【CSUST 7041】: lazy tree 线段树

本文主要是介绍【CSUST 7041】: lazy tree 线段树，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

传送门

题意

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;
}

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