BZOJ--1012: [JSOI2008] Maximum number of MaxNumber | | Rokua--p1198 [JSOI2008] Maximum number

http://www.lydsy.com/JudgeOnline/problem.php?id=1012| |https://www.luogu.org/problem/show?pid=1198Time limit:3 Sec Memory limit:162 MB
submit:10824 solved:4735
[Submit] [Status] [Discuss] Description

Now ask you to maintain a sequence, the following two operations are required: 1, query operation. Syntax: Q l Function: Query the end of the current sequence of L
The largest number in the number, and outputs the value of the number. Limit: l does not exceed the length of the current series. 2, insert operation. Syntax: a n function: N Plus
On 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 taken to a fixed constant d
The resulting answer is inserted at the end of the sequence. Limit: N is a non-negative integer and is within a long range. Note: The initial sequence is empty and there is no
Number.

Input

The first line is two integers, M and D, where m represents the number of operations (M <= 200,000), D, as described above, satisfies D within Longint. Next
M row, query operation, or insert operation.

Output

For each query operation, output a line. The line has only one number, which is the maximum number of the last L in the sequence.

Sample Input5 100
A 96
Q 1
A 97
Q 1
Q 2Sample Output96
93
theHINT

The data is as follows Http://pan.baidu.com/s/1i4JxCH3

Source

① interval Query Most values consider the segment tree,

1#include <algorithm>2#include <iostream>3#include <cstdio>4 5 using namespacestd;6 7 Const intN200005);8 intT,mod,n,x[n],ans;9 CharCh,s[n];Ten  One structTree A { -     intl,r,sum; -}tr[n<<2]; the #defineLC (NOW&LT;&LT;1) - #defineRC (now<<1|1) - #defineMid (Tr[now].l+tr[now].r>>1) -InlinevoidTREE_UP (intNow ) + { -tr[now].sum=Max (tr[lc].sum,tr[rc].sum); + } A voidTree_build (intNowintLintR) at { -Tr[now].l=l; Tr[now].r=R; -     if(L==R)return ; - Tree_build (lc,l,mid); -Tree_build (rc,mid+1, R); - } in voidTree_change (intNowintTo,intx) - { to     if(tr[now].l==TR[NOW].R) +     { -tr[now].sum=x; the         return ; *     } $     if(to<=mid) Tree_change (lc,to,x);Panax Notoginseng     ElseTree_change (rc,to,x); - tree_up (now); the } + intTree_query (intNowintLintR) A { the     if(TR[NOW].L==L&AMP;&AMP;R==TR[NOW].R)returntr[now].sum; +     if(R<=mid)returntree_query (lc,l,r); -     Else if(L>mid)returntree_query (rc,l,r); $     Else returnMax (Tree_query (Lc,l,mid), Tree_query (rc,mid+1, R)); $ } -  - intMain () the { -scanf"%d%d",&t,&MoD);Wuyi      for(intI=1; i<=t;i++) the     { -cin>>ch>>x[i]; s[i]=ch; Wu         if(ch=='A') n++; -     } AboutTree_build (1,1, n); $     intCnt=0; -      for(intI=1; i<=t;i++) -     { -         if(s[i]=='A') ATree_change (1, ++cnt, (Ans+x[i])%MoD); +         Else the         { -Ans=tree_query (1, cnt-x[i]+1, CNT); $printf"%d\n", ans); the         } the     } the     return 0; the}

Line segment Tree AC

② monotonically Descending Stack, two-point lookup value at l

1#include <algorithm>2#include <cstdio>3 4 using namespacestd;5 6 Const intN200005);7 intn,mod,ans,x;8 intStack[n],top,num[n],len;9 Ten intMain () One { Ascanf"%d%d",&n,&MoD); -      for(Charch[2];n--;) -     { thescanf"%s%d",ch,&x); -         if(ch[0]=='A') -         { -x= (X+ans)%MoD; +num[++len]=x; -              for(;num[stack[top]]<=x&&stack[top]>0&&top;) top--; +stack[++top]=Len; A         } at         Else -         { -             intL=1, r=top; -              for(intmid;l<=R;) -             { -Mid=l+r>>1; in                 if(stack[mid]<len-x+1) l=mid+1; -                 Elser=mid-1, ans=Num[stack[mid]]; to             } +printf"%d\n", ans); -         } the     } *     return 0; $}

monotonic Stack ac

BZOJ--1012: [JSOI2008] Maximum number of MaxNumber | | Rokua--p1198 [JSOI2008] Maximum number