HDU 4578 Transformation（线段树+有技巧的懒标记下放）

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
解题思路：
要是按照直接记录的放法，这个题的懒标记肯定非常复杂，但是这个题的时间给了8秒，数据也很特别，起始数据都为0。现在我们每次更改过后，那段区间的值也应该是相同的，所以，我们直接可以计算一段区间相同的数的操作的和，
sum=(r-l+1)*a[rt]^p;注意：每次更新和求和时都应该要注意懒标记。
代码：

#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));
        }
}
}