2021ccpc东北四省D思维线段树

高效动态维护区间加法操作

最新推荐文章于 2022-09-09 08:42:19 发布

原创最新推荐文章于 2022-09-09 08:42:19 发布 · 1.3k 阅读

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#线段树

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这篇博客介绍了一种高效处理区间加法操作的方法，通过log次更新就能将数值变为10000(2)。文章详细阐述了如何使用二进制位运算和树状数组（区间更新和查询）来实现这一操作，并提供了C++代码实现。

每个数最多取log次就会编程10000(2)，如果是这种数字的话直接可以打一个标记，我们直接乘2，这样不必每个数都加lowbit
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/*
 * @Author: 0iq_love_zy
 * @LastEditTime: 2021-06-16 17:55:19
 * @优快云 blog: https://blog.youkuaiyun.com/acm_durante
 * @E-mail: 1055323152@qq.com
 * @ProbTitle: 
 */
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define lowbit(x) ((x) & -(x))
const ll mod = 998244353;
ll powmod(ll a, ll b)
{
    ll res = 1;
    a %= mod;
    assert(b >= 0);
    for (; b; b >>= 1)
    {
        if (b & 1)
            res = res * a % mod;
        a = a * a % mod;
    }
    return res;
}
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
template <class T>
inline void read(T &res)
{
    char c;
    T flag = 1;
    while ((c = getchar()) < '0' || c > '9')
        if (c == '-')
            flag = -1;
    res = c - '0';
    while ((c = getchar()) >= '0' && c <= '9')
        res = res * 10 + c - '0';
    res *= flag;
}
const int N = 1e5 + 50;
int n, m;
struct node
{
    int l, r;
    ll x, sum;
    int lazy;
    int flag;
} tree[N << 2];

void push_up(int p)
{
    tree[p].sum = (tree[p << 1].sum + tree[p << 1 | 1].sum) % mod;
    tree[p].flag = (tree[p << 1].flag & tree[p << 1 | 1].flag);
}

void push_down(int p)
{
    if (tree[p].lazy == 0)
        return;
    tree[p << 1].lazy += tree[p].lazy;
    tree[p << 1 | 1].lazy += tree[p].lazy;
    tree[p << 1].sum = (tree[p << 1].sum * powmod(2, tree[p].lazy)) % mod;
    tree[p << 1 | 1].sum = (tree[p << 1 | 1].sum * powmod(2, tree[p].lazy)) % mod;
    tree[p].lazy = 0;
}

void build_tree(int l, int r, int p)
{
    tree[p].l = l, tree[p].r = r;
    tree[p].lazy = tree[p].flag = 0;
    if (l == r)
    {
        scanf("%lld", &tree[p].x);
        tree[p].sum = tree[p].x;
        if (lowbit(tree[p].x) == tree[p].x)
            tree[p].flag = 1;
        return;
    }
    int mid = l + r >> 1;
    build_tree(l, mid, p << 1);
    build_tree(mid + 1, r, p << 1 | 1);
    push_up(p);
}

void update_tree(int l, int r, int p)
{
    if (l <= tree[p].l && tree[p].r <= r && tree[p].flag)
    {
        tree[p].sum = (tree[p].sum * 2) % mod;
        tree[p].lazy++;
        return;
    }
    if (tree[p].l == tree[p].r)
    {
        tree[p].x = tree[p].sum = tree[p].sum + lowbit(tree[p].sum);
        if (lowbit(tree[p].x) == tree[p].x)
            tree[p].flag = 1;
        return;
    }
    push_down(p);
    int mid = (tree[p].l + tree[p].r) >> 1;
    if (l <= mid)
        update_tree(l, r, p << 1);
    if (mid < r)
        update_tree(l, r, p << 1 | 1);
    push_up(p);
}

ll ask_sum(int l, int r, int p)
{
    if (l <= tree[p].l && tree[p].r <= r)
    {   
        return tree[p].sum;
    }
    ll res = 0;
    push_down(p);
    int mid = (tree[p].l + tree[p].r) >> 1;
    if (l <= mid)
        res = (res + ask_sum(l, r, p << 1)) % mod;
    if (mid < r)
        res = (res + ask_sum(l, r, p << 1 | 1)) % mod;
    return res;
}
void solve()
{
    scanf("%d", &n);
    build_tree(1, n, 1);
    scanf("%d", &m);
    int op, l, r;
    for (int i = 1; i <= m; i++)
    {
        scanf("%d%d%d", &op, &l, &r);
        if (op == 1)
        {
            update_tree(l, r, 1);
        }
        else
        {
            printf("%lld\n", ask_sum(l, r, 1));
        }
    }
}
int main()
{
    int t;
    while (~scanf("%d", &t))
    {
        while (t--)
        {
            solve();
        }
    }

    return 0;
}