BZOJ3922 Karin的弹幕 【线段树】

题目链接

BZOJ3922

题解

考虑暴力，修改\(O(1)\)，查询\(O(\frac{n}{d})\)

考虑线段树，如果对每种差值建一棵线段树，修改\(O(nlogn)\)，查询\(O(logn)\)

能不能均摊？

线段树查询快，修改慢

暴力修改快，查询慢

容易发现当\(d\)比较大时暴力还是可以接受的，所以我们不需要建那么多线段树

我们设我们对前\(T\)种差值建线段树

那么暴力查询总共\(O(m\frac{n}{T})\)，线段树修改总共\(O(mTlogn)\)

我们令

\[m\frac{n}{T} = mTlogn
\]

解得

\[T = \sqrt{\frac{n}{logn}}
\]

总复杂度\(O(n\sqrt{nlogn})\)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
#define ls (u << 1)
#define rs (u << 1 | 1)
using namespace std;
const int maxn = 70005,maxm = 100005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
int B,n,m,id[70][maxn],A[maxn],C[maxn];
struct Segmnet_Tree{
	int mx[maxn << 2];
	void upd(int u){mx[u] = max(mx[ls],mx[rs]);}
	void build(int u,int l,int r){
		if (l == r) {mx[u] = C[l]; return;}
		int mid = l + r >> 1;
		build(ls,l,mid);
		build(rs,mid + 1,r);
		upd(u);
	}
	void modify(int u,int l,int r,int pos,int v){
		if (l == r){mx[u] += v; return;}
		int mid = l + r >> 1;
		if (mid >= pos) modify(ls,l,mid,pos,v);
		else modify(rs,mid + 1,r,pos,v);
		upd(u);
	}
	int query(int u,int l,int r,int L,int R){
		if (l >= L && r <= R) return mx[u];
		int mid = l + r >> 1;
		if (mid >= R) return query(ls,l,mid,L,R);
		if (mid < L) return query(rs,mid + 1,r,L,R);
		return max(query(ls,l,mid,L,R),query(rs,mid + 1,r,L,R));
	}
}T[70];

int main(){
	n = read(); B = (int)sqrt(n / (log(n) / (log(2))));
	REP(i,n) A[i] = read();
	for (int t = 1; t <= B; t++){
		int cnt = 0;
		for (int i = 1; i <= t; i++){
			for (int j = i; j <= n; j += t){
				id[t][j] = ++cnt;
				C[cnt] = A[j];
			}
		}
		T[t].build(1,1,n);
	}
	m = read();
	int opt,v,x,d,l,r;
	while (m--){
		opt = read();
		if (!opt){
			x = read(); v = read(); A[x] += v;
			for (int t = 1; t <= B; t++)
				T[t].modify(1,1,n,id[t][x],v);
		}
		else {
			x = read(); d = read();
			if (d <= B){
				l = id[d][x]; r = id[d][x + (n - x) / d * d];
				printf("%d\n",T[d].query(1,1,n,l,r));
			}
			else {
				v = A[x];
				for (int i = x + d; i <= n; i += d)
					v = max(v,A[i]);
				printf("%d\n",v);
			}
		}
	}
	return 0;
}