HDU 4578 transformation (segment tree + skillful lazy tag placement)

Yuanfang is puzzled with the question below:
There are n integers, a 1, a 2, …, a n. The initial values of them are 0. There are four kinds of operations.
Operation 1: Add c to each number between a x and a y inclusive. In other words, do transformation a k<—a k+c, k = x,x+1,…,y.
Operation 2: Multiply c to each number between a x and a y inclusive. In other words, do transformation a k<—a k×c, k = x,x+1,…,y.
Operation 3: Change the numbers between a x and a y to c, inclusive. In other words, do transformation a k<—c, k = x,x+1,…,y.
Operation 4: Get the sum of p power among the numbers between a x and a y inclusive. In other words, get the result of a x p+a x+1 p+…+a y 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
Their thinking ：
If you follow the direct recording method , The lazy mark of this question must be very complicated , But the time for this question is 8 second , The data is also very special , Starting data is 0. Now after every change we make , The value of that interval should also be the same , therefore , We can directly calculate the sum of operations of the same number in a range ,
sum=(r-l+1)*a[rt]^p; Be careful ： You should pay attention to lazy tags every time you update and sum .
Code ：

#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<map>
#include<queue>
#include<set>
#include<cmath>
#include<stack>
#include<string>
const int maxn=1e5+5;
const int mod=10007;
const int inf=1e9;
const long long onf=1e18;
#define me(a,b) memset(a,b,sizeof(a))
#define lowbit(x) x&(-x)
#define lson p*2,l,mid
#define rson p*2+1,mid+1,r
#define PI 3.14159265358979323846
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
int n,m;
struct segmenttree{
    int l,r;
    int dat,lasy;
}t[maxn<<2];
void pushup(int p){
    if(!t[p*2].lasy||!t[p*2+1].lasy)
        t[p].lasy=0;
    else if(t[p*2].dat!=t[p*2+1].dat)
        t[p].lasy=0;
    else {
        t[p].dat=t[p*2].dat=t[p*2+1].dat;
        t[p].lasy=1;
    }
}
void pushdown(int p){
  
    if(t[p].lasy){

        t[p*2].lasy=t[p*2+1].lasy=1;
         t[p*2].dat=t[p*2+1].dat=t[p].dat;
        t[p].lasy=0;
        
    }
}
void build(int p,int l,int r){
    t[p].l=l,t[p].r=r;
    if(l==r){
         t[p].dat=0,t[p].lasy=1;
        return ;
    }
    int mid=(l+r)>>1;
    build(lson);
    build(rson);
    pushup(p);
}
void change(int p,int l,int r,int op,int c){
    if(l<=t[p].l&&r>=t[p].r&&t[p].lasy){
        if(op==1) t[p].dat=(t[p].dat+c)%mod;
        if(op==2) t[p].dat=(t[p].dat*c)%mod;
        if(op==3) t[p].dat=c%mod;
        return ;

    }
   pushdown(p);
    int mid=(t[p].l+t[p].r)>>1;
    if(l<=mid) change(p*2,l,r,op,c);
    if(r>mid) change(p*2+1,l,r,op,c);
    pushup(p);

}
int ask(int p,int l,int r,int x){
    if(l<=t[p].l&&r>=t[p].r&&t[p].lasy)
       { int ans=1;
       for(int i=1;i<=x;i++)
            ans=(ans*t[p].dat)%mod;
        ans=(ans*(t[p].r-t[p].l+1))%mod;
        return ans;
       }
    int mid=(t[p].l+t[p].r)>>1;
    pushdown(p);
    int ans=0;
    if(l<=mid) ans=(ans+ask(p*2,l,r,x))%mod;
    if(r>mid) ans=(ans+ask(p*2+1,l,r,x))%mod;
    return ans%mod;

}
int main()
{   while(~scanf("%d %d",&n,&m),n+n){
        build(1,1,n);
        int p,x,y,z;
        for(int i=1;i<=m;i++){
            scanf("%d %d %d %d",&p,&x,&y,&z);
            if(p<=3)change(1,x,y,p,z);
            else printf("%d\n",ask(1,x,y,z));
        }
}
}