HDU4578 线段树(区间更新 + 多种操作)和平方，立方

参考：https://www.cnblogs.com/H-Vking/p/4297973.html

题意：

虽然是比较裸的线段树，但是比较麻烦，并且有很多细节需要考虑，对着别人的ac代码debug了一个晚上。纪念一下

　这道题坑在有三种询问：set , add , mul。所以lazy标记要有三个，如果三个标记同时出现的处理方法——当更新set操作时，就把add标记和mul标记全部取消；当更新mul操作时，如果当前节点add标记存在，就把add标记改为：add * mul。这样的话就可以在PushDown()操作中先执行set，然后mul，最后add。

　　麻烦在有三种询问：和 , 平方和 , 立方和。对于set和mul操作来说，这三种询问都比较好弄。

　　对于add操作，和的话就比较好弄，按照正常方法就可以；

　　平方和这样来推：(a + c)^{2 =} a^{2 +} c² + 2ac  , 即sum2[rt] = sum2[rt] + (r - l + 1) * c * c + 2 * sum1[rt] * c；

　　立方和这样推：(a + c)^{3 =} a³ + c³ + 3a(a² + ac) , 即sum3[rt] = sum3[rt] + (r - l + 1) * c * c * c + 3 * c * (sum2[rt] + sum1[rt] * c)；

　　几个注意点：add标记取消的时候是置0，mul标记取消的时候是置1；在PushDown()中也也要注意取消标记，如set操作中取消add和mul，mul操作中更新add； 在add操作中要注意sum3 , sum2 , sum1的先后顺序，一定是先sum3 , 然后sum2 , 最后sum1； int容易爆，还是用LL要保险一点； 最后就是运算较多，不要漏掉东西。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <list>
#include <iterator>

using namespace std;
#define lson l , mid , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define debug(x) cerr << #x << " = " << x << "\n";

typedef long long ll;
const int maxn = 1e5+;
const int mdd = ;

struct node {
    int l,r;
    ll sum1,sum2,sum3;
    ll lazy1,lazy2,lazy3;
}st[maxn<<];

void allmdd(int rt)
{
    st[rt].sum1 = st[rt].sum1%mdd;
    st[rt].sum2 = st[rt].sum2%mdd;
    st[rt].sum3 = st[rt].sum3%mdd;
}

void pushup(int rt)
{
    st[rt].sum1  = (st[rt<<].sum1 + st[rt<<|].sum1)%mdd;
    st[rt].sum2  = (st[rt<<].sum2 + st[rt<<|].sum2)%mdd;
    st[rt].sum3  = (st[rt<<].sum3 + st[rt<<|].sum3)%mdd;
}

void pushdown(int rt)
{
    int len  = st[rt].r - st[rt].l + ;
    int llen = len - (len>>);
    int rlen = len>>;
    if(st[rt].lazy3)
    {
        ll c = st[rt].lazy3%mdd;
        st[rt<<].lazy3 = st[rt<<|].lazy3 = st[rt].lazy3;
        st[rt<<].lazy1 = st[rt<<|].lazy1 =  ;
        st[rt<<].lazy2 = st[rt<<|].lazy2 =  ;

        st[rt<<].sum1 = (llen * (c%mdd))%mdd;
        st[rt<<|].sum1 = (rlen * c)%mdd;

        st[rt<<].sum2 = (llen * (c * c)%mdd ) %mdd;
        st[rt<<|].sum2 = ((rlen * ((c * c)%mdd))%mdd)%mdd;

        st[rt<<].sum3 = (llen * ((c * c)%mdd * (c % mdd))%mdd)%mdd;
        st[rt<<|].sum3 = (rlen * ((c * c)%mdd * (c % mdd))%mdd)%mdd;
        st[rt].lazy3 = ;
    }

    if(st[rt].lazy2 !=  )
    {
        ll c = st[rt].lazy2 % mdd;
        st[rt<<].lazy2 = (st[rt<<].lazy2 * c) % mdd;
        st[rt<<|].lazy2 = (st[rt<<|].lazy2 * c) % mdd;
        if(st[rt<<].lazy1){
            st[rt<<].lazy1 = (st[rt<<].lazy1 * c)%mdd;
        }
        if(st[rt<<|].lazy1){
            st[rt<<|].lazy1 = (st[rt<<|].lazy1 * c)%mdd;
        }
        st[rt<<].sum1 = (st[rt<<].sum1 * c)%mdd;
        st[rt<<|].sum1 = (st[rt<<|].sum1 * c)%mdd;

        st[rt<<].sum2 = ((st[rt<<].sum2 * c%mdd)*c)%mdd;
        st[rt<<|].sum2 = ((st[rt<<|].sum2 * c%mdd)*c)%mdd;

