P1471 学习笔记

太闲了，找点毒瘤数据结构写。

题目传送门

这不一眼线段树吗？

对于我们的操作 \(2\)，我们知道平均数

\[d=\frac1n \sum_{i=1}^{n}a_i \]

由于给定了区间长度 \(n\)，所以这个相当于查询区间和，数据结构基本操作。

对于我们的操作 \(3\)，我们知道方差

\[s^2=\frac1n \sum_{i=1}^{n}(a_i-d)^2 \]

展开

\[s^2=\frac1n \sum_{i=1}^{n} (a_i^2-2a_id+d^2) \]

再展开

\[s^2=\frac1n (\sum_{i=1}^{n}a_i^2-2d\sum_{i=1}^{n}a_i+n\cdot d^2) \]

注意到 \(\dfrac12\) 和 \(nd^2\) 还有 \(2d\) 均为定值，我们只需要在操作二的基础上维护一个平方和就行了。

其实可以只开一个线段树，但是我开了两个，这个实现还是很考细节的，害的我调了一下午。

code

#include <iostream>
#include <cmath>
#include <iomanip>
#define DEBUG
#define Ofile(s) freopen(s".in", "r", stdin), freopen (s".out", "w", stdout)
#define Cfile(s) fclose(stdin), fclose(stdout)
#define fast ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
using namespace std;

using ll = long long;
using ull = unsigned long long;
using lb = long double;

constexpr int mod = 998244353;
constexpr int maxn = 100005;

ll n, q, opt, x, y;
lb k;

lb a[maxn], tree[maxn << 2], ptree[maxn << 2], tag[maxn << 2], lazy[maxn << 2];//开两个线段树进行维护

ll ls(ll p){
	return p << 1;
}

ll rs(ll p){
	return p << 1 | 1;
}

void pushup1(ll p){
	tree[p] = tree[ls(p)] + tree[rs(p)];
}

void pushup2(ll p){
	ptree[p] = ptree[ls(p)] + ptree[rs(p)];
}

void build1(ll p, ll pl, ll pr){
	if (pl == pr){
		tree[p] = a[pl];
		return;
	}
	ll mid = (pl + pr) >> 1;
	build1(ls(p), pl, mid);
	build1(rs(p), mid + 1, pr);
	pushup1(p);
}

void build2(ll p, ll pl, ll pr){
	if (pl == pr){
		ptree[p] = a[pl] * a[pl];
		return;
	}
	ll mid = (pl + pr) >> 1;
	build2(ls(p), pl, mid);
	build2(rs(p), mid + 1, pr);
	pushup2(p);
}

void addtag1(ll p, ll pl, ll pr, lb d){
	tag[p] += d;
	tree[p] += (pr - pl + 1) * d;
}

void addtag2(ll p, ll pl, ll pr, lb d){
	lb sum = tree[p];
	lazy[p] += d;
	ptree[p] += 2 * d * sum + (pr - pl + 1) * d * d;//根据刚刚的推导
}

void pushdown1(ll p, ll pl, ll pr){
	if (!tag[p])
		return;
	ll mid = (pl + pr) >> 1;
	addtag1(ls(p), pl, mid, tag[p]);
	addtag1(rs(p), mid + 1, pr, tag[p]);
	tag[p] = 0;
}

void pushdown2(ll p, ll pl, ll pr){
	if (!lazy[p])
		return;
	ll mid = (pl + pr) >> 1;
	addtag2(ls(p), pl, mid, lazy[p]);
	addtag2(rs(p), mid + 1, pr, lazy[p]);
	lazy[p] = 0;
}
lb query1(ll L, ll R, ll p, ll pl, ll pr){
	if (L <= pl && pr <= R)
		return tree[p];
	pushdown2(p, pl, pr);
	pushdown1(p, pl, pr);//顺序很重要，不然你会像我一样调一下午，因为平方和的修改要用到原来的区间和
	ll mid = (pl + pr) >> 1;
	lb res = 0;
	if (L <= mid)
		res += query1(L, R, ls(p), pl, mid);
	if (mid < R)
		res += query1(L, R, rs(p), mid + 1, pr);
	return res;
}

lb query2(ll L, ll R, ll p, ll pl, ll pr){
	if (L <= pl && pr <= R)
		return ptree[p];
	pushdown2(p, pl, pr);
	pushdown1(p, pl, pr);//同理
	ll mid = (pl + pr) >> 1;
	lb res = 0;
	if (L <= mid)
		res += query2(L, R, ls(p), pl, mid);
	if (mid < R)
		res += query2(L, R, rs(p), mid + 1, pr);
	return res;
}

void update(ll L, ll R, ll p, ll pl, ll pr, lb d){
	if (L <= pl && pr <= R){
		addtag2(p, pl, pr, d);
		addtag1(p, pl, pr, d);//顺序
		return;
	}
	pushdown2(p, pl, pr);
	pushdown1(p, pl, pr);
	ll mid = (pl + pr) >> 1;
	if (L <= mid)
		update(L, R, ls(p), pl, mid, d);
	if (mid < R)
		update(L, R, rs(p), mid + 1, pr, d);
	pushup1(p);
	pushup2(p);
}

int main() {
	fast;
	cin >> n >> q;
	for (ll i = 1; i <= n; i++)
		cin >> a[i];
	build1(1, 1, n);
	build2(1, 1, n);
	while (q--){
		cin >> opt;
		if (opt == 1){
			cin >> x >> y >> k;
			update(x, y, 1, 1, n, k);
		}
		else if (opt == 2){
			cin >> x >> y;
			cout << fixed << setprecision(4) << query1(x, y, 1, 1, n) / (y - x + 1) << endl;//按照要求保留四位小数输出
		}
		else {
			cin >> x >> y;
			lb pingjun = query1(x, y, 1, 1, n) / (y - x + 1);
			lb sqf = (query2(x, y, 1, 1, n) - 2 * pingjun * query1(x, y, 1, 1, n) + (y - x + 1) * pingjun * pingjun) / (y - x + 1);
			cout << fixed << setprecision(4) << sqf << endl;//根据方差计算公式
		}
	}
	return 0;
}