P4145 上帝造题的七分钟 2 / 花神游历各国 题解

题目链接：上帝造题的七分钟2/花神游历各国

差不多的题：[Ynoi Easy Round 2023] TEST_69

注意到对某个点来说暴力单点即为反复的：\(x=\sqrt{x}\)，最终为 \(1\)，根据 \(master\) 主定理可知，跟 \(veb\) 树分析差不多的，复杂度为：\(O(\log{\log{V_{max}}})\)。不懂的可以去学学 这篇文章。

那么考虑到如果每个点都做有效的暴力修改，总复杂度也就 \(n\log{\log{V_{max}}}\)，所以只需要解决无效修改的快速判断就行。很好想的是，\(1=\sqrt{1}\)，这就是无效修改，抽象到区间上，一个区间快速判断是否不需要修改就是看这个区间是否全是 \(1\)，随便咋做都行，比如维护区间最大值，判断是否是 \(1\)，或者区间和是否等于区间长度。这里用区间最大值维护，判断如果不为 \(1\) 向下递归就行了，类似 \(O(n)\) 遍历修改有效位置。

参照代码

#include <bits/stdc++.h>

// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")

#define isPbdsFile

#ifdef isPbdsFile

#include <bits/extc++.h>

#else

#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>

#endif

using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef __int128 i128;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用于Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};

template <typename T>
int disc(T* a, int n)
{
    return unique(a + 1, a + n + 1) - (a + 1);
}

template <typename T>
T lowBit(T x)
{
    return x & -x;
}

template <typename T>
T Rand(T l, T r)
{
    static mt19937 Rand(time(nullptr));
    uniform_int_distribution<T> dis(l, r);
    return dis(Rand);
}

template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
    return (a % b + b) % b;
}

template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
    a %= c;
    T1 ans = 1;
    for (; b; b >>= 1, (a *= a) %= c)if (b & 1)(ans *= a) %= c;
    return modt(ans, c);
}

template <typename T>
void read(T& x)
{
    x = 0;
    T sign = 1;
    char ch = getchar();
    while (!isdigit(ch))
    {
        if (ch == '-')sign = -1;
        ch = getchar();
    }
    while (isdigit(ch))
    {
        x = (x << 3) + (x << 1) + (ch ^ 48);
        ch = getchar();
    }
    x *= sign;
}

template <typename T, typename... U>
void read(T& x, U&... y)
{
    read(x);
    read(y...);
}

template <typename T>
void write(T x)
{
    if (typeid(x) == typeid(char))return;
    if (x < 0)x = -x, putchar('-');
    if (x > 9)write(x / 10);
    putchar(x % 10 ^ 48);
}

template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
    write(x), putchar(c);
    write(c, y...);
}

template <typename T11, typename T22, typename T33>
struct T3
{
    T11 one;
    T22 tow;
    T33 three;

    bool operator<(const T3 other) const
    {
        if (one == other.one)
        {
            if (tow == other.tow)return three < other.three;
            return tow < other.tow;
        }
        return one < other.one;
    }

    T3() { one = tow = three = 0; }

    T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
    {
    }
};

template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
    if (x < y)x = y;
}

template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
    if (x > y)x = y;
}

constexpr int N = 1e5 + 10;

struct Node
{
    ll mx, sum;
} node[N << 2];

#define mx(x) node[x].mx
#define sum(x) node[x].sum

inline void push_up(const int curr)
{
    mx(curr) = max(mx(ls(curr)),mx(rs(curr)));
    sum(curr) = sum(ls(curr)) + sum(rs(curr));
}

int n, q;
ll a[N];

inline void build(const int curr = 1, const int l = 1, const int r = n)
{
    const int mid = l + r >> 1;
    if (l == r)
    {
        sum(curr) = mx(curr) = a[l];
        return;
    }
    build(ls(curr), l, mid);
    build(rs(curr), mid + 1, r);
    push_up(curr);
}

inline void update(const int curr, const int l, const int r, const int s = 1, const int e = n)
{
    if (mx(curr) == 1)return;
    const int mid = s + e >> 1;
    if (s == e)
    {
        mx(curr) = sum(curr) = sqrt(sum(curr));
        return;
    }
    if (l <= mid)update(ls(curr), l, r, s, mid);
    if (r > mid)update(rs(curr), l, r, mid + 1, e);
    push_up(curr);
}

inline ll query(const int curr, const int l, const int r, const int s = 1, const int e = n)
{
    if (l <= s and e <= r)return sum(curr);
    const int mid = s + e >> 1;
    ll ans = 0;
    if (l <= mid)ans += query(ls(curr), l, r, s, mid);
    if (r > mid)ans += query(rs(curr), l, r, mid + 1, e);
    return ans;
}

inline void solve()
{
    cin >> n;
    forn(i, 1, n)cin >> a[i];
    build();
    cin >> q;
    while (q--)
    {
        int op, l, r;
        cin >> op >> l >> r;
        if (l > r)swap(l, r);
        if (op == 0)update(1, l, r);
        else cout << query(1, l, r) << endl;
    }
}

signed int main()
{
    // MyFile
    Spider
    //------------------------------------------------------
    // clock_t start = clock();
    int test = 1;
    //    read(test);
    // cin >> test;
    forn(i, 1, test)solve();
    //    while (cin >> n, n)solve();
    //    while (cin >> test)solve();
    // clock_t end = clock();
    // cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
}

\[时间复杂度上界为:\ O(n\log{\log{V_{max}}}+q\log{n})
\]