今天剛試了一種新式寫法:
好處是複用性比較好
壞處是程式碼稍微長點, 效率稍微差點把.
不過這種寫法的優勢大概沒怎麼體現吧.
做題的時候直接改模板方便些.

// luogu-judger-enable-o2
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
const int N = 500005;

template<typename Int>
struct BaseNode {
    Int val, len, f;
    BaseNode *ls, *rs;
    BaseNode(Int _v = 0, BaseNode *_ls = nullptr, 
             BaseNode *_rs = nullptr, Int _f = 0, Int _len = 0):
        val(_v), ls(_ls), rs(_rs), f(_f), len(_len) { }
    void update() {
        val = ls->val + rs->val;
    }
    void Merge(Int k) {
        val += k * len, f += k;
    }
    void Down() {
        if (f) ls->Merge(f), rs->Merge(f), f = 0;
    }
};
#define new_Node() new Node()

template<typename Node, typename Int>
class Tree {
    int n;
    Node *root;
#define LS l, mid, node->ls
#define RS mid + 1, r, node->rs
    template<typename Integar>
    void build(int l, int r, Node *node, Integar *A) {
        node->len = r - l + 1;
        if (l == r) {
            node->val = A[l];
            return;
        }
        int mid = l + r >> 1;
        node->ls = new_Node();
        node->rs = new_Node();
        build(LS, A), build(RS, A);
        node->update();
    }
    void addition(int l, int r, Node *node, int L, int R, Int k) {
        if (l >= L and r <= R) {
            return node->Merge(k);
        }
        node->Down();
        int mid = l + r >> 1;
        if (L <= mid) addition(LS, L, R, k);
        if (R >  mid) addition(RS, L, R, k);
        node->update();
    }
    Int Query(int l, int r, Node *node, int L, int R) {
        if (l >= L and r <= R) {
            return node->val;
        }
        node->Down();
        int mid = l + r >> 1;
        Int res = 0;
        if (L <= mid) res += Query(LS, L, R);
        if (R >  mid) res += Query(RS, L, R);
        return res;
    }
  public:
    Tree(int _n) : n(_n), root(new_Node()) {}
    template<typename Integar>
    void build(Integar *A) {
        build(1, n, root, A);
    }
    void addition(int L, int R, Int k) {
        addition(1, n, root, L, R, k);
    }
    Int Query(int L, int R) {
        return Query(1, n, root, L, R);
    }
};

int A[N];
int main () {
    int n, m;
    scanf("%d%d", &n, &m);
    Tree<BaseNode<long long>, long long>* T = 
        new Tree<BaseNode<long long>, long long>(n);
    for (int i = 1; i <= n; i += 1) scanf("%d", &A[i]);
    T->build(A);
    while (m--) {
        int opt, x, y, k;
        scanf("%d%d%d", &opt, &x, &y);
        if (opt == 1) {
            scanf("%d", &k);
            T->addition(x, y, k);
        } else {
            printf("%lld\n", T->Query(x, y));
        }
    }
    return 0;
}