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