c ), mul ( pwr[1], c ), mul ( pwr[2], c )

		};

	}

	inline Cube operator * ( Cube c ) const {

		return {

			mul ( pwr[0], c[0] ), mul ( pwr[1], c[1] ), mul ( pwr[2], c[2] )

		};

	}

};

struct NRTreap {

	int node, root, ch[MAXN + 5][2], key[MAXN + 5], siz[MAXN + 5];

	int a[MAXN + 5], b[MAXN + 5], zb[MAXN + 5];

	Cube sum[MAXN + 5], lad[MAXN + 5];

	int asgt[MAXN + 5], mult[MAXN + 5], addt[MAXN + 5];

	bool revt[MAXN + 5];

	NRTreap (): zb { -INF } { srand ( 20120712 ); }

	inline int crtnd ( const int va, const int vb ) {

		int u = ++node;

		key[u] = rand (), siz[u] = 1;

		lad[u] = sum[u] = a[u] = va;

		zb[u] = b[u] = vb;

		asgt[u] = INF, mult[u] = 1, addt[u] = revt[u] = 0;

		return u;

	}

	inline void pushrv ( const int x ) {

		revt[x] ^= 1, ch[x][0] ^= ch[x][1] ^= ch[x][0] ^= ch[x][1];

		lad[x] = sum[x] * ( siz[x] + 1 ) - lad[x];

	}

	inline void pushas ( const int x, const int v ) {

		a[x] = asgt[x] = v, mult[x] = 1, addt[x] = 0;

		sum[x] = Cube ( v ) * siz[x], lad[x] = Cube ( v ) * sqr ( siz[x] );

	}

	inline void pushmu ( const int x, const int v ) {

		muleq ( a[x], v ), muleq ( mult[x], v ), muleq ( addt[x], v );

		sum[x] = sum[x] * Cube ( v );

		lad[x] = lad[x] * Cube ( v );

	}

	inline void pushad ( const int x, const int v1 ) {

		int v2 = mul ( v1, v1 ), v3 = mul ( v2, v1 );

		addeq ( a[x], v1 ), addeq ( addt[x], v1 );

		addeq ( lad[x][2], mul ( sqr ( siz[x] ), v3 ) );

		addeq ( lad[x][2], mul ( mul ( 3, v2 ), lad[x][0] ) );

		addeq ( lad[x][2], mul ( mul ( 3, v1 ), lad[x][1] ) );

		addeq ( lad[x][1], mul ( sqr ( siz[x] ), v2 ) );

		addeq ( lad[x][1], mul ( mul ( 2, v1 ), lad[x][0] ) );

		addeq ( lad[x][0], mul ( sqr ( siz[x] ), v1 ) );

		addeq ( sum[x][2], mul ( siz[x], v3 ) );

		addeq ( sum[x][2], mul ( mul ( 3, v2 ), sum[x][0] ) );

		addeq ( sum[x][2], mul ( mul ( 3, v1 ), sum[x][1] ) );

		addeq ( sum[x][1], mul ( siz[x], v2 ) );

		addeq ( sum[x][1], mul ( mul ( 2, v1 ), sum[x][0] ) );

		addeq ( sum[x][0], mul ( siz[x], v1 ) );

	}

	inline void pushdn ( const int x ) {

		if ( revt[x] ) {

			if ( ch[x][0] ) pushrv ( ch[x][0] );

			if ( ch[x][1] ) pushrv ( ch[x][1] );

			revt[x] = 0;

		}

		if ( asgt[x] != INF ) {

			if ( ch[x][0] ) pushas ( ch[x][0], asgt[x] );

			if ( ch[x][1] ) pushas ( ch[x][1], asgt[x] );

			asgt[x] = INF;

		}

		if ( mult[x] != 1 ) {

			if ( ch[x][0] ) pushmu ( ch[x][0], mult[x] );

			if ( ch[x][1] ) pushmu ( ch[x][1], mult[x] );

			mult[x] = 1;

		}

		if ( addt[x] ) {

			if ( ch[x][0] ) pushad ( ch[x][0], addt[x] );

			if ( ch[x][1] ) pushad ( ch[x][1], addt[x] );

			addt[x] = 0;

		}

	}

	inline void pushup ( const int x ) {

		siz[x] = siz[ch[x][0]] + siz[ch[x][1]] + 1;

		zb[x] = imax ( b[x], imax ( zb[ch[x][0]], zb[ch[x][1]] ) );

		sum[x] = sum[ch[x][0]] + a[x] + sum[ch[x][1]];

		lad[x] = lad[ch[x][0]] + Cube ( a[x] ) * ( siz[ch[x][0]] + 1 )

			+ lad[ch[x][1]] + sum[ch[x][1]] * ( siz[ch[x][0]] + 1 );