        st[rt<<].sum3 = ((((st[rt<<].sum3 * c)%mdd)*c%mdd)*c)%mdd;
        st[rt<<|].sum3 = ((((st[rt<<|].sum3 * c)%mdd)*c%mdd)*c)%mdd;
        st[rt].lazy2 = ;
    }

    if(st[rt].lazy1)
    {
        ll c = st[rt].lazy1;
        st[rt<<].lazy1  = (st[rt<<].lazy1  + c)%mdd;
        st[rt<<|].lazy1  = (st[rt<<|].lazy1  +c)%mdd;
        //注意sum3,sum2,sum1 的顺序;
        st[rt<<].sum3  = st[rt<<].sum3 + ((c%mdd)*c%mdd)*c%mdd*llen + *c*((st[rt<<].sum2 + st[rt<<].sum1 * c)%mdd);
        st[rt<<|].sum3  = st[rt<<|].sum3 + ((c%mdd)*c%mdd)*c%mdd*rlen +*c*((st[rt<<|].sum2 + st[rt<<|].sum1 * c)%mdd);

        st[rt<<].sum2 =  st[rt<<].sum2 + (((llen *c)%mdd) * c)%mdd + *(st[rt<<].sum1)%mdd*c;
        st[rt<<|].sum2 =  st[rt<<|].sum2 + (((rlen *c)%mdd) * c)%mdd + *(st[rt<<|].sum1*c)%mdd;

        st[rt<<].sum1 = st[rt<<].sum1 + (llen*c)%mdd;
        st[rt<<|].sum1 = st[rt<<|].sum1 + (rlen*c)%mdd;

        allmdd(rt<<);allmdd(rt<<|);
        st[rt].lazy1 = ;
    }
}

void build(int l,int r,int rt)
{
    st[rt].sum1 = st[rt].sum2 = st[rt].sum3 = ;
    st[rt].lazy1 = st[rt].lazy3 = ;
    st[rt].lazy2 = ;
    st[rt].l = l,st[rt].r = r;
    int mid = (l + r) >> ;
    if(l == r)return;
    build(lson);
    build(rson);
    pushup(rt);
}
void update(int l,int r,int rt,int L,int R,int c,int op)
{
    if(l >= L&&r <= R)
    {
        if(op==)
        {
            st[rt].lazy1 += c;
            st[rt].sum3  =  (st[rt].sum3 + ((c*c%mdd)*c)%mdd*(r-l+)%mdd + (*c*(st[rt].sum2 + st[rt].sum1*c)%mdd))%mdd;
            st[rt].sum2 =    (st[rt].sum2 + (c*c)%mdd*(r-l+)%mdd + ( * st[rt].sum1 * c)%mdd)%mdd;
            st[rt].sum1 =   (st[rt].sum1 + (r-l+)*c%mdd)%mdd;
        }
        else if(op==)
        {
            st[rt].lazy2 = (st[rt].lazy2 * c)%mdd;
            if(st[rt].lazy1)
                st[rt].lazy1 = (st[rt].lazy1 * c)%mdd;
            st[rt].sum1 = (st[rt].sum1 * c %mdd)%mdd;
            st[rt].sum2 = (st[rt].sum2 * ((c * c) %mdd))%mdd;
            st[rt].sum3 = (st[rt].sum3 * (((c * c)%mdd  * c) % mdd))%mdd;
        }
        else if(op==)
        {
            st[rt].lazy3 = c;
            st[rt].lazy1 = ;
            st[rt].lazy2 = ;
            st[rt].sum1 = (((r-l+)*c )% mdd);
            st[rt].sum2 = (((((r-l+)* c)%mdd )* c %mdd))%mdd;
            st[rt].sum3 = ((r-l+)* (((c * c)%mdd  * c) % mdd));
        }
        return;
    }

    int mid = (l+r)>>;
    pushdown(rt);
    if(L<=mid)update(lson,L,R,c,op);
    if(R>mid)update(rson,L,R,c,op);
    pushup(rt);
}
ll query(int l,int r,int rt,int L,int R,int c)
{
    if(l>=L&&r<=R)
    {
        if(c == )return st[rt].sum1%mdd;
        else if(c == )return st[rt].sum2%mdd;
        else return st[rt].sum3%mdd;
    }
    int mid = (l+r)>>;
    ll ans = ;
    pushdown(rt);
    if(L<=mid)ans = (ans + query(lson,L,R,c))%mdd;
    if(R>mid)ans = (ans + query(rson,L,R,c))%mdd;
    return ans;
}
int main(){
    // freopen("input","r",stdin);
     int n,m;
     while(scanf("%d%d", &n, &m)==&&n+m)
     {
         // debug(n);
         build(,n,);
         while(m--)
         {
             int op;
             int a,b,c;
             scanf("%d%d%d%d", & op, & a,& b, & c);
             if(op==){
                 printf("%lld\n", query(,n,,a,b,c)%mdd);
             }
             else {
                 update(,n,,a,b,c,op);
             }
         }
     }
    return ;
}