HFUUOJ1024 动态开点线段树+标记永久化

Posted 2022-10-17 xyq0220

题意

分析

动态加点线段树，标记永久化好写常数小

Code

#include<bits/stdc++.h>
#define fi first
#define se second
#define lson l,mid,p<<1
#define rson mid+1,r,p<<1|1
#define pb push_back
#define ll long long
using namespace std;
const ll inf=1e18;
const int mod=1e9+7;
const int maxn=1e7+10;
int n,m;
ll tr[maxn];
int ls[maxn],rs[maxn],tag[maxn],tot;
int rt;
void up(ll x,ll y,ll l,ll r,int &p,ll k)
    tr[++tot]=(tr[p]+(y-x+1)%mod*k%mod)%mod;
    ls[tot]=ls[p],rs[tot]=rs[p],tag[tot]=tag[p],p=tot;
    if(x==l&&y==r)
        tag[p]=(tag[p]+k)%mod;
        return;
    ll mid=l+r>>1;
    if(y<=mid) up(x,y,l,mid,ls[p],k);
    else if(x>mid) up(x,y,mid+1,r,rs[p],k);
    else up(x,mid,l,mid,ls[p],k),up(mid+1,y,mid+1,r,rs[p],k);

ll qy(ll x,ll y,ll l,ll r,int p,ll k)
    if(x==l&&y==r)
        return (tr[p]+(y-x+1)%mod*k%mod)%mod;
    ll mid=l+r>>1;
    if(y<=mid) return qy(x,y,l,mid,ls[p],(k+tag[p])%mod);
    else if(x>mid) return qy(x,y,mid+1,r,rs[p],(k+tag[p])%mod);
    else return (qy(x,mid,l,mid,ls[p],(k+tag[p])%mod)+qy(mid+1,y,mid+1,r,rs[p],(k+tag[p])%mod))%mod;

int main()
    //ios::sync_with_stdio(false);
    //freopen("in","r",stdin);
    cin>>m;
    for(int i=1,op;i<=m;i++)
        ll x,l,r;scanf("%d%lld%lld",&op,&l,&r);
        if(op==1)
            scanf("%lld",&x);
            x%=mod;
            up(l,r,1,inf,rt,x);
        else
            printf("%lld\n",qy(l,r,1,inf,rt,0));
        
    
    return 0;