HDU4578-代码一点都不长的线段树

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题意：传送门

原题目描述在最下面。
4种操作，1：区间加法，2：区间乘法，3：区间的所有数都变成一个数，4：访问区间每个数的p次方和(1 <= p <= 3)。

思路：

三个lazy标记：lazy1表示区间加上的数的延迟，lazy2表示区间乘上的数的延迟，lazy3表示区间变成的那个数字。初始化lazy1 = lazy3 = 0, lazy2 = 1.
如果进行3号操作，则将lazy1和lazy2全部恢复初始值。
对于1，2号操作，假设原数是x，乘上a，加上b。则 (a Rightarrow a imes x + b) 。
$$sum1 = sum x Rightarrow sum (a imes x + b) = sum1 imes a + b imes length$$
$$sum2 = sum x^2 Rightarrow sum (a imes x + b)^2 Rightarrow sum (a^2 imes x^2+b2+2 imes a imes b imes x)=sum2 imes a^2+b2 imes length+2 imes a imes b imes sum1

[&emsp;$$sum3 = sum x^3 Rightarrow sum (a imes x + b)^3 Rightarrow sum (a^3 imes x^3 + b^3 + 3 imes a^2 imes x^2 imes b+3 imes a imes x imes b^2)= sum3 imes a^3 + b^3 imes length + 3 imes sum2 imes a^2 imes b+ 3 imes sum1 imes a imes b^2 ]

注意了，再进行更新sum操作时，要先更新sum3再更新sum2最后更新sum1。比如sum3式子中的sum2的含义是原始数列的sum2，不是更新后的sum2。

然后就是对于push_down中的lazy1和lazy2的处理。
(lazy1 Rightarrow lazy1 imes a + b)
$lazy2 Rightarrow lazy2 imes a $

(因为update哪里忘了写return;，导致debug一天)

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
#include<assert.h>
#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) (x)&(-(x))
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const int N = (int)1e5+7;
const int mod = 10007;
int n, q;
int ar[N];
struct lp{
  int l,r;
  LL sum1,sum2,sum3,d;
  LL lazy1,lazy2,lazy3;
}cw[N<<2];
//pushdown先乘除后加减
//加法 lazy1 lazy2 乘法
void push_up(int rt){
  cw[rt].sum1 = (cw[lson].sum1+cw[rson].sum1)%mod;
  cw[rt].sum2 = (cw[lson].sum2+cw[rson].sum2)%mod;
  cw[rt].sum3 = (cw[lson].sum3+cw[rson].sum3)%mod;
}
void build(int l,int r,int rt){
  int m=(l+r)>>1;
  cw[rt].l=l;cw[rt].r=r;
  cw[rt].d=r-l+1;
  cw[rt].sum1 = cw[rt].sum2 = cw[rt].sum3=0;
  cw[rt].lazy1 = cw[rt].lazy3 = 0;
  cw[rt].lazy2 = 1;
  if(l==r){return;}
  build(l,m,lson); build(m+1,r,rson);
  push_up(rt);
}
//pushdown先乘除后加减
void push_down(int rt){
  if(cw[rt].lazy3){
    LL tmp = cw[rt].lazy3*cw[rt].lazy3%mod*cw[rt].lazy3%mod;
    cw[lson].lazy1=cw[rson].lazy1=0;
    cw[lson].lazy2=cw[rson].lazy2=1;
    cw[lson].lazy3=cw[rson].lazy3=cw[rt].lazy3;

    cw[lson].sum3 = cw[lson].d*tmp%mod;
    cw[rson].sum3 = cw[rson].d*tmp%mod;

    cw[lson].sum2 = cw[lson].d*cw[rt].lazy3%mod*cw[rt].lazy3%mod;
    cw[rson].sum2 = cw[rson].d*cw[rt].lazy3%mod*cw[rt].lazy3%mod;

    cw[lson].sum1 = cw[lson].d*cw[rt].lazy3%mod;
    cw[rson].sum1 = cw[rson].d*cw[rt].lazy3%mod;

    
    cw[rt].lazy3 = 0;
  }
  if(cw[rt].lazy1!=0||cw[rt].lazy2!=1){
    LL add = cw[rt].lazy1, mul = cw[rt].lazy2;
    LL l1=cw[lson].sum1,l2=cw[lson].sum2,l3=cw[lson].sum3;
    LL r1=cw[rson].sum1,r2=cw[rson].sum2,r3=cw[rson].sum3;
    LL tmp = mul*mul%mod*mul%mod;

    cw[lson].lazy1=(cw[lson].lazy1*mul%mod+add)%mod;
    cw[rson].lazy1=(cw[rson].lazy1*mul%mod+add)%mod;
    cw[lson].lazy2=cw[lson].lazy2*mul%mod;
    cw[rson].lazy2=cw[rson].lazy2*mul%mod;

    cw[lson].sum3=(cw[lson].sum3*tmp%mod + add*add%mod*add%mod*cw[lson].d%mod + 3*cw[lson].sum2*mul%mod*mul%mod*add%mod + 3*cw[lson].sum1*mul%mod*add%mod*add%mod)%mod;   
    cw[rson].sum3=(cw[rson].sum3*tmp%mod + add*add%mod*add%mod*cw[rson].d%mod + 3*cw[rson].sum2*mul%mod*mul%mod*add%mod + 3*cw[rson].sum1*mul%mod*add%mod*add%mod)%mod;

