P1198 [JSOI2008] Maximum number

Title Description

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

1, query operation.

Grammar: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: lL does not exceed the length of the current series. (L > 0)(l>0)

2, insert operation.

Grammar:A n

Function: WillNN Plust t , where  t t   is the answer to the most recent query operation ( t=0 t =0 ), and the resulting result to a fixed constant  d d  modulo, inserting the resulting answer into the end of the sequence.

Limit: nN 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 row of two integers,MM andDD, whereMM indicates the number of operations(M \le 200,000)(M≤200,000) ，d d  to meet   (0<d<2,000,000,000) (0<d<2,0 00,00 0,000

The next m-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 sample #: Copy

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

Output Example # #: Replication

969396

Description

[JSOI2008]

The data has been strengthened

Single-point modification + interval query

#include <bits/stdc++.h>#defineLL Long Long#defineN 800005using namespacestd;structnode{intL,r,w;} Tr[n];intn,d,t,l;voidBuildintKintLintR) {TR[K].L=l;tr[k].r=R; if(L==R)return; intMid= (L+R)/2; Build (k*2, l,mid); Build (k*2+1, mid+1, R);}voidChangeintKintPintX) {    if(TR[K].L==TR[K].R) {tr[k].w=p;return;} if(x<= (TR[K].L+TR[K].R)/2) Change (k*2, p,x); ElseChange (k*2+1, p,x); TR[K].W=max (tr[k*2].w,tr[k*2+1].W); }voidAskintKintLintR) {    intLl=tr[k].l,rr=tr[k].r,mid= (LL+RR)/2; if(ll>=l&&rr<=R) {T=Max (tr[k].w,t); return; }if(L<=mid) Ask (k*2, L,r); if(R>mid) Ask (k*2+1, l,r);}intMain () {scanf ("%d%d",&n,&d); Build (1,1, N);  for(intX,i=1; i<=n;i++){        CharP;cin>>p>>x; if(p=='A') Change (1, (x+t)%d,++L); Else{T=-101010; Ask (1, l-x+1, L); printf ("%d\n", T); }    }return 0;}

P1198 [JSOI2008] Maximum number