【ybt】【数据结构 线段树 课过 例4】维护序列

维护序列

题目链接：YbtOJ/Luogu

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解题思路

线段树模板题

code

#include<iostream>
#include<cstdio>
#define int long long
using namespace std;

int n,m,mod;

struct abc{
	int s;
	int flg1,flg2;
}tree[400010];

void build(int now,int l,int r)
{
	tree[now].flg1=1;
	if(l==r)
	{
		scanf("%lld",&tree[now].s);
		return;
	}
	int mid=(l+r)/2;
	build(now*2,l,mid);
	build(now*2+1,mid+1,r);
	tree[now].s=(tree[now*2].s+tree[now*2+1].s)%mod;
}

void down(abc &now,abc &l,abc &r,int x,int y)
{
	int mid=(x+y)/2;
	l.s=(l.s*now.flg1+(mid-x+1)*now.flg2)%mod;
	r.s=(r.s*now.flg1+(y-mid)*now.flg2)%mod;
	l.flg1=(l.flg1*now.flg1)%mod;
	l.flg2=(l.flg2*now.flg1+now.flg2)%mod;
	r.flg1=(r.flg1*now.flg1)%mod;
	r.flg2=(r.flg2*now.flg1+now.flg2)%mod;
	now.flg1=1;
	now.flg2=0;
}

void change1(int now,int l,int r,int x,int y,int w)
{
	if(x<=l&&r<=y)
	{
		tree[now].s=(tree[now].s*w)%mod;
		tree[now].flg1=(tree[now].flg1*w)%mod;
		tree[now].flg2=(tree[now].flg2*w)%mod;
		return;
	}
	int mid=(l+r)/2;
	down(tree[now],tree[now*2],tree[now*2+1],l,r);
	if(x<=mid)
		change1(now*2,l,mid,x,y,w);
	if(y>mid)
		change1(now*2+1,mid+1,r,x,y,w);
	tree[now].s=(tree[now*2].s+tree[now*2+1].s)%mod; 
}

void change2(int now,int l,int r,int x,int y,int w)
{
	if(x<=l&&r<=y)
	{
		tree[now].s=(tree[now].s+(r-l+1)*w)%mod;
		tree[now].flg2=(tree[now].flg2+w)%mod;
		return;
	}
	int mid=(l+r)/2;
	down(tree[now],tree[now*2],tree[now*2+1],l,r);
	if(x<=mid)
		change2(now*2,l,mid,x,y,w);
	if(y>mid)
		change2(now*2+1,mid+1,r,x,y,w);
	tree[now].s=(tree[now*2].s+tree[now*2+1].s)%mod; 
}

int fd(int now,int l,int r,int x,int y)
{
	if(x<=l&&r<=y)
		return tree[now].s;
	int mid=(l+r)/2,ans=0;
	down(tree[now],tree[now*2],tree[now*2+1],l,r);
	if(x<=mid)
		ans=(ans+fd(now*2,l,mid,x,y))%mod;
	if(y>mid)
		ans=(ans+fd(now*2+1,mid+1,r,x,y))%mod;
	return ans;
}

signed main()
{
	cin>>n>>mod;
	build(1,1,n);
	cin>>m;
	while(m--)
	{
		int k,t,g,c;
		scanf("%lld%lld%lld",&k,&t,&g);
		if(k==1)
		{
			scanf("%lld",&c);
			change1(1,1,n,t,g,c);
		}
		if(k==2)
		{
			scanf("%lld",&c);
			change2(1,1,n,t,g,c);
		}
		if(k==3)
			cout<<fd(1,1,n,t,g)<<endl;
	}
}