luoguP2781 传教

https://www.luogu.org/problemnew/show/P2781

简化版题意：有 n 个数，初始值为 0，进行 m 次操作，每次操作支持将 [l, r] 加 v 和查询 [l, r] 中所有的数的和

n <= 1e9，m <= 1e3

博主 zz 的打了一个支持分裂节点的 splay，AC 后发现可以 m 方暴力过

方法和方伯伯的OJ这题类似，可以参考它的做法

#include <bits/stdc++.h>
using namespace std;

typedef unsigned long long ull;
typedef long long ll;

template <typename T>
inline void read(T &f) {
	f = 0; T fu = 1; char c = getchar();
	while(c < '0' || c > '9') {if(c == '-') fu = -1; c = getchar();}
	while(c >= '0' && c <= '9') {f = (f << 3) + (f << 1) + (c & 15); c = getchar();}
	f *= fu;
}

struct Node {
	ll val, tag, sum;
	int l, r, size;
	Node *ch[2];
	Node () {
		val = tag = l = r = size = 0;
		ch[0] = ch[1] = NULL;
	}
}*root;

int n, m;

void update(Node *u) {
	u -> size = u -> r - u -> l + 1;
	u -> val = u -> sum;
	if(u -> ch[0]) u -> size += u -> ch[0] -> size, u -> val += u -> ch[0] -> val;
	if(u -> ch[1]) u -> size += u -> ch[1] -> size, u -> val += u -> ch[1] -> val;
}

void pushdown(Node *u) {
	if(u -> tag) {
		if(u -> ch[0]) {
			u -> ch[0] -> tag += u -> tag;
			u -> ch[0] -> sum += (ll)(u -> ch[0] -> r - u -> ch[0] -> l + 1) * u -> tag;
			u -> ch[0] -> val += (ll)u -> ch[0] -> size * u -> tag; 
		}
		if(u -> ch[1]) {
			u -> ch[1] -> tag += u -> tag;
			u -> ch[1] -> sum += (ll)(u -> ch[1] -> r - u -> ch[1] -> l + 1) * u -> tag; 
			u -> ch[1] -> val += (ll)u -> ch[1] -> size * u -> tag;
		}
		u -> tag = 0;
	}
}

void rotate(Node *&u, int d) {
	Node *tmp = u -> ch[d];
	u -> ch[d] = tmp -> ch[d ^ 1];
	tmp -> ch[d ^ 1] = u;
	update(u); update(tmp);
	u = tmp;
}

void splay(Node *&u, int k) {
	if(u == NULL) return;
	pushdown(u);
	int ltree = u -> ch[0] ? u -> ch[0] -> size : 0;
	if(k > ltree && ltree + (u -> r - u -> l + 1) >= k) return;
	int d = k > ltree; 
	splay(u -> ch[d], d ? k - ltree - (u -> r - u -> l + 1) : k);
	rotate(u, d);
}

void split(Node *&u, int x) {
	splay(u, x); ll sum = u -> sum; int l = u -> l, r = u -> r;
	if(u -> l != x) {
		Node *tmp = new Node();
		tmp -> sum = sum / (ll)(r - l + 1) * (x - l);
		tmp -> l = l, tmp -> r = x - 1;
		tmp -> ch[0] = u -> ch[0]; update(tmp);
		u -> ch[0] = tmp; u -> l = x;
	}
	if(u -> r != x) {
		Node *tmp = new Node();
		tmp -> sum = sum / (ll)(r - l + 1) * (ll)(r - x);
		tmp -> l = x + 1, tmp -> r = r;
		tmp -> ch[1] = u -> ch[1]; update(tmp);
		u -> ch[1] = tmp; u -> r = x;
	}
	u -> sum /= (ll)(r - l + 1); update(u);
}

int main() {
	cin >> n >> m;
	root = new Node();
	root -> val = root -> tag = root -> sum = 0;
	root -> l = 1, root -> r = n; root -> size = n;
	root -> ch[0] = root -> ch[1] = NULL;
	for(int i = 1; i <= m; i++) {
//		printf("root -> size = %d\n", root -> size);
		int t; read(t);
		if(t == 1) {
			int a, b; ll c;
			read(a); read(b); read(c);
			if(a == 1 && b == n) {
				root -> val += (ll)n * c;
				root -> tag += c;
				root -> sum += (ll)(root -> r - root -> l + 1) * c;
			} else if(a == 1) {
				split(root, b + 1);
				root -> ch[0] -> val += (ll)root -> ch[0] -> size * c;
				root -> ch[0] -> tag += c;
				root -> ch[0] -> sum += (ll)(root -> ch[0] -> r - root -> ch[0] -> l + 1) * c;
				update(root);
			} else if(b == n) {
				split(root, a - 1);
				root -> ch[1] -> val += (ll)root -> ch[1] -> size * c;
				root -> ch[1] -> tag += c;
				root -> ch[1] -> sum += (ll)(root -> ch[1] -> r - root -> ch[1] -> l + 1) * c;
				update(root);
			} else {
				split(root, b + 1);
				split(root -> ch[0], a - 1);
				root -> ch[0] -> ch[1] -> val += (ll)root -> ch[0] -> ch[1] -> size * c;
				root -> ch[0] -> ch[1] -> tag += c;
				root -> ch[0] -> ch[1] -> sum += (ll)(root -> ch[0] -> ch[1] -> r - root -> ch[0] -> ch[1] -> l + 1) * c;
				update(root -> ch[0]); update(root);
			}
		}
		if(t == 2) {
			int a, b; read(a); read(b);
			if(a == 1 && b == n) {
				printf("%lld\n", root -> val);
			} else if(a == 1) {
				split(root, b + 1);
				printf("%lld\n", root -> ch[0] -> val);
			} else if(b == n) {
				split(root, a - 1);
				printf("%lld\n", root -> ch[1] -> val);
			} else {
				split(root, b + 1);
				split(root -> ch[0], a - 1);
				printf("%lld\n", root -> ch[0] -> ch[1] -> val);
			}
		} 
	}
	return 0;
}