HDU4578-代码一点都不长的线段树

（有任何问题欢迎留言或私聊 && 欢迎交流讨论哦

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 \(a \Rightarrow a \times x + b\) 。

 $$sum1 = \sum x \Rightarrow \sum (a\times x + b) = sum1 \times a + b \times length$$

 $$sum2 = \sum x^2 \Rightarrow \sum (a\times x + b)^2 \Rightarrow \sum (a^2\times x^2+b2+2\times a\times b\times x)=sum2\times a^2+b2 \times length+2\times a\times b\times sum1

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

 \(lazy1 \Rightarrow lazy1\times a + b\)

 $lazy2 \Rightarrow lazy2\times 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\n", 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 akInput

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

Source

2013ACM-ICPC杭州赛区全国邀请赛