【CSUST 7041】: lazy tree 线段树_lazytree-优快云博客

本文链接：https://blog.youkuaiyun.com/tlyzxc/article/details/120296117

这篇博客主要介绍了如何使用线段树进行区间修改和查询操作，特别是在处理区间价值和区间和的问题上，通过维护节点的附加信息来优化算法。文章详细解释了代码逻辑，并展示了如何处理细节，确保在O(logn)的时间复杂度内完成操作。

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传送门

题意

在这里插入图片描述

分析

这题思路不是很难，难在细节的处理上
首先我们去维护区间的价值和区间和，区间的价值为两个子区间的价值和+两个子区间和的乘积
然后推出来区间+x对价值的改变公式即可

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
	char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
	while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int a[N],n,m;
struct Node{
	int l,r;
	ll sum;
	ll res;
	ll add;
}tr[N * 4];

struct node{
	ll x,y;
};

void push(int u){
	tr[u].sum = (tr[u << 1].sum + tr[u << 1 | 1].sum) % mod;
	tr[u].res = ((tr[u << 1].res + tr[u << 1 | 1].res) % mod + tr[u << 1].sum * tr[u << 1 | 1].sum % mod) % mod;
}

void down(int u){
	if(tr[u].add){
		ll &k = tr[u].add;
		tr[u << 1].add = (tr[u << 1].add + k) % mod;
		tr[u << 1 | 1].add = (tr[u << 1 | 1].add + k) % mod;
		int l = tr[u << 1].r - tr[u << 1].l + 1;
		int r = tr[u << 1 | 1].r - tr[u << 1 | 1].l + 1;
		tr[u << 1].res = (tr[u << 1].res + (l - 1) * tr[u << 1].sum % mod * k + l * (l - 1) / 2 % mod * k % mod * k % mod) % mod;
		tr[u << 1 | 1].res = (tr[u << 1 | 1].res + (r - 1) * tr[u << 1 | 1].sum % mod * k + r * (r - 1) / 2 % mod * k % mod * k % mod) % mod;
		tr[u << 1].sum = (tr[u << 1].sum + l * k % mod) % mod;
		tr[u << 1 | 1].sum = (tr[u << 1 | 1].sum + r * k % mod) % mod;
		k = 0;
	}
}

void build(int u,int l,int r){
	tr[u] = {l,r};
	tr[u].add = 0;
	if(l == r){
		tr[u].sum = a[l];
		tr[u].res = 0;
		return;
	}
	int mid = (l + r) >> 1;
	build(u << 1,l,mid),build(u << 1 | 1,mid + 1,r);
	push(u);
}

node query(int u,int l,int r){
	if(tr[u].l >= l && tr[u].r <= r) {
		node a;
		a.x = tr[u].sum,a.y = tr[u].res;
		return a;
	}
	down(u);
	int mid = (tr[u].l + tr[u].r) >> 1;
	node a,b,c;
	int x = 0,y = 0;
	if(l <= mid) a = query(u << 1,l,r),x = 1;
	if(r > mid) b = query(u << 1 | 1,l,r),y = 1;
	if(x + y == 2){
		c.x = (a.x + b.x) % mod;
		c.y = ((a.y + b.y) % mod + (a.x * b.x % mod)) % mod;
		return c;
	}
	if(x) return a;
	return b;
}

void modify(int u,int l,int r,ll x){
	if(tr[u].l >= l && tr[u].r <= r) {
		tr[u].add += x;
		int p = tr[u].r - tr[u].l + 1;
		tr[u].res = (tr[u].res + (p - 1) * tr[u].sum % mod * x + p * (p - 1) / 2 % mod * x % mod * x % mod) % mod;
		tr[u].sum = (tr[u].sum + p * x % mod) % mod;
		return;
	}
	down(u);
	int mid = (tr[u].l + tr[u].r) >> 1;
	if(l <= mid) modify(u << 1,l,r,x);
	if(r > mid) modify(u << 1 | 1,l,r,x);
	push(u);
}

int main() {
	int T;
	read(T);
	while(T--){
		read(n),read(m);
		for(int i = 1;i <= n;i++) read(a[i]);
		build(1,1,n);
		while(m--){
			int op,x,y,z;
			read(op),read(x),read(y);
			if(op == 1){
				read(z);
				modify(1,x,y,z);
			}
			else{
				dl(query(1,x,y).y);
			}
		}
	}
	return 0;
}