	}

	inline int merge ( const int x, const int y ) {

		if ( !x || !y ) return x | y;

		pushdn ( x ), pushdn ( y );

		if ( key[x] < key[y] ) {

			ch[x][1] = merge ( ch[x][1], y ), pushup ( x );

			return x;

		} else {

			ch[y][0] = merge ( x, ch[y][0] ), pushup ( y );

			return y;

		}

	}

	inline void rsplit ( const int r, const int k, int& x, int& y ) {

		if ( !r ) return void ( x = y = 0 );

		pushdn ( r );

		if ( k <= siz[ch[r][0]] ) y = r, rsplit ( ch[r][0], k, x, ch[r][0] );

		else x = r, rsplit ( ch[r][1], k - siz[ch[r][0]] - 1, ch[r][1], y );

		pushup ( r );

	}

	inline void insert ( const int p, const int va, const int vb ) {

		int x, y; rsplit ( root, p, x, y );

		root = merge ( x, merge ( crtnd ( va, vb ), y ) );

	}

#define extract() ( rsplit ( root, l - 1, x, y ), rsplit ( y, r - l + 1, y, z ) )

	inline void reverse ( const int l, const int r ) {

		int x, y, z; extract ();

		if ( y ) pushdn ( y ), pushrv ( y );

		root = merge ( x, merge ( y, z ) );

	}

	inline void addsec ( const int l, const int r, const int v ) {

		int x, y, z; extract ();

		if ( y ) pushdn ( y ), pushad ( y, v );

		root = merge ( x, merge ( y, z ) );

	}

	inline void mulsec ( const int l, const int r, const int v ) {

		int x, y, z; extract ();

		if ( y ) pushdn ( y ), pushmu ( y, v );

		root = merge ( x, merge ( y, z ) );

	}

	inline void asgsec ( const int l, const int r, const int v ) {

		int x, y, z; extract ();

		if ( y ) pushdn ( y ), pushas ( y, v );

		root = merge ( x, merge ( y, z ) );

	}

	inline int query ( const int l, const int r ) {

		int x, y, z, ret; extract ();

		ret = lad[y][2], root = merge ( x, merge ( y, z ) );

		return ret;

	}

#undef extract

	inline int maxrk ( const int r ) {

		int u = r, mx = zb[r], ret = 0;

		while ( u ) {

			pushdn ( u );

			if ( zb[ch[u][0]] == mx ) u = ch[u][0];

			else {

				ret += siz[ch[u][0]] + 1;

				if ( b[u] == mx ) return ret;

				else u = ch[u][1];

			}

		}

		return assert ( false ), -1;

	}

	inline int smaxrk ( const int l, const int r ) {

		int p, x, y, z, q, ret;

		rsplit ( root, l - 1, p, x );

		rsplit ( x, r - l + 1, x, q );

		rsplit ( x, maxrk ( x ) - 1, x, y );

		rsplit ( y, 1, y, z );

		assert ( siz[y] == 1 && ( zb[x] > -INF || zb[z] > -INF ) );

		if ( zb[x] >= zb[z] ) ret = l - 1 + maxrk ( x );

		else ret = l - 1 + siz[x] + 1 + maxrk ( z );

		root = merge ( p, merge ( x, merge ( y, merge ( z, q ) ) ) );

		#ifdef RYBY

			printf ( "s(%d,%d)=%d\n", l, r, ret );

		#endif

		return ret;

	}

} trp;

int main () {

	n = rint (), q = rint ();

	rep ( i, 1, n ) a[i] = rint () % MOD;

	rep ( i, 1, n ) {

		trp.root = trp.merge ( trp.root, trp.crtnd ( a[i], rint () ) );

	}

	for ( int tp, op, l, r; q--; ) {

		tp = rint (), op = rint (), l = rint (), r = rint ();

		if ( op == 1 ) { trp.insert ( l, ( r + MOD ) % MOD, rint () ); continue; }

		if ( tp ) l = trp.smaxrk ( l, r );

		if ( op == 2 ) { trp.reverse ( l, r ); continue; }

		if ( op == 3 ) { trp.addsec ( l, r, ( rint () + MOD ) % MOD ); continue; }

		if ( op == 4 ) { trp.mulsec ( l, r, ( rint () + MOD ) % MOD ); continue; }

		if ( op == 5 ) { trp.asgsec ( l, r, ( rint () + MOD ) % MOD ); continue; }

		wint ( trp.query ( l, r ) ), putchar ( '\n' );

	}

	return 0;

}

/*

	1 2 3 14 15 16 7 8 9 10 ||| 3 2 5 7 9 1 4 2 7 10

 */

巴特西

Solution -「LOCAL」ZB 平衡树

\(\mathcal{Description}\)

\(\mathcal{Solution}\)

\(\mathcal{Code}\)

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