2015
05-23

# To the moon

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

n m
A1 A2 … An
… (here following the m operations. )

n m
A1 A2 … An
… (here following the m operations. )

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

4
55
9
15

0
1

/*****************************************
USER: a180285
PROB:
LANG: C++
2013,UESTC
*****************************************/

# include <math.h>
# include <stdio.h>
# include <string.h>
# include <stdlib.h>
# include <algorithm>
# include <iostream>
# include <string>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <vector>
# include <cstring>
# include <list>
# include <ctime>
# include <sstream>

# define For(i,a)   for((i)=0;i<(a);(i)++)
# define MAX(x,y)   ((x)>(y)? (x):(y))
# define MIN(x,y)   ((x)<(y)? (x):(y))
# define sz(a)      (sizeof(a))
# define MEM(a)     (memset((a),0,sizeof(a)))
# define MEME(a)    (memset((a),-1,sizeof(a)))
# define MEMX(a)    (memset((a),0x7f,sizeof(a)))
# define pb(a)      push_back(a)

using namespace std;

typedef long long           ll      ;
typedef unsigned long long  ull     ;
typedef unsigned int        uint    ;
typedef unsigned char       uchar   ;

template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}
template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;}

const int oo=1<<30          ;
const double eps=1e-7       ;
const int N=100000 + 128              ;
const int M=300               ;
const int MAXQ = N * 95	;
const int MAXB = N * 16		;
const ll P=10000000097ll    ;

#pragma 	pack(1)

struct Block
{
int* vp;
int last;
ll sum;
};

int Q[MAXQ];
int* Qp[N];
Block B[MAXB];

int num, qn;
int len, bn;
int now;

void outputlimit()
{
while( 1 )
puts("RE");
}

int get_length(int kth)
{
return MIN(num-kth*len, len);
}

int* new_vp(int* old,int n)
{
int* p = Qp[now];
Qp[now] += n;
if(Qp[now] - Q >= MAXQ)
outputlimit();
for(int i=0; i<n; i++)
p[i] = old[i];
return p;
}

{
for(len=1; len*len<num; len++)
;
bn = (num+len-1) / len;
for(int i=0; i<num; i++)
scanf("%d", &Q[i]);

for(int i=0; i<bn; i++)
{
B[i].vp = Q + len * i;
B[i].last = 0;
B[i].sum = 0;
int n = get_length(i);
for(int j=0; j<n; j++)
B[i].sum += B[i].vp[j];
}

int sum = 0;
for(int i=0; i<bn; i++)
sum += get_length(i);
if(sum != num)
while( 1 )
puts("sum != num");
now = 0;

Qp[now] = Q + num;
}

{
l--;
now++;
Qp[now] = Qp[now-1];
Block* b   = B + now*bn;
Block* old = b - bn;

if(now*bn + bn >= MAXB)
outputlimit();

for(int i=0; i<bn; i++)
b[i] = old[i];
for(int i=l/len; i*len<r; i++)
{
int st = i * len;
int ed = st + get_length(i);
if(l<=st && ed<=r)
{
b[i].sum += d * get_length(i);
b[i].last += d;
}
else
{
int mn = MAX(st, l) - st;
int mx = MIN(ed, r) - st;

b[i].vp = new_vp(b[i].vp, get_length(i));
b[i].sum += d * (ll)(mx - mn);
for(int j=mn; j<mx; j++)
b[i].vp[j] += d;
}
}
}

void print()
{

}

ll query(int t,int l,int r)
{
l--;
Block* b = B + t*bn;

if(now*bn + bn >= MAXB)
outputlimit();

ll ans = 0;
for(int i=l/len; i*len<r; i++)
{
int st = i * len;
int ed = st + get_length(i);
if(l<=st && ed<=r)
{
ans += b[i].sum;
}
else
{
int mn = MAX(st, l) - st;
int mx = MIN(ed, r) - st;
ans += b[i].last * (ll)(mx - mn);
for(int j=mn; j<mx; j++)
ans += b[i].vp[j];
}
}
return ans;
}

void doit()
{
// 1. C l r d: Adding a constant d for every {Ai | 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 {Ai | l <= i <= r}.
// 3. H l r t: Querying a history sum of {Ai | l <= i <= r} in time t.
// 4. B t: Back to time t. And once you decide return to a past, y
int l, r, d, t;
ll ans;
for(int i=0; i<qn; i++)
{
char q;
scanf(" %c", &q);
switch(q)
{
case 'C':
scanf("%d %d %d", &l, &r, &d);
break;
case 'Q':
scanf("%d %d", &l, &r);
ans = query(now, l, r);
printf("%I64d\n", ans);
break;
case 'H':
scanf("%d %d %d", &l, &r, &t);
ans = query(t, l, r);
printf("%I64d\n", ans);
break;
case 'B':
scanf("%d", &t);
now = t;
break;
}
}
}

int main()
{
#ifdef Hw
freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);
double use = 0;
use += sizeof(Q);
use += sizeof(Qp);
use += sizeof(B);
printf("%f MB\n", use / (1<<20));
#endif
while( 2==scanf("%d %d", &num, &qn) )
{
doit();
print();
}

return 0;
}

/**

10 15
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
Q 1 1
Q 1 2
Q 1 3
Q 1 4
Q 1 5
Q 1 6
Q 1 7
Q 1 8
Q 1 9
Q 1 10

2 4
0 0
C 1 1 1
C 2 2 -1
Q 1 2
H 1 2 1

Sample Output
4
55
9
15

0
1

**/