[JSOI2008] [BZOJ1012] Maximum number (dynamic open Point segment Tree)

Title Description

Now you are asked to maintain a sequence that requires the following two actions:

1, Query Operation.

Syntax: Q L

Function: Query the maximum number of L in the number of the end of the current sequence and output the value of this number.

Limit: L does not exceed the length of the current Series.

2, Insert Operation.

Syntax: A N

Function: n plus t, where T is the answer to the most recent query operation (t=0 if the query operation has not been performed), and the resulting result is modeled on a fixed constant d and the resulting answer is inserted at the end of the Sequence.

Limit: n is an integer (possibly negative) and within a long range.

Note: The initial sequence is empty and does not have a number.

Input/output format

Input Format:

The first line is two integers, m and d, where m represents the number of operations (m <= 200,000), and D is satisfied (0<d<2,000,000,000) as described above

The next m-line, one string per line, describes a specific operation. The syntax is as described Above.

Output format:

For each query operation, you should output the results sequentially, with each result in one ROW.

Input and Output sample input example # #:

5 100A 96Q 1 a 97Q 1Q 2

Sample # # of Output:

969396

• Jiangsu 2008 Provincial topic, in fact, is not difficult .
• Maintain the maximum number of sequence intervals, first think of the segment tree, and the line segment tree is suitable for the data range (O (nlogn)).
• But the interval is not long, how to build it?
• By Test instructions, the longest segment tree will be no more than 200000, then build a 200000 tree directly.
• Make a note of the current number of insertions, and insert the data dynamically behind the Sequence.
• The segment tree is used to maintain the maximum interval Value.
• Expect to score 100 Points.

1#include <cstdio>2#include <iostream>3#include <algorithm>4#include <cstring>5 using namespacestd;6 7 structtree{8     intl,r,maxx;9} t[1000050];Ten  one intn,mod,tt,now; a  - voidBuildintXintLintR) { -t[x].l=l; T[x].r=r; the     if(t[x].l==t[x].r)return; -     intMid= (t[x].l+t[x].r) >>1; -Build (x*2, l,mid); -Build (x*2+1, mid+1, r); + } -  + voidChangeintXintLintY) { a     if(t[x].l==T[x].r) { att[x].maxx=y; -         return; -     } -     intMid= (t[x].l+t[x].r) >>1; -     if(l>mid) Change (x*2+1, l,y);ElseChange (x*2, l,y); -T[x].maxx=max (t[x*2].maxx,t[x*2+1].maxx); in } -  to intFindintXintLintR) { +     if(t[x].l==l && T[x].r==r)returnt[x].maxx; -     intMid= (t[x].l+t[x].r) >>1; the     if(l>mid)returnFind (x*2+1, l,r);Else *     if(r<=mid)returnFind (x*2, l,r);Else $         returnMax (find (x*2, l,mid), find (x*2+1, mid+1, r));Panax Notoginseng } -  the intmain () { +scanf"%d%d\n",&n,&mod); aBuild1,1, n); the      for(intI=1; i<=n; i++) { +         Charopt; -         intx; $scanf"%c%d\n",&opt,&x); $         if(opt=='A') { -now++; -Change1, now, (x+tt)%mod); the}Else  -         if(opt=='Q') {WuyiTt=find (1, now-x+1, now); theprintf"%d\n", tt); -         } wu     } -     return 0; about}

[JSOI2008] [BZOJ1012] Maximum number (dynamic open Point segment Tree)