最大数(线段树)

思路: 先整体看成全是0的一个线段树,然后按照题目要求进行修改和查询。

#include <iostream>
#include <string.h>
#include <string>
#include<cstdio>
#include <algorithm>
#define ll long long
using namespace std;
ll tree[800010],add[800010];
int a[800010];
int N=0;
const int M=2e5+10;
void push_up(ll rt)
{
    tree[rt] = tree[rt*2+1] + tree[rt*2+2];
}
void push_down(ll rt,ll m)
{
    if(add[rt])
    {
        add[rt*2+1]+=add[rt];
        add[rt*2+2]+=add[rt];
        tree[rt*2+1]+=(m-(m/2))*add[rt];
        tree[rt*2+2]+= (m/2)*add[rt];
        add[rt] = 0;
    }
}       //板子直接套(注意该树是从节点0开始的)
void update_tree(int node,int start,int end,int idx,int val)//点更新
{

    if(start==end){
        a[idx]=val;
        tree[node]=val;
        //cout<<"node "<<node<<"  "<<start<<"  "<<end<<" "<<idx<<endl;
    }
    else
    {
        int mid=(start+end)/2;
        int left_node=2*node+1;
        int right_node=2*node+2;
        if(idx>=start&&idx<=mid) update_tree(left_node,start,mid,idx,val);
        else update_tree(right_node,mid+1,end,idx,val);
        tree[node]=max(tree[left_node],tree[right_node]);
       // cout<<"node= "<<node<<endl;
    }
}
ll query_tree(int L, int R, int l, int r, int rt)  //L R 要查询的区间,l r 起始点,tr 根节点0
{
    if(L <= l && R >= r)
        return tree[rt];
    //push_down(rt,r-l+1);   //区间修改特有的一步
    ll mid=(l+r)/2;
    ll left_node=2*rt+1;
    ll right_node=2*rt+2;
    ll ans1= -0x3f3f3f,ans2=-0x3f3f3f;
    if(L <= mid) ans1= query_tree(L, R,l,mid,left_node);
    if(R > mid) ans2= query_tree(L, R,mid+1,r,right_node);
    return max(ans1,ans2);
}
int main()
{
    int m,p;
    ll tp=0;
    scanf("%d%d",&m,&p);
    memset(a,0,sizeof(a));
    memset(tree,0,sizeof(tree));
    while(m--){
        char c;
        int b;
        cin>>c>>b;
        if(c=='Q'){
            tp=query_tree( N-b, N-1,0,M, 0);
            cout<<tp<<endl;
        }
        if(c=='A')
        {
            N++;
            ll tp1=(b+tp)%p;
            update_tree(0,0,M,N-1,tp1);  //注意这里的区间是(0~M)M=2e5+10,不是进行添加操作的次数。
        }

    }
    return 0;
}
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posted @ 2020-11-11 19:31  Swelsh-corgi  阅读(130)  评论(0编辑  收藏  举报