E. Paimon Segment Tree

传送门

题意:
首先给出n, m, q, 一个长度为n的数组，m次修改操作，每次修改操作，l, r, x, 对l, r区间里面的数加上k, q次询问操作，每次询问操作，问
$\sum\limits_{i = lk}^{rk}\sum\limits_{j = xk}^{yk}a_{i, j}^2$

思路:
等过段时间在写，先贴代码，提醒下线段树维护矩阵

总结:
线段树维护矩阵元素，重载运算符，细节写法

点击查看代码

#include <bits/stdc++.h>
#pragma gcc optimize("O2")
#pragma g++ optimize("O2")
#define endl '\n'
#define ls rt << 1
#define rs rt << 1 | 1
#define lson ls, l, mid
#define rson rs, mid + 1, r
using namespace std;

typedef long long ll;

const ll MAXN = 1E5 + 10, MOD = 1e9 + 7;
ll n, m, q;
ll a[MAXN], ans[MAXN];

struct Mat {
	ll v[4][4];
	Mat() { memset(v, 0, sizeof(v)); }
	Mat(ll k) {
		v[0][0] = v[1][1] = v[2][2] = v[3][3] = v[2][3] = 1;
		v[0][1] = k;
		v[0][2] = v[0][3] = k * k % MOD;
		v[1][2] = v[1][3] = 2 * k % MOD;
		v[1][0] = v[2][0] = v[2][1] = v[3][0] = v[3][1] = v[3][2] = 0;
	}

	void init()
	{
		for (int i = 0; i < 4; ++i)
			for (int j = 0; j < 4; ++j)
				v[i][j] = (i == j ? 1 : 0);
	}

	ll* operator [] (int x) { return v[x]; }
	const ll* operator [] (int x) const { return v[x]; }

	Mat operator * (const Mat& b)
	{
		const Mat& a = *this;
		Mat res;
		for (int i = 0; i < 4; ++i)
			for (int j = 0; j < 4; ++j)
				for (int k = 0; k < 4; ++k)
					res.v[i][j] = (res.v[i][j] + a.v[i][k] * b.v[k][j]) % MOD;
		return res;
	}
};

struct Node {
	int val[4];
	Node() { memset(val, 0, sizeof val); }
	Node(ll x) { val[0] = 1, val[1] = x, val[2] = val[3] = x * x % MOD; }
	Node operator + (Node b)
	{
		Node res;
		for (int i = 0; i < 4; ++i)
			res.val[i] = (val[i] + b.val[i]) % MOD;
		return res;
	}

	Node operator * (Mat b)
	{
		ll res[4] = { 0 };
		for (int i = 0; i < 4; ++i)
			for (int j = 0; j < 4; ++j)
				res[j] = (res[j] + val[i] * b.v[i][j]) % MOD;
		for (int i = 0; i < 4; ++i)
			val[i] = res[i];
		return *this;
	}

} tree[MAXN << 2];

Mat lazy[MAXN << 2];

inline void push_up(int rt) 
{
	tree[rt] = tree[ls] + tree[rs]; 
}

inline void push(int rt, Mat k) 
{
	tree[rt] = tree[rt] * k, lazy[rt] = lazy[rt] * k; 
}

inline void push_down(int rt) {
	push(ls, lazy[rt]), push(rs, lazy[rt]);
	lazy[rt].init();
	return;
}

void build(int rt, int l, int r)
{
	lazy[rt].init();
	if (l == r)
	{
		tree[rt] = Node(a[l]);
		return;
	}
	int mid = l + r >> 1;
	build(lson), build(rson);
	push_up(rt);
}

void update(int rt, int l, int r, int L, int R, ll val) 
{
	if (r < L || l > R || (l >= L && r <= R))	//这里直接少调用了两个函数
		return push(rt, Mat((l >= L && r <= R) * val));
	push_down(rt);
	int mid = l + r >> 1;
	update(lson, L, R, val), update(rson, L, R, val);
	push_up(rt);
}

Node query(int rt, int l, int r, int x, int y)
{
	if (x <= l && r <= y)
		return tree[rt];
	push_down(rt);
	int mid = l + r >> 1;
	Node ans;
	if (x <= mid)
		ans = (ans + query(ls, l, mid, x, y));
	if (y >= mid + 1)
		ans = (ans + query(rs, mid + 1, r, x, y));
	return ans;
}

struct modify 
{
	int l, r, val; 
} op[MAXN];

struct query 
{
	int op, id, x, l, r;
	const bool operator< (const query& s) const { return x < s.x; }
} que[MAXN];

signed main() {
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n >> m >> q;
	for (int i = 1; i <= n; i++)
		cin >> a[i], a[i] = (a[i] + MOD) % MOD;
	build(1, 1, n);
	for (int i = 1, l, r, k; i <= m; ++i)
		cin >> op[i].l >> op[i].r >> op[i].val, op[i].val = (op[i].val + MOD) % MOD;
	for (int i = 1, l, r, x, y; i <= q; i++) 
	{
		cin >> l >> r >> x >> y;
		que[i * 2 - 1] = { -1, i, x - 1, l, r };
		que[i * 2] = { 1, i, y, l, r };
	}
	sort(que + 1, que + 1 + 2 * q);
	int pos = 1; 
	while (pos <= 2 * q && que[pos].x < 0)	//如果当前的x小于0，那只可能是减的操作，但其实，根本不用考虑
		++pos;
	for (int i = 0; i <= m; i++) 
	{
		if (i)
		{
			update(1, 1, n, op[i].l, op[i].r, op[i].val);
			//update(1, 1, n, 1, op[i].l - 1, 0);	//边上的两个版本也要更新，就因为调用了这两个函数，直接超时了
			//update(1, 1, n, op[i].r + 1, n, 0);
		}
		while (pos <= 2 * q && que[pos].x == i)
			ans[que[pos].id] += que[pos].op * query(1, 1, n, que[pos].l, que[pos].r).val[3] % MOD, ++pos;
	}
	for (int i = 1; i <= q; ++i)
		cout << (ans[i] + MOD) % MOD << endl;
	return 0;
}