洛谷 P2023 维护序列——线段树

先上一波题目 https://www.luogu.org/problem/P2023

复习了一波线段树 题目涉及的操作有区间加 区间乘以及区间求和

tips：线段树在传标记的时候 优先传乘法标记再传加法标记 规定好顺序就不会错了

#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
using namespace std;
const int M=;
LL read(){
    LL ans=,f=,c=getchar();
    while(c<''||c>''){if(c=='-') f=-; c=getchar();}
    while(c>=''&&c<=''){ans=ans*+(c-''); c=getchar();}
    return ans*f;
}
int n,L,R;
LL C,mod;
struct node{
    int l,r;
    LL ad,mx,sum;
}e[M*];
void up(int x){e[x].sum=(e[x<<].sum+e[x<<^].sum)%mod;}
void build(int x,int l,int r){
    e[x].l=l; e[x].r=r;
    e[x].ad=; e[x].mx=;
    if(l==r){e[x].sum=read()%mod; return ;}
    int mid=(l+r)>>;
    build(x<<,l,mid);
    build(x<<^,mid+,r);
    up(x);
}
void down(int x){
    LL ad=e[x].ad,mx=e[x].mx;
    LL ls=x<<,rs=x<<^,len1=e[ls].r-e[ls].l+,len2=e[rs].r-e[rs].l+;
    if(mx!=){
        e[ls].sum=e[ls].sum*mx%mod;
        e[ls].ad=e[ls].ad*mx%mod;
        e[ls].mx=e[ls].mx*mx%mod;
        e[rs].sum=e[rs].sum*mx%mod;
        e[rs].ad=e[rs].ad*mx%mod;
        e[rs].mx=e[rs].mx*mx%mod;
        e[x].mx=;
    }
    if(ad){
        e[ls].sum=(e[ls].sum+ad*len1%mod)%mod;
        e[ls].ad=(e[ls].ad+ad)%mod;
        e[rs].sum=(e[rs].sum+ad*len2%mod)%mod;
        e[rs].ad=(e[rs].ad+ad)%mod;
        e[x].ad=;
    }
}
void dmx(int x){
    if(L<=e[x].l&&e[x].r<=R){
        e[x].sum=e[x].sum*C%mod;
        e[x].ad=e[x].ad*C%mod;
        e[x].mx=e[x].mx*C%mod;
        return ;
    }
    down(x);
    int mid=(e[x].l+e[x].r)>>;
    if(L<=mid) dmx(x<<);
    if(R>mid)  dmx(x<<^);
    up(x);
}
void dadd(int x){
    if(L<=e[x].l&&e[x].r<=R){
        LL len=e[x].r-e[x].l+;
        e[x].sum=(e[x].sum+C*len%mod)%mod;
        e[x].ad=(e[x].ad+C)%mod;
        return ;
    }
    down(x);
    int mid=(e[x].l+e[x].r)>>;
    if(L<=mid) dadd(x<<);
    if(R>mid)  dadd(x<<^);
    up(x);
}
LL push(int x){
    if(L<=e[x].l&&e[x].r<=R) return e[x].sum;
    down(x);
    int mid=(e[x].l+e[x].r)>>;
    LL ans=;
    if(L<=mid) ans=(ans+push(x<<))%mod;
    if(R>mid) ans=(ans+push(x<<^))%mod;
    up(x);
    return ans;
}
int k,m;
int main(){
    n=read(); mod=read();
    build(,,n);
    m=read();
    for(int i=;i<=m;i++){
        k=read(); L=read(); R=read();
        if(k==) C=read()%mod,dmx();
        else if(k==) C=read()%mod,dadd();
        else printf("%lld\n",push());
    }
    return ;
}