[hdoj4578][多延迟标记的线段树]

Transformation

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 65535/65536 K (Java/Others)
Total Submission(s): 9392    Accepted Submission(s): 2408

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

题意：有4种操作，区间加，区间乘，区间修改以及区间幂之和。

题解：区间加==(mul==1,add==x) 区间乘==(mul==x,add==0)区间修改==(mul==0,add==x)区间幂通过展开式转化为区间乘和加，lazy标记用来标记当前点已经被更新过，pushdown(rt)用于更新子节点

#include<bits/stdc++.h>
using namespace std;
#define debug(x) cout<<"["<<#x<<"]"<<" is "<<x<<endl
#define forp(x) for(int i=1;i<=x;i++)
#define scai(x) scanf("%d",&x)
#define scal(x) scanf("%lld",&x)
#define pri(x) printf("%d
",x)
#define prl(x) printf("%lld
",x)
typedef long long ll;
const int maxn=1e5+5;
const ll mod=1e4+7;
struct node{
    int l;
    int r;
    ll ad;
    ll mu;
    ll a[4];
}N[maxn<<2];
void pushup(int rt){
    N[rt].a[1]=N[rt<<1].a[1]+N[(rt<<1)|1].a[1];
    N[rt].a[2]=N[rt<<1].a[2]+N[(rt<<1)|1].a[2];
    N[rt].a[3]=N[rt<<1].a[3]+N[(rt<<1)|1].a[3];
    N[rt].mu%=mod;
    N[rt].ad%=mod;
    N[rt].a[1]%=mod;
    N[rt].a[2]%=mod;
    N[rt].a[3]%=mod;
}
void pushdown(int rt){
    if(N[rt].mu!=1){
        ll m=N[rt].mu;
        N[rt].mu=1;
        N[rt<<1].ad=N[rt<<1].ad*m%mod;
        N[(rt<<1)|1].ad=N[(rt<<1)|1].ad*m%mod;
        N[rt<<1].mu=N[rt<<1].mu*m%mod;
        N[(rt<<1)|1].mu=N[(rt<<1)|1].mu*m%mod;
        N[rt<<1].a[1]=N[rt<<1].a[1]*m%mod;
        N[rt<<1].a[2]=N[rt<<1].a[2]*m%mod*m%mod;
        N[rt<<1].a[3]=N[rt<<1].a[3]*m%mod*m%mod*m%mod;
        N[(rt<<1)|1].a[1]=N[(rt<<1)|1].a[1]*m%mod;
        N[(rt<<1)|1].a[2]=N[(rt<<1)|1].a[2]*m%mod*m%mod;
        N[(rt<<1)|1].a[3]=N[(rt<<1)|1].a[3]*m%mod*m%mod*m%mod;
    }
    if(N[rt].ad!=0){
        ll m=N[rt].ad;
        N[rt].ad=0;
        N[rt<<1].ad=N[rt<<1].ad+m%mod;
        N[(rt<<1)|1].ad=N[(rt<<1)|1].ad+m%mod;
        N[rt<<1].a[3]=N[rt<<1].a[3]+m%mod*m%mod*m%mod*(N[rt<<1].r-N[rt<<1].l+1)%mod+3*m*N[rt<<1].a[2]%mod+3*m%mod*m%mod*N[rt<<1].a[1]%mod;
        N[rt<<1].a[3]%=mod;
        N[rt<<1].a[2]=N[rt<<1].a[2]+m%mod*m%mod*(N[rt<<1].r-N[rt<<1].l+1)%mod+2*m*N[rt<<1].a[1]%mod;
        N[rt<<1].a[2]%=mod;
        N[rt<<1].a[1]=N[rt<<1].a[1]+m*(N[rt<<1].r-N[rt<<1].l+1)%mod;
        N[rt<<1].a[1]%=mod;
        N[(rt<<1)|1].a[3]=N[(rt<<1)|1].a[3]%mod+m%mod*m%mod*m%mod*(N[(rt<<1)|1].r-N[(rt<<1)|1].l+1)%mod+3*m*N[(rt<<1)|1].a[2]%mod+3*m%mod*m%mod*N[(rt<<1)|1].a[1]%mod;
        N[(rt<<1)|1].a[3]%=mod;
        N[(rt<<1)|1].a[2]=N[(rt<<1)|1].a[2]%mod+m%mod*m%mod*(N[(rt<<1)|1].r-N[(rt<<1)|1].l+1)%mod+2*m*N[(rt<<1)|1].a[1]%mod;
        N[(rt<<1)|1].a[2]%=mod;
        N[(rt<<1)|1].a[1]=N[(rt<<1)|1].a[1]+m*(N[(rt<<1)|1].r-N[(rt<<1)|1].l+1)%mod;
        N[(rt<<1)|1].a[1]%=mod;
    }
}
void build(int L,int R,int rt){
    N[rt].l=L;
    N[rt].r=R;
    N[rt].mu=1;
    N[rt].ad=0;
    if(L==R){
        N[rt].a[1]=0;
        N[rt].a[2]=0;
        N[rt].a[3]=0;
        return;
    }
    int mid=(L+R)/2;
    build(L,mid,rt<<1);
    build(mid+1,R,(rt<<1)|1);
    pushup(rt);
}
void update(int L,int R,int rt,int l1,int r1,ll add,ll mul){
    N[rt].mu%=mod;
    N[rt].ad%=mod;
    N[rt].a[1]%=mod;
    N[rt].a[2]%=mod;
    N[rt].a[3]%=mod;
    if(l1<=L&&r1>=R){
    if(mul!=1){
        ll m=mul;
        N[rt].ad=N[rt].ad*m%mod;
        N[rt].mu=N[rt].mu*m%mod;
        N[rt].a[1]=N[rt].a[1]*m%mod;
        N[rt].a[2]=N[rt].a[2]*m%mod*m%mod;
        N[rt].a[3]=N[rt].a[3]*m%mod*m%mod*m%mod;
    }
    if(add!=0){
        ll m=add;
        N[rt].ad=N[rt].ad+m%mod;
        N[rt].a[3]=N[rt].a[3]+m%mod*m%mod*m%mod*(N[rt].r-N[rt].l+1)%mod+3*m*N[rt].a[2]%mod+3*m%mod*m%mod*N[rt].a[1]%mod;
        N[rt].a[3]%=mod;
        N[rt].a[2]=N[rt].a[2]+m%mod*m%mod*(N[rt].r-N[rt].l+1)%mod+2*m%mod*N[rt].a[1]%mod;
        N[rt].a[2]%=mod;
        N[rt].a[1]=N[rt].a[1]+m*(N[rt].r-N[rt].l+1)%mod;
        N[rt].a[1]%=mod;
    }
        return;
    }
    pushdown(rt);
    int mid=(L+R)/2;
    if(mid>=l1){
        update(L,mid,rt<<1,l1,r1,add,mul);
    }
    if(mid<r1){
        update(mid+1,R,(rt<<1)|1,l1,r1,add,mul);
    }
    pushup(rt);
}
ll query(int L,int R,int rt,int l1,int r1,int p){
    N[rt].mu%=mod;
    N[rt].ad%=mod;
    N[rt].a[1]%=mod;
    N[rt].a[2]%=mod;
    N[rt].a[3]%=mod;
    if(l1<=L&&r1>=R){
        return N[rt].a[p]%mod;
    }
    int mid=(L+R)/2;
    ll ak=0;
    pushdown(rt);
    if(mid>=l1){
        ak+=query(L,mid,rt<<1,l1,r1,p);
    }
    if(mid<r1){
        ak+=query(mid+1,R,(rt<<1)|1,l1,r1,p);
    }


    return ak%mod;
}
int main(){
    int n,m;
    scai(n);
    scai(m);
    while(n||m){
        build(1,n,1);
        forp(m){
            ll a,b,c,d;
            scal(a);
            scal(b);
            scal(c);
            scal(d);
            if(a==1){
                update(1,n,1,b,c,d,1);
            }
            else if(a==2){
                update(1,n,1,b,c,0,d);
            }
            else if(a==3){
                update(1,n,1,b,c,d,0);
            }
            else{
                prl(query(1,n,1,b,c,d));
            }
        }
        scai(n);
        scai(m);
    }
    return 0;
}