[HDU 4348]To the moon

题目

Background

To The Moon is a independent game released in November 2011, it is a role-playing adventure game powered by RPG Maker.

The premise of To The Moon is based around a technology that allows us to permanently reconstruct the memory on dying man. In this problem, we'll give you a chance, to implement the logic behind the scene.

You‘ve been given N integers A_[1], A_[2],..., A_[N]. On these integers, you need to implement the following operations:

1. C l r d: Adding a constant d for every {A_i | l <= i <= r}, and increase the time stamp by 1, this is the only operation that will cause the time stamp increase. 

2. Q l r: Querying the current sum of {A_i | l <= i <= r}.

3. H l r t: Querying a history sum of {A_i | l <= i <= r} in time t.

4. B t: Back to time t. And once you decide return to a past, you can never be access to a forward edition anymore.

.. N, M ≤ 10⁵, |A_[i]| ≤ 10⁹, 1 ≤ l ≤ r ≤ N, |d| ≤ 10⁴ .. the system start from time 0, and the first modification is in time 1, t ≥ 0, and won't introduce you to a future state.

INPUT

n m

A₁ A₂ ... A_n

... (here following the m operations. )

OUTPUT

... (for each query, simply print the result. )

SAMPLE

INPUT

10 5

1 2 3 4 5 6 7 8 9 10

Q 4 4

Q 1 10

Q 2 4

C 3 6 3

Q 2 4

 

2 4

0 0

C 1 1 1

C 2 2 -1

Q 1 2

H 1 2 1

OUTPUT

4

55

9

15

 

0

1

解题报告

可。。。可。。。可持久化？

初识可持久化线段树+区间修改

1 #include
      
         2 #include
       
          3 #include
        
           4 using namespace std;  5 typedef long long L;  6 inline int read(){  7     int sum(0),f(1);  8     char ch(getchar());  9     for(;ch<'0'||ch>'9';ch=getchar()) 10         if(ch=='-') 11             f=-1; 12     for(;ch>='0'&&ch<='9';sum=sum*10+(ch^48),ch=getchar()); 13     return sum*f; 14 } 15 int n,m,t,cnt; 16 int v[100005]; 17 int rt[100005],lch[4000005],rch[4000005]; 18 L sum[4000005],add[4000005]; 19 char op[2]; 20 inline int build(int l,int r){ 21     int x(++cnt); 22     add[x]=0; 23     if(l==r){ 24         sum[x]=v[l]; 25         lch[x]=rch[x]=0; 26 //        cout<<"pos "<
         
          <<" "<
          
           <<' '<
           
            <<" "<
            
             <<' '<
             
              <
              
               >1); 30 lch[x]=build(l,mid); 31 // cout<<"left child"<
               
                <<" "<
                
                 <<' '<
                 
                  <
                  
                   >1))+add[rch[x]]*(len>>1); 42 } 43 inline int update(int rt,int ll,int rr,L w,int l,int r){ 44 int newrt(++cnt); 45 add[newrt]=add[rt]; 46 if(ll<=l&&r<=rr){ 47 sum[newrt]=sum[rt]; 48 add[newrt]=add[rt]+w; 49 lch[newrt]=lch[rt]; 50 rch[newrt]=rch[rt]; 51 return newrt; 52 } 53 int mid((l+r)>>1); 54 if(ll<=mid) 55 lch[newrt]=update(lch[rt],ll,rr,w,l,mid); 56 else 57 lch[newrt]=lch[rt]; 58 if(mid
                   
                    >1); 70 L ret(0); 71 if(ll<=mid) 72 ret+=query(lch[rt],ll,rr,l,mid,add[rt]+add1); 73 if(mid