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为了解决这个问题，我们需要处理一系列数组的操作，包括区间更新和区间查询。为了高效地处理这些操作，我们使用Segment Tree数据结构，它能够在O(log N)的时间复杂度下支持区间更新和查询。

方法思路

Segment Tree是一种适合处理区间查询和更新的数据结构。它通过分割区间并存储子区间的信息来实现高效的操作。每个节点存储其区间的和以及一个懒标记，用来记录待传递的更新操作。懒标记的作用是避免在每次操作时重复更新所有相关节点，从而优化了性能。

具体步骤如下：

解决代码

#include 
   
    #include 
    
     using namespace std;struct Node {    int l, r;    long long val, lazy;    int len;};void pushdown(Node* cur) {    if (cur->lazy != 0) {        cur->lchild->lazy += cur->lazy;        cur->rchild->lazy += cur->lazy;        cur->lchild->val += cur->lchild->len * cur->lazy;        cur->rchild->val += cur->rchild->len * cur->lazy;        cur->lazy = 0;    }}void build(Node* tree, int l, int r, long long* arr) {    tree->l = l;    tree->r = r;    if (l == r) {        tree->val = arr[l];        tree->len = 1;        return;    }    int mid = (l + r) / 2;    build(tree->lchild, l, mid, arr);    build(tree->rchild, mid + 1, r, arr);    pushdown(tree);    tree->val = tree->lchild->val + tree->rchild->val;}void update_range(Node* tree, int a, int b, long long c) {    if (tree->r < a || tree->l > b) {        return;    }    if (a <= tree->l && tree->r <= b) {        tree->val += tree->len * c;        tree->lazy += c;        return;    }    pushdown(tree);    mid = (tree->l + tree->r) / 2;    update_range(tree->lchild, a, b, c);    update_range(tree->rchild, a, b, c);    pushdown(tree);    tree->val = tree->lchild->val + tree->rchild->val;}long long query_range(Node* tree, int a, int b) {    if (tree->r < a || tree->l > b) {        return 0;    }    if (a <= tree->l && tree->r <= b) {        return tree->val;    }    pushdown(tree);    int mid = (tree->l + tree->r) / 2;    long long left = query_range(tree->lchild, a, b);    long long right = query_range(tree->rchild, a, b);    return left + right;}int main() {    // 读取输入    int n, q;    long long* arr = new long long[n + 1];    for (int i = 1; i <= n; ++i) {        scanf("%lld", &arr[i]);    }    // 初始化Segment Tree    Node* tree = new Node;    tree->l = 1;    tree->r = n;    tree->len = n;    // build left and right children    Node* lchild = new Node;    Node* rchild = new Node;    tree->lchild = lchild;    tree->rchild = rchild;    build(tree, 1, n, arr);    // 处理每个查询    for (int i = 0; i < q; ++i) {        char op;        int a, b;        scanf("%s%d%d", &op, &a, &b);        if (op == 'C') {            long long c;            scanf("%lld", &c);            update_range(tree, a, b, c);        } else {            long long res = query_range(tree, a, b);            printf("%lld\n", res);        }    }    delete[] arr;    delete lchild;    delete rchild;    delete tree;    return 0;}
    
   

代码解释

通过这种方法，我们能够高效地处理大规模的区间更新和查询操作。

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