P9989 [Ynoi Easy Round 2023] TEST_69 题解

题目链接: [Ynoi Easy Round 2023] TEST_69

首先GCD有比较良好的一些性质。我们观察到一次 \(GCD(a_i,x)\) 操作，会有以下两种变化。

1. 如果 \(x \bmod a_i == 0\)，那么很显然 \(\gcd(a_i,x)==a_i\)，不会发生任何改变。
2. 如果不是情况 \(1\),那么很显然 \(a_i\) 至少会变为原来的一半，因为容易知道 \(a_i\) 的最小因子不会小于 \(2\)，所以由于公因子是成对出现的，另一个对应的公因子则是最大为 \(\frac{a_i}{2}\)，所以最大因子显然不超过 \(\frac{a_i}{2}\)。不断减小，最终为 \(1\)，这个过程至多进行 \(\log{a_i}\) 次。

考虑暴力一点的做法，为每个 \(a_i\) 进行反复的有效操作，容易发现，最终也至多执行 \(\sum_{i=1}^{n}{\log{a_i}}\) 次，很显然，如果每次操作我们都保证有效的，那么这个次数也是完全接受的。

瓶颈点：如何快速判断出无效的状态。这里有个技巧性质：第一种操作情况可以拓展到 \(LCM\) 上。

\[\text{如果有 } x \bmod lcm_{区间}==0 \text{，则这个区间内所有数都不会变化}
\]

很显然这个是对的，第一条其实可以翻译成如果 \(x\) 是一个数的倍数，那么这个操作不会发生变化，一堆数的公倍数，显然就是说的最小公倍数了。

算法框架

用线段树维护维护区间最小倍数，当遇到可以 \(x \bmod lcm_{区间} ==0\) 时，则停止往下递归，否则往下递归一直到单点修改。

\[\text{最终复杂度为建树+查询与修改：} O(n\log{V_{max}}+m\log{n}\log{V_{max}})
\]

参考代码

#include <bits/stdc++.h>

//#pragma GCC optimize("Ofast,unroll-loops")

#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 = 2e5 + 10;
constexpr ll MOD = 1ll << 32;
constexpr i128 INF = 1e18 + 10;

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

auto GCD(const ll x, const ll y) -> ll
{
    if (!y)return x;
    return GCD(y, x % y);
}

auto LCM(const ll x, const ll y) -> ll
{
    return min(static_cast<i128>(x) / GCD(x, y) * y, INF);
}

inline void push_up(const int curr)
{
    node[curr].sum = (node[ls(curr)].sum + node[rs(curr)].sum) % MOD;
    node[curr].lcm = LCM(node[ls(curr)].lcm, node[rs(curr)].lcm), INF;
}

inline void update(const int curr, const int l, const int r, const int s, const int e, const ll x)
{
    if (l <= s and e <= r)
    {
        if (x % node[curr].lcm == 0)return;
        if (s == e)
        {
            node[curr].lcm = GCD(node[curr].lcm, x);
            node[curr].sum = node[curr].lcm;
            return;
        }
    }
    const int mid = (s + e) >> 1;
    if (l <= mid)update(ls(curr), l, r, s, mid, x);
    if (r > mid)update(rs(curr), l, r, mid + 1, e, x);
    push_up(curr);
}

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

inline void build(const int curr, const int l, const int r)
{
    if (l == r)
    {
        read(node[curr].sum);
        node[curr].lcm = node[curr].sum;
        return;
    }
    const int mid = (l + r) >> 1;
    build(ls(curr), l, mid);
    build(rs(curr), mid + 1, r);
    push_up(curr);
}

int n, q;

inline void solve()
{
    read(n, q);
    build(1, 1, n);
    while (q--)
    {
        int op;
        read(op);
        if (op == 1)
        {
            int l, r;
            ll x;
            read(l, r, x);
            update(1, l, r, 1, n, x);
        }
        else
        {
            int l, r;
            read(l, r);
            write(endl, query(1, l, r, 1, n));
        }
    }
}

signed int main()
{
    Spider
    //------------------------------------------------------
    int test = 1;
    //    read(test);
    // cin >> test;
    forn(i, 1, test)solve();
    //    while (cin >> n, n)solve();
    //    while (cin >> test)solve();
}