    cw[lson].sum2=(cw[lson].sum2*mul%mod*mul%mod + add*add%mod*cw[lson].d%mod + 2*mul*add*cw[lson].sum1)%mod;
    cw[rson].sum2=(cw[rson].sum2*mul%mod*mul%mod + add*add%mod*cw[rson].d%mod + 2*mul*add*cw[rson].sum1)%mod;

    cw[lson].sum1=(cw[lson].sum1*mul+add*cw[lson].d)%mod;
    cw[rson].sum1=(cw[rson].sum1*mul+add*cw[rson].d)%mod;

    cw[rt].lazy1 = 0;cw[rt].lazy2 = 1;
  }
}
//op==1 加法, op==2 乘法, op==3 改变值
void update(int L,int R,int op,int z,int rt){
  int l=cw[rt].l, r=cw[rt].r, mid=(l+r)>>1;
  if(cw[rt].l>R||cw[rt].r<L)return;
  if(L<=l&&r<=R){
    LL l1 = cw[rt].sum1, l2 = cw[rt].sum2;
    if(op==1){//加法
      cw[rt].lazy1 = (z + cw[rt].lazy1)%mod;
      cw[rt].sum3 = (cw[rt].sum3 + (z*z%mod)*z%mod*cw[rt].d%mod + 3*z*(cw[rt].sum2+(cw[rt].sum1*z)%mod)%mod)%mod;
      cw[rt].sum2 = (cw[rt].sum2 + cw[rt].d*z%mod*z%mod + 2*z*cw[rt].sum1%mod)%mod;
      cw[rt].sum1 = (cw[rt].sum1 + cw[rt].d*z%mod)%mod;
    }else if(op==2){//乘法
      cw[rt].lazy1 = cw[rt].lazy1*z%mod;
      cw[rt].lazy2 = cw[rt].lazy2*z%mod;
      cw[rt].sum1 = cw[rt].sum1*z%mod;
      cw[rt].sum2 = (cw[rt].sum2*z%mod)*z%mod;
      cw[rt].sum3 = ((cw[rt].sum3*z%mod)*z%mod)*z%mod;
    }else{
      cw[rt].lazy1=0;cw[rt].lazy2=1;
      cw[rt].lazy3=z;
      cw[rt].sum3 = cw[rt].d*z%mod*z%mod*z%mod;
      cw[rt].sum2 = cw[rt].d*z%mod*z%mod;
      cw[rt].sum1 = cw[rt].d*z%mod;
    }
    return;
  }
  if(cw[rt].l==cw[rt].r)return;
  push_down(rt);
  if(L>mid)update(L,R,op,z,rson);
  else if(R<=mid)update(L,R,op,z,lson);
  else{
    update(L,mid,op,z,lson);
    update(mid+1,R,op,z,rson);
  }
  push_up(rt);
}
LL query(int L,int R,int op,int rt){
  int l=cw[rt].l, r=cw[rt].r, mid=(l+r)>>1;
  if(L<=l&&r<=R){
    if(op==1)return cw[rt].sum1;
    else if(op==2)return cw[rt].sum2;
    return cw[rt].sum3;
  }
  if(cw[rt].l==cw[rt].r)return 0;
  push_down(rt);
  LL sum = 0;
  if(L>mid) sum = query(L,R,op,rson);
  else if(R<=mid) sum = query(L,R,op,lson);
  else {
    sum = query(L,mid,op,lson);
    sum += query(mid+1,R,op,rson);
  }
  return (sum%mod);
}
int main(){
  while(~scanf("%d%d", &n,&q)&&(n+q)){
    build(1,n,1);
    while(q--){
      int op,l,r,z;
      scanf("%d%d%d%d",&op,&l,&r,&z);
      if(op<=3){
        z%=mod;
        update(l,r,op,z,1);
      }else{
        printf("%lld
", query(l,r,z,1));
      }
    }
  }
  return 0;
}

####原题目描述： Problem Description Yuanfang is puzzled with the question below: There are n integers, a1, a2, …, an. The initial values of them are 0. There are four kinds of operations. Operation 1: Add c to each number between ax and ay inclusive. In other words, do transformation ak<---ak+c, k = x,x+1,…,y. Operation 2: Multiply c to each number between ax and ay inclusive. In other words, do transformation ak<---ak×c, k = x,x+1,…,y. Operation 3: Change the numbers between ax and ay to c, inclusive. In other words, do transformation ak<---c, k = x,x+1,…,y. Operation 4: Get the sum of p power among the numbers between ax and ay inclusive. In other words, get the result of axp+ax+1p+…+ay p. Yuanfang has no idea of how to do it. So he wants to ask you to help him.

Input
There are no more than 10 test cases.
For each case, the first line contains two numbers n and m, meaning that there are n integers and m operations. 1 <= n, m <= 100,000.
Each the following m lines contains an operation. Operation 1 to 3 is in this format: "1 x y c" or "2 x y c" or "3 x y c". Operation 4 is in this format: "4 x y p". (1 <= x <= y <= n, 1 <= c <= 10,000, 1 <= p <= 3)
The input ends with 0 0.

Output
For each operation 4, output a single integer in one line representing the result. The answer may be quite large. You just need to calculate the remainder of the answer when divided by 10007.

Sample Input
5 5
3 3 5 7
1 2 4 4
4 1 5 2
2 2 5 8
4 3 5 3
0 0

Sample Output
307
7489