[Tree line] [template] seeking maximum range P1198 [JSOI2008] The maximum number of

Title Description

Now request you to maintain a number of columns, it requires the following two actions:

1, the query operation.

grammar:Q L

Function: the maximum number of queries in the end of the number of columns in the current number L, and outputs the value of this number.

Restrictions: L L does not exceed the length of the current sequence. (L> 0) ( L > 0 )

2, the insertion operation.

grammar:A n

Function: n- n-plus T T, where T T is the answer to the last query operation (if not already performed a query operation, then T = 0 T = 0), and the result of a fixed constant D D taken die, the resultant answer inserted at the end of the number of columns.

Restrictions: n- n-integer (possibly negative) and in the entire length range.

Note: The number of columns is initially empty, not a number.

Input and output formats

Input formats:

 

The first line of two integers, M M and D D, where M M represents the number of operations (M \ Le 200,000) ( M ≤ 2 0 0 , 0 0 0 ), D D described hereinabove, satisfying (0 < D <2,000,000,000) ( 0 < D < 2 , 0 0 0 , 0 0 0 , 0 0 0 )

The next M M rows, each row a string describing a specific operation. Syntax as described above.

 

Output formats:

 

For each query, you should turn the output in the order, and each result row.

 

Topics requirements too straight ball

 

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=2e5+5;
int n;
ll a[maxn],d;
struct node
{
    int l,r;
    ll id;
}t[maxn*4];
void build(int p,int l,int r)
{
    t[p].l=l,t[p].r=r;
    if(l==r)
    {
        t[p].id=a[l];
        return;
    }
    int mid=(l+r)>>1;
    build(p*2,l,mid);
    build(p*2+1,mid+1,r);
    t[p].id=max(t[p*2].id,t[p*2+1].id);
}
void change(int p,int x,ll v,int l,int r)
{
    if(l==r)
    {
        t[p].id=v;
        return;
    }
    int mid=(l+r)>>1;
    if(x<=mid)
        change(p*2,x,v,l,mid);
    else
        change(p*2+1,x,v,mid+1,r);
    t[p].id=max(t[p*2].id,t[p*2+1].id);
}
ll ask(int l0,int r0,int p,int l,int r)
{
    if(l<=l0&&r>=r0)
    {
        return t[p].id;
    }
    int mid=(l0+r0)>>1;
    ll val=-(1<<30);
    if(mid>=l)
        val=max(val,ask(l0,mid,p*2,l,r));
    if(mid<r)
        val=max(val,ask(mid+1,r0,p*2+1,l,r));
    return val;
}
int main()
{
    scanf("%d %lld",&n,&d);
    int len=0;
    ll t=0;
    for(int i=1;i<=n;i++)
    {
        char ch;
        ll x;
        scanf(" %c %lld",&ch,&x);
        //cout<<ch<<" "<<x<<endl;
        if(ch=='A')
        {
            change(1,++len,(x+t)%d,1,n);
        }
        else
        {
            if(x==0)
                t=0;
            else
                t=ask(1,n,1,len-x+1,len);
            printf("%lld\n",t);
        }
    }
    return 0;
}