Rokua P1198 [JSOI2008] Maximum number Label: line 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

Description

[JSOI2008]

Code

1#include <algorithm>2#include <iostream>3#include <cstring>4#include <cstdio>5#include <vector>6#include <Set>7#include <map>8 #defineMAXN 30000059 #defineINF 0x3f3f3f3fTen #defineM 262144 One using namespacestd; A  - intTr[maxn],n,d,t,p;//P Point to the tail of the team -  the voidAddintXintv) { -      for(tr[x+=m]=v,x>>=1; x;x>>=1){ -Tr[x]=max (tr[x<<1],tr[x<<1|1]); -     } + } -  + intAskintLintR) {//closed Interval A     intsum=0; at      for(l=l+m-1, r=r+m+1; l^r^1; l>>=1, r>>=1){ -         if(~l&1) Sum=max (sum,tr[l^1]); -         if(r&1) Sum=max (sum,tr[r^1]); -     } -     returnsum; - } in  - intMain () { to //freopen ("01.in", "R", stdin); +scanf"%d%d",&n,&D); -      for(intI=0; i<=3000000; i++){ thetr[i]=-INF; *     } $      while(n--){Panax Notoginseng         CharOpintx; -Cin>>op>>x; the         if(op=='A'){ +Add (++p, (x+t)%D); A         } the         Else{ +T=ask (p-x+1, p); -printf"%d\n", T); $         } $     } -     return 0; -}

80 min, Qaq

Originally wanted to do a simple problem practice practiced hand, the result blew up

Rokua P1198 [JSOI2008] Maximum number Label: line segment Tree