hdu 4578 Transformation（线段树）

Problem Description

Yuanfang is puzzled with the question below: 
There are n integers, a₁, a₂, …, 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

思路：

题目要实现多个区间更新以及一种区间查询 我们对于次方我们可以维护三棵线段树 注意lazy标记的相互影响 以及跟新的顺序

#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const int N = 1e5+7;
const int inf = 0x3f3f3f3f;
const double eps = 1e-6;
typedef long long ll;
const ll mod = 10007;
struct tree{
    int l,r;
    ll add,mul,cha;
    ll sum1,sum2,sum3;
}t[N<<2];
void pushup(int p){
    t[p].sum1=(t[p<<1].sum1+t[p<<1|1].sum1)%mod;
    t[p].sum2=(t[p<<1].sum2+t[p<<1|1].sum2)%mod;
    t[p].sum3=(t[p<<1].sum3+t[p<<1|1].sum3)%mod;
}
void build(int p,int l,int r){
    t[p].sum1=t[p].sum2=t[p].sum3=t[p].add=t[p].cha=0;
    t[p].l=l; t[p].r=r;
    t[p].mul=1;
    if(l==r){
        return ;
    }
    int mid=(l+r)>>1;
    build(p<<1,l,mid);
    build(p<<1|1,mid+1,r);
}
void pushdown(int p,int l,int r){
    int mid=(l+r)>>1;
    if(t[p].cha){
        t[p<<1].cha=t[p].cha;
        t[p<<1].sum1=(mid-l+1)*t[p].cha%mod;
        t[p<<1].sum2=t[p<<1].sum1*t[p].cha%mod;
        t[p<<1].sum3=t[p<<1].sum2*t[p].cha%mod;
        t[p<<1|1].cha=t[p].cha;
        t[p<<1|1].sum1=(r-mid)*t[p].cha%mod;
        t[p<<1|1].sum2=t[p<<1|1].sum1*t[p].cha%mod;
        t[p<<1|1].sum3=t[p<<1|1].sum2*t[p].cha%mod;
        t[p<<1].add=t[p<<1|1].add=0;
        t[p<<1].mul=t[p<<1|1].mul=1;
        t[p].cha=0;
    }
    if(t[p].mul!=1){
        t[p<<1].mul=t[p<<1].mul*t[p].mul%mod;
        t[p<<1|1].mul=t[p<<1|1].mul*t[p].mul%mod;
        t[p<<1].sum3=t[p<<1].sum3*t[p].mul%mod*t[p].mul%mod*t[p].mul%mod;
        t[p<<1].sum2=t[p<<1].sum2*t[p].mul%mod*t[p].mul%mod;
        t[p<<1].sum1=t[p<<1].sum1*t[p].mul%mod;
        t[p<<1|1].sum3=t[p<<1|1].sum3*t[p].mul%mod*t[p].mul%mod*t[p].mul%mod;
        t[p<<1|1].sum2=t[p<<1|1].sum2*t[p].mul%mod*t[p].mul%mod;
        t[p<<1|1].sum1=t[p<<1|1].sum1*t[p].mul%mod;
        t[p<<1].add=t[p<<1].add*t[p].mul%mod;
        t[p<<1|1].add=t[p<<1|1].add*t[p].mul%mod;
        t[p].mul=1;
    }
    if(t[p].add){
        t[p<<1].add=(t[p<<1].add+t[p].add)%mod;
        t[p<<1|1].add=(t[p<<1|1].add+t[p].add)%mod;
        ll c=t[p].add*t[p].add%mod*t[p].add%mod;
        t[p<<1].sum3=(t[p<<1].sum3+(mid-l+1)*c%mod+3*t[p].add*(t[p<<1].sum1*t[p].add%mod+t[p<<1].sum2)%mod)%mod;
        t[p<<1|1].sum3=(t[p<<1|1].sum3+(r-mid)*c%mod+3*t[p].add*(t[p<<1|1].sum1*t[p].add%mod+t[p<<1|1].sum2)%mod)%mod;
        t[p<<1].sum2=(t[p<<1].sum2+2*t[p<<1].sum1*t[p].add%mod+t[p].add*t[p].add%mod*(mid-l+1)%mod)%mod;
        t[p<<1|1].sum2=(t[p<<1|1].sum2+2*t[p<<1|1].sum1*t[p].add%mod+t[p].add*t[p].add%mod*(r-mid)%mod)%mod;
        t[p<<1].sum1=(t[p<<1].sum1+(mid-l+1)*t[p].add)%mod;
        t[p<<1|1].sum1=(t[p<<1|1].sum1+(r-mid)*t[p].add)%mod;
        t[p].add=0;
    }    
}
void update(int p,int l,int r,ll v,int op){
    //cout<<t[p].l<<" "<<t[p].r<<endl;
    if(l<=t[p].l&&t[p].r<=r){
        if(op==3){
            t[p].cha=v;
            t[p].add=0;
            t[p].mul=1;
            t[p].sum1=(t[p].r-t[p].l+1)*v%mod;
            t[p].sum2=t[p].sum1*v%mod;
            t[p].sum3=t[p].sum2*v%mod;
        }else if(op==2){
            t[p].mul=t[p].mul*v%mod;
            t[p].add=t[p].add*v%mod;
            t[p].sum1=t[p].sum1*v%mod;
            t[p].sum2=t[p].sum2*v%mod*v%mod;
            t[p].sum3=t[p].sum3*v%mod*v%mod*v%mod;
        }else{
            t[p].add=(t[p].add+v)%mod;
            t[p].sum3=(t[p].sum3+v*v%mod*v%mod*(t[p].r-t[p].l+1)%mod+3*v*(t[p].sum1*v+t[p].sum2)%mod)%mod;
            t[p].sum2=(t[p].sum2+v*v%mod*(t[p].r-t[p].l+1)%mod+2*v*t[p].sum1%mod)%mod;
            t[p].sum1=(t[p].sum1+v*(t[p].r-t[p].l+1))%mod;    
        }
        return ;
    }
    pushdown(p,t[p].l,t[p].r);
    int mid=(t[p].l+t[p].r)>>1;
    if(l<=mid) update(p<<1,l,r,v,op);
    if(r>mid) update(p<<1|1,l,r,v,op);
    pushup(p);
}
ll query(int p,int l,int r,int op){
//    cout<<p<<" "<<t[p].l<<" "<<t[p].r<<endl;
    if(l<=t[p].l&&t[p].r<=r){
        if(op==1){
            return t[p].sum1;
        }else if(op==2){
            return t[p].sum2;
        }else{
            return t[p].sum3;
        }
    }
    pushdown(p,t[p].l,t[p].r);
    int mid=(t[p].l+t[p].r)>>1;
    ll res=0;
    if(l<=mid) res=(res+query(p<<1,l,r,op))%mod;
    if(r>mid) res=(res+query(p<<1|1,l,r,op))%mod;
    return res; 
}
int main(){
//    ios::sync_with_stdio(false);
//    cin.tie(0); cout.tie(0);
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        if(!n&&!m) break;
        build(1,1,n);
        for(int i=1;i<=m;i++){
            int op,x,y,c;
            scanf("%d%d%d%d",&op,&x,&y,&c);
            if(op==4){
                printf("%lld
",query(1,x,y,c));
            }else{
                update(1,x,y,c,op);
            }
        }
    }
}