BZOJ 1798 Ahoi 2009 维护序列seq

标签：bzoj1798   bzoj   线段树   ahoi2009   

题目大意：维护一个序列，能够区间加，区间乘，然后取区间和模一个数的值。

思路：线段树维护一个有两个域的标记，一个表示加，一个表示乘。下传的时候一起下传，先乘后加。

CODE：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 100010
#define MO p
#define LEFT (pos << 1)
#define RIGHT (pos << 1|1)
#define CNT (r - l + 1)
using namespace std;

long long p;

struct Mark{
	long long p_mark,m_mark;
	
	Mark() {
		m_mark = 1;
		p_mark = 0;
	}
	void operator +=(long long c) {
		p_mark = (p_mark + c) % MO;
	}
	void operator *=(long long c) {
		p_mark = (p_mark * c) % MO;
		m_mark = (m_mark * c) % MO;
	}
	void operator +=(const Mark &a) {
		m_mark = (m_mark * a.m_mark) % MO;
		p_mark = (p_mark * a.m_mark + a.p_mark) % MO;
	}
};
struct Complex{
	long long sum;
	Mark mark;
}tree[MAX << 2];

int cnt,asks;
long long src[MAX];

void BuildTree(int l,int r,int pos);
void Mulitiply(int l,int r,int x,int y,int pos,long long c);
void Plus(int l,int r,int x,int y,int pos,long long c);
int Ask(int l,int r,int x,int y,int pos);

inline void PushDown(int pos,int cnt);

int main()
{
	cin >> cnt >> p;
	for(int i = 1;i <= cnt; ++i)
		scanf("%lld",&src[i]);
	BuildTree(1,cnt,1);
	cin >> asks;
	for(int flag,i = 1;i <= asks; ++i) {
		scanf("%d",&flag);
		int x,y,z;
		switch(flag) {
			case 1:
				scanf("%d%d%d",&x,&y,&z);
				Mulitiply(1,cnt,x,y,1,z);
				break;
			case 2:
				scanf("%d%d%d",&x,&y,&z);
				Plus(1,cnt,x,y,1,z);
				break;
			case 3:
				scanf("%d%d",&x,&y);
				printf("%d\n",(int)Ask(1,cnt,x,y,1));
				break;
		}
	}
	return 0;
}

void BuildTree(int l,int r,int pos)
{	
	tree[pos].mark.p_mark = 0;
	tree[pos].mark.m_mark = 1;
	if(l == r) {
		tree[pos].sum = src[l] % MO;
		return ;
	}
	int mid = (l + r) >> 1;
	BuildTree(l,mid,LEFT);
	BuildTree(mid + 1,r,RIGHT);
	tree[pos].sum = (tree[LEFT].sum + tree[RIGHT].sum) % MO;
}

void Mulitiply(int l,int r,int x,int y,int pos,long long c)
{
	if(l == x && y == r) {
		tree[pos].sum = (tree[pos].sum * c) % MO;
		tree[pos].mark *= c;
		return ;
	}
	PushDown(pos,CNT);
	int mid = (l + r) >> 1;
	if(y <= mid)	Mulitiply(l,mid,x,y,LEFT,c);
	else if(x > mid)	Mulitiply(mid + 1,r,x,y,RIGHT,c);
	else {
		Mulitiply(l,mid,x,mid,LEFT,c);
		Mulitiply(mid + 1,r,mid + 1,y,RIGHT,c);
	}
	tree[pos].sum = (tree[LEFT].sum + tree[RIGHT].sum) % MO;
}

void Plus(int l,int r,int x,int y,int pos,long long c)
{
	if(l == x && y == r) {
		tree[pos].sum = (tree[pos].sum + (c * CNT) % MO) % MO;
		tree[pos].mark += c;
		return ;
	}
	PushDown(pos,CNT);
	int mid = (l + r) >> 1;
	if(y <= mid)	Plus(l,mid,x,y,LEFT,c);
	else if(x > mid)	Plus(mid + 1,r,x,y,RIGHT,c);
	else {
		Plus(l,mid,x,mid,LEFT,c);
		Plus(mid + 1,r,mid + 1,y,RIGHT,c);
	}
	tree[pos].sum = (tree[LEFT].sum + tree[RIGHT].sum) % MO;
}

int Ask(int l,int r,int x,int y,int pos)
{
	if(l == x && y == r)
		return tree[pos].sum;
	PushDown(pos,CNT);
	int mid = (l + r) >> 1;
	if(y <= mid)	return Ask(l,mid,x,y,LEFT);
	if(x > mid)		return Ask(mid + 1,r,x,y,RIGHT);
	int left = Ask(l,mid,x,mid,LEFT);
	int right = Ask(mid + 1,r,mid + 1,y,RIGHT);
	return (left + right) % MO;
}

inline void PushDown(int pos,int cnt)
{
	if(tree[pos].mark.m_mark == 1 && !tree[pos].mark.p_mark)
		return ;
	tree[LEFT].mark += tree[pos].mark;
	tree[RIGHT].mark += tree[pos].mark;
	if(tree[pos].mark.m_mark != 1) {
		tree[LEFT].sum = (tree[LEFT].sum * tree[pos].mark.m_mark) % MO;
		tree[RIGHT].sum = (tree[RIGHT].sum * tree[pos].mark.m_mark) % MO;
		tree[pos].mark.m_mark = 1;
	}
	if(tree[pos].mark.p_mark) {
		tree[LEFT].sum = (tree[LEFT].sum + ((cnt - (cnt >> 1)) * tree[pos].mark.p_mark) % MO) % MO;
		tree[RIGHT].sum = (tree[RIGHT].sum + ((cnt >> 1) * tree[pos].mark.p_mark) % MO) % MO;
		tree[pos].mark.p_mark = 0;
	}
}

BZOJ 1798 Ahoi 2009 维护序列seq

标签：bzoj1798   bzoj   线段树   ahoi2009   

原文地址：http://blog.csdn.net/jiangyuze831/article/details/39576307