【COGS】1272 [AHOI2009] 行星序列线段树

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本文详细介绍了线段树在处理乘法和加法运算时的结合律应用，并通过实例演示了如何使用线段树进行序列操作。包括更新节点值和查询区间和的操作，适用于解决复杂的数据查询问题。

传送门：【COGS】1272 [AHOI2009] 行星序列

题目分析：线段树的乘法运算以及加法运算满足结合律。

1 t g c

sum[ o ] = sum[ o ] * c ;

add[ o ] = add[ o ] * c ;

mul[ o ] = mul[ o ] * c ;

2 t g c

sum[ o ] = sum[ o ] + c * ( r - l + 1 )

add[ o ] = add[ o ] + c ;

pushdown时

mul[ ls ] = mul[ ls ] * mul[ o ] ;

mul[ rs ] = mul[ rs ] * mul[ o ] ;

sum[ ls ] = sum[ ls ] * mul[ o ] + add[ o ] * ( m - l + 1 ) ;

sum[ rs ] = sum[ rs ] * mul[ o ] + add[ o ] * ( r - m ) ;

add[ ls ] = add[ ls ] + add[ o ] ;

add[ rs ] = add[ rs ] + add[ o ] ;

代码如下：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

#define REP( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )
#define FOR( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define REV( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )
#define CLR( a , x ) memset ( a , x , sizeof a )
#define CPY( a , x ) memcpy ( a , x , sizeof a )
#define ls ( o << 1 )
#define rs ( o << 1 | 1 )
#define lson ls , l , m
#define rson rs , m + 1 , r
#define mid ( ( l + r ) >> 1 )
#define root 1 , 1 , n
#define rt o , l , r

typedef long long LL ;

const int MAXN = 100005 ;

LL sum[MAXN << 2] , add[MAXN << 2] , mul[MAXN << 2] ;
int n , q , mod ;

void pushup ( int o ) {
	sum[o] = sum[ls] + sum[rs] ;
	if ( sum[o] >= mod ) sum[o] -= mod ;
}

void pushdown ( int o , int l , int r ) {
	if ( mul[o] != 1 ) {
		sum[ls] = sum[ls] * mul[o] % mod ;
		sum[rs] = sum[rs] * mul[o] % mod ;
		mul[ls] = mul[ls] * mul[o] % mod ;
		mul[rs] = mul[rs] * mul[o] % mod ;
		add[ls] = add[ls] * mul[o] % mod ;
		add[rs] = add[rs] * mul[o] % mod ;
		mul[o] = 1 ;
	}
	if ( add[o] ) {
		int m = mid ;
		sum[ls] = ( sum[ls] + add[o] * ( m - l + 1 ) ) % mod ;
		sum[rs] = ( sum[rs] + add[o] * ( r - m ) ) % mod ;
		add[ls] += add[o] ; if ( add[ls] >= mod ) add[ls] -= mod ;
		add[rs] += add[o] ; if ( add[rs] >= mod ) add[rs] -= mod ;
		add[o] = 0 ;
	}
}

void update ( int L , int R , int op , int v , int o , int l , int r ) {
	if ( L <= l && r <= R ) {
		if ( op == 1 ) {
			mul[o] = mul[o] * v % mod ;
			add[o] = add[o] * v % mod ;
			sum[o] = sum[o] * v % mod ;
		} else {
			add[o] += v ;
			if ( add[o] >= mod ) add[o] -= mod ;
			sum[o] = ( sum[o] + ( r - l + 1 ) * v ) % mod ;
		}
		return ;
	}
	int m = mid ;
	pushdown ( rt ) ;
	if ( L <= m ) update ( L , R , op , v , lson ) ;
	if ( m <  R ) update ( L , R , op , v , rson ) ;
	pushup ( o ) ;
}

int query ( int L , int R , int o , int l , int r ) {
	if ( L <= l && r <= R ) return sum[o] ;
	int m = mid ;
	pushdown ( rt ) ;
	if ( R <= m ) return query ( L , R , lson ) ;
	if ( m <  L ) return query ( L , R , rson ) ;
	return ( query ( L , R , lson ) + query ( L , R , rson ) ) % mod ;
}

void scanf ( int& x , char c = 0 ) {
	while ( ( c = getchar () ) < '0' || c > '9' ) ;
	x = c - '0' ;
	while ( ( c = getchar () ) >= '0' && c <= '9' ) x = x * 10 + c - '0' ;
}

void build ( int o , int l , int r ) {
	add[o] = 0 ;
	mul[o] = 1 ;
	if ( l == r ) {
		int x ;
		scanf ( x ) ;
		sum[o] = x ;
		return ;
	}
	int m = mid ;
	build ( lson ) ;
	build ( rson ) ;
	pushup ( o ) ;
}

void solve () {
	int op , L , R , v ;
	scanf ( n ) , scanf ( mod ) ;
	build ( root ) ;
	scanf ( q ) ;
	while ( q -- ) {
		scanf ( op ) , scanf ( L ) , scanf ( R ) ;
		if ( op < 3 ) {
			scanf ( v ) ;
			update ( L , R , op , v , root ) ;
		} else printf ( "%d\n" , query ( L , R , root ) ) ;
	}
}

int main () {
	freopen ( "seqb.in" , "r" , stdin ) ;
	freopen ( "seqb.out" , "w" , stdout ) ;
	solve () ;
	return 0 ;